chapter 0aplustestbank.eu/sample/solution-manual-for-applied... · 2017. 4. 7. · 1 chapter 0...

1

Chapter 0 Chapter 0 Review Questions 1. d 2. b 3. c 4. b 5. d 6. Archimedes conducted and documented his physical theories. 7. Science is a system of knowledge while technology uses that knowledge to develop material products

or processes that satisfy human needs and desires. 8. Behavior of light – Fiber Optics ; Electricity – light bulb. 9. The scientific method is used to discover facts about the natural world. The problem solving method

uses the scientific method to create something useful. 10. Knowledgeable of the physical world. It allows one to understand and answer new questions about

everyday occurrences. It is also important in many technical fields.

Chapter 1

1.2

1. kilo

2. centi

3. hecto

4. deci

5. milli

6. deka

7. mega

8. micro

9. h

10. k

11. m

12. d

13. M

14. da

15. c

16. μ

17. 135 mm

18. 83 dag

19. 28 kl

20. 52 cm

21. 49 cg

22. 85 mg

23. 75 hm

24. 15 dL

25. 24 metres

26. 185 litres

27. 59 grams

28. 125 kilograms

29. 27 millimeters

30. 25 dicilitres

31. 45 dekametres

32. 27 milligrams

33. 26 megametres

34. 275 micrograms

35. metre

36. kilogram

37. litre and cubic metre

38. ampere

39. second

40. watt

1.3

1. 3.26 × 102 2. 7.98 × 102 3. 2.65 × 103 4. 1.45 ×10 4 5. 8.264 × 102 6. 2.497 × 101 7. 4.13× 10−3 8. 5.3× 10−4 9. 6.43×100 10. 4.823× 105

11. 6.5× 10−5 12. 2.24 × 10−3 13. 5.4 × 105 14. 1.4 × 106 15. 7.5× 10−6 16. 9× 10−7 17. 5× 10−8 18. 3.5× 109 19. 7.32 × 1017 20. 6.18 × 10−16

21. 86,200 22. 867 23. 0.000631 24. 5410 25. 0.768 26. 99.4 27. 777,000,000 28. 0.00000419 29. 69.3 30. 0.0378 31. 96,100

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2

32. 7330 33. 1.4 34. 0.000096 35. 0.0000084 36. 900,000,000

37. 700,000,000,000 38. 4.05 39. 0.00000072 40. 0.000000008 41. 4,500,000,000,000

42. 150,000,000,000 43. 0.000000000055 44. 0.000000000872

1.4

1. 1 metre

2. 1 metre

3. 1 kilometre

4. 1 centimetre

5. 1 kilometre

6. 1 kilometre

7. cm

8. mm

9. m

10. cm

11. mm

12. km

13. km

14. m

15. mm

16. m

17. m

18. cm

19. km

20. mm

21. cm

22. m

23. km

24. mm

25. cm

26. cm

27. km

28. cm

29. mm

30. cm

31. cm

32. cm,cm

33. 1000

34. 0.001

35. 100

36. 0.01

37. 0.1

38. 10

39. 1000

40. 10

41. 100

42. 0.01

43. 0.001

44. 0.1

45. 10

46. 25,000 cm

47. 0.25 km

48. 54.6 cm

49. 178,000 m

50. 0.35 dam

51. 8.3 m

52. 7.5 km

53. 3750 mm

54. 7500 μm

55. 4,000,000μm

56. Answers vary

57. (a) 9 ft (b) 3 yd

58. (a) 8.2 mi (b)

14,432 yd

59. (a) 11,000 yd (b)

33,000 ft

60. 17.78 cm

61. 412.16 km

62. 19.4 ft

63. 2.80 in.

64. 3.048 cm

65. 6.1 m

66. Smaller

67. 20 cm = 200 mm, ea. cut = 3mm

203mm, therefore 6 ft ×

305mm

1ft×

reamercut

203mm= 9 reamers

68. 18

in. × 2.54 cm

1 in.= 0.318cm waste, therefore (214 pcs)(47 cm + 0.318 cm) = 10,126 cm=

101.26m


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3

1.5

1. (8cm) (5 cm) = 40cm 2

2. (28 cm) (15 cm)= 420cm 2

3. (8 in.)(3in.) + (3 in.)(5in) = 39in 2

4. (15cm)(12cm) – (6cm)(7cm) = 138cm 2

5. (8 in.)(1 in.) + (1 in.)(6 in.) + (8 in.)(1 in.) =

22in 2

6. (40 cm)(35 cm) – (15 cm)(20 cm) = 1100cm 2

7. (6 in.)(3 in.)(4 in.) = 72in 3

8. (7cm)(7cm) − (3cm)(3cm)[ ](3cm) = 120cm3

9. (5cm)(2cm)(4cm)= 40cm 3

10. (20 in.)(20 in.)(20 in.) = 8000in 3

11. 1 litre

12. 1 kilolitre

13. 1 cubic centimeter

14. 1m3

15. 1 square kilometre

16. 1dm2

17. L

18. L

19. m2

20. cm2

21. m3

22. L

23. ha

24. cm2

25. mL

26. ha

27. m3

28. L

29. L

30. L

31. mL

32. ha

33. L

34. mL

35. m2

36. cm2

37. L

38. L

39. ha

40. cm2

41. m3

42. m3

43. m2

44. ha

45. L

46. mL

47. 1000

48. 1000

49. 0.1

50. 0.001

51. 0.01

52. 10

53. 10

54. 0.001

55. 1

56. 1000

57. 1,000,000

58. 1

59. 0.001

60. 1

61. 10,000

62. 106

63. 100

64. 10−6

65. 0.01

66. 10,000

67. 100

68. 0.01

69. 7.5 L

70. 850 mL



4

71. 1600 mL

72. 0.009 L

73. 275,000 mm3

74. 5× 106cm 3

75. 4 × 109 mm3

76. 0.52 cm3

77. 275 mL

78. 0.125 L

79. 1000 L

80. 1.5× 10−4 L

81. 7500 cm3

82. 0.45m3

83. 50cm 2

84. 1.75× 106m2

85. 50,000 cm2

86. 25,000 mm2

87. 400 km2

88. 5× 103 m2

89. 45 ft 2

90. 20.9m2

91. 13,935cm2

92. 1944 ft 2

93. 0.75 ft 2

94. 7.91in 2

95. 156,816in 2

96. 1.858 m2

97. 513 ft 3

98. 1.769in 3

99. 30.1 yd 3

100. 1442 cm3

101. 13,824 in 3

102. 732,000 in 3

103. 623.3 cm3

104. 12 ft 3 ×2.83× 104 cm3

1 ft 3 ÷ 14cm3 = 24,257

105. 2(5cm)(2cm) + 2(4cm)(2cm) = 36cm 2

106. 4(20in.)(20in.) = 1600in2

107. 36cm 2 + 2(5cm)(4cm) = 76cm 2

108. 6(20in.)(20in.) = 2400in 2

109. (5cm)(4cm)(2cm) ×1mLcm 3 = 40mL

110. 120cm 3 ×1mLcm3 = 120mL

1.6

1. 1 gram

2. 1 gram

3. 1 kilogram

4. 1 centigram

5. 1 kilogram

6. 1 kilogram

7. kg

8. mg

9. kg

10. g

11. metric ton

12. mg

13. g

14. kg

15. mg

16. kg

17. kg

18. kg

19. g

20. metric ton

21. kg

22. g

23. g

24. kg



5

25. g

26. g

27. kg

28. g

29. kg

30. g

31. metric ton

32. g

33. kg

34. mg

35. 1000

36. 0.001

37. 100

38. 0.01

39. 0.1

40. 10

41. 1000

42. 10

43. 100

44. 0.01

45. 0.001

46. 0.1

47. 1,000,000

48. 1000

49. 575,000 mg

50. 0.575 kg

51. 0.65 g

52. 375,000 g

53. 5 g

54. 48,500 dg

55. 30,000,000 mg

56. 4000 kg

57. 2.5 kg

58. 5.8 × 10−5 g

59. 0.4 mg

60. 30 metric tons

61. 750mL ×1gmL

= 750g

62. 1m3 ×1g

cm3 ×106cm3

1m3 ×1kg

103 g= 1000kg

63. 3500 lb ×4.45N1lb

= 15,575N

64. 150,00 lb ×4.45N1lb

= 667, 500N

65. 200lb×4.45N

1lb= 890N

66. 80lb ×4.45N1lb

= 356N

67. 2000N ×0.225lb

1N= 450lb

68. 2000lb×4.45N

1lb= 8900N

69. 120oz ×1lb

16oz= 7.5lb

70. 3.5lb×16oz1lb

= 56oz

71. 10N ×0.225lb

1N×

16oz1lb

= 36oz

72. 25oz ×1lb

16oz×

4.45N1lb

= 6.95N

73. 94lb ×4.45N1lb

= 418.3N

74. 500× 3lb×4.45N

1lb= 6675N

75. second, s

76. kilogram, kg

77. newton, N

78. 1 second

79. 1 millisecond

80. 1 μs

81. 1 ms

82. 8.6 μs

83. 45 ns

84. 75 ps

85. 345μs×1s

106μs= 3.45× 10−4 s

86. 1h ×60 min

1h+ 25min = 85min



6

87. 4h ×3600s

1h+ 25min×

60s1min

+ 15s = 15,915s

88. 7× 106s ×1h

3600s= 1944h

89. 4s ×1ns

10−9 s= 4 × 109 ns

90. 1h ×3600s

1h×

1012 ps1s

= 3.6× 1015 ps

1.7

1. 3

2. 4

3. 4

4. 3

5. 2

6. 4

7. 3

8. 3

9. 5

10. 2

11. 3

12. 4

13. 5

14. 3

15. 4

16. 4

17. 3

18. 3

19. 2

20. 3

21. 1

22. 2

23. 4

24. 1

25. 2

26. 3

27. 3

28. 2

29. 4

30. 1

1.8

1. 1 ft.

2. 0.1 mi

3. 1 m

4. 0.01cm

5. 0.0001 in.

6. 0.001 cm

7. 10km

8. 10 ft

9. 0.01 m

10. 10g

11. 0.00001 in.

12. 0.01 cm

13. 0.01m

14. 0.01mi.

15. 1 kg

16. 0.01 mm

17. 0.0001 in.

18. 1 lb

19. 1000 N

20. 10 ft 3

21. 1 N

22. 1000 tons

23. 0.01 m 2

24. 0.001m

25. 100 kg

26. 1× 10−7 ms

27. 0.000001A or 1× 10−6 kg

28. 1× 105 ft

29. 10,000 V or 1× 104 kg

30. 1× 109 m

31. (a) 15.7 in. (b) 0.018 in.

32. (a) 368 ft (b) 368 ft

33. (a) 16.01 cm (b) 0.734 cm

34. (a) All have 3 significant digits (b) All have

precision of 0.01 m.

35. (a) 0.0350 s (b) 0.00040 s

36. (a) 125.00g (b) 9.000 g

37. (a) 27, 00 0L (b) 4.75 L

38. (a) All have 2 significant digits (b) 0.40 m

39. (a) All have one significant digit (b) 50 N

40. (a) 120.0 ms (b) All have precision of 0.1

ms

41. (a) 0.05 in. (b) 16.4 in.



7

42. (a) 300 ft (b) 300 ft

43. (a) 0.65 m (b) 27.5 m

44. (a) All have 2 significant digits (b) 120 kg

45. (a) 0.00005 g (b) 0.75 g

46. (a) 0.09 m (b) 185.0 m

47. (a) 3 N (b) 45,000 N

48. (a) All have 3 significant digits. (b) 150 kg

49. (a) 20 kg (b) 400 ,000kg

50. (a) 80 ft (b) All have precision of 10 ft

1.9

1. 14,200 ft

2. 124,000 m

3. 83.3 cm

4. 2.83 g

5. 70 ,000N

6. 15,600 ft

7. 802 m or 80,200 cm

8. 166.2 m or 16,620 cm

9. 18 s

10. 50 0,000N

11. 500 kg

12. 5.35 m

13. 41.0 g

14. 25,00 kg

15. 3200 km

16. 55.24 g

17. 900,000 kg

18. 7 cm or 70 mm

19. 0.40 m or 40 cm

20. 0.69 ms

21. 4900 m2

22.

560, 000 ft2or5.6 × 105 ft 2

23.

1,40 0,000km 2or1.40× 106km 2

24.

5,500, 000m2or5.5× 106 m2

25. 737.7m2

26.

10 0,000cm 2or1.0 × 105 cm2

27. 5560 cm3

28. 6.4 × 105m3

29. 2.91× 107in 3

30. 4060 kg m

31. 30 ft

32. 42 m

33. 3.06 cm

34. 41 mi/h

35. 75 km/h

36. 3.0 m

37. 10 00mi / h

38. 1680 ft/s

39. 1100 ft lb/s

40. 4.41 kg m/s

41. 370 mi/h

42. 10 kg / m2

43. 43.2 m

44. 36, 000kgm 2 / s2

45. 4530 kg m/s 2

46. 88.2 kg m/s 2

47. 10,300 m 3

48. 2.6 g/cm 3

49. 6100 m 2

50. 560 ft 3

51. 28,800 ft

52. 62,700 m 2

Chapter 1 Review Questions

1. c

2. b

3. b

4. c

5. a

6. (1) Pieces made separately may not fit together (2) Workers could not communicate directions to each

other (3) Workers could not tell each other how much material to buy.

7. It is based on the decimal system

8. (1)The distance to the moon (2) The thickness of the aluminum foil.



8

9. Yes

10. The surface that would be seen by cutting a geometric solid with a thin plate.

11. Yes

12. hectare

13. litre

14. (1) medicines (2) perfumes (3) wine

15. Mass measures the quantity of matter. Weight measures the gravitational pull on an object.

16. Newton (N)

17. millionth

18. Because nearly all measurements are approximate numbers rather than exact numbers.

19. No

20. Yes

Chapter 1 Review Problems

1. k 2. m 3. μ 4. M 5. 45 mg 6. 138 cm 7. 1 L 8. 1 kg

9. 1 m3 10. 250m ×1km

103 m= 0.25km 11. 850mL ×

1L103mL

= 0.85L

12. 5.4kg ×103 g

1kg= 5400g 13. 0.55s×

1μs10−6s

= 550,000μs 14. 25kg ×103 g1kg

= 25, 000g

15. 75μs×10−6 s1μs

×1ns

10−9 s= 75,000ns 16. 275cm 2×

10mm1cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 27,500mm2

17. 350cm 2 ×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 0.035m2 18. 0.15m3 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 150, 000cm3

19. 500cm 3 ×1mL1cm3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 500mL 20. 150lb×

1kg2.20lb

= 68.2kg

21. 36ft×1m

3.28 ft= 11.0m 22. 250cm ×

1in.2.54cm

= 98.4 in.

23. 150in 2 ×2.54cm

1in.⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 968cm2 24. 24yd 2 ×3 ft1yd

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 216 ft 2

25. 6m3 ×3.28 ft

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 212 ft 3 26. 16lb ×4.45N1lb

= 71.2N

27. 15,600s×1min60s

= 260 min 260min×1hr

60 min= 4

13

h = 4h20 min 28. 3 29. 4

30. 2

31. 3

32. 0.1ft

33. 0.0001s

34. 1000mi

35. 100,000N



9

36. (a) 12.00 m (b) 0.008m (c) 0.150m (d)

2600m

37. (a) 18,050 L (b) 0.75L (c) 0.75 L (d)

18.050 L

38. 0.125s

39. 63,000 N

40. 1,800,000 cm 3

41. 150m 2

42. 9.73kgm / s 2

43. 4.50m( ) 2.20cm( ) = 9.90m2

44. 9.0cm( ) 6.0cm( ) 13cm( ) = 70 0cm 3



10

Chapter 2

2.1

1. s = vt 2. v = at 3. m =wg

4. a =Fm

5. R =EI

6. w =V1h

7. g =PEmh

8. h =PEmg

9. h =v2

2g 10. f =

XL

2πL 11. w = Pt 12. F = pA

13. t =WP

14. A =FP

15. m =2 KE( )

v 2 16. v2 =2 KE( )

m 17. s =

WF

18. a =v f − vi

t

19. I =V − E

−r or I =

E − Vr

20. t =v2 − v1

a 21. P =

π2R

22. L =Rd 2

k

23. C =5F −160

9 or C =

5 F − 32( )9

24. F =95

C + 32 25. f =1

2πCXC

26. L =RAρ

27. R3 = RT − R1 − R2 − R4 28. Q2 =Q1 + Q1P

P or Q2 =

Q1

P+ Q1

29. IP =I sNs

NP 30. NS =

VsN P

VP 31. vi = 2vavg − v f 32. a =

v2 − vi2

2 s − si( ) orv 2 − vi

2

2s − 2si

33. s =v 2 − vi

2 + 2asi

2a 34. V1 = V2 −

Ftm

or V1 =mV2 − Ft

m 35. R =

QJI 2t

36. xi = x − vit −12

at 2 37. r =Aπ

38. r =Vπh

39. d =kLR

40. r =3Vπh

41. I = +QJRt

42. I = +Frm

2.2

1. (a) A= bh (b) 162 cm 2 2. (a) V= lwh (b) 4420m3 3. (a) b =Ah

(b) 7.50 cm

4. (a) b =P4

(b) 105 in. 5. (a) c = P – a – b (b) 6.0 cm 6. (a) d =Cπ

(b) 158 ft 7. (a) r =C2π

(b) 10.9 yd 8. (a) h =2Ab

(b) 26.0 m

9. (a) b =P − 2a

2 (b) b =

P2

− a (b) 33.2 km 10. (a) V = πr 2h (b) 1,460,000 m3

11. (a) h =V

πr 2 (b) 6.11 m 12. (a) h =A

2πr (b) 5.80 cm 13. (a) B =

Vh

(b) 154 m2

14. (a) r =Aπ

(b) 12.15 m 15. (a) b = A (b) 21.6 in. 16. (a) r =3Vπh

(b) 13.2 m

17. (a) C = 2πr (b) 121.6 m 18. (a) V =43

πr 3 (b) 70,690 m3 19. (a) B =3Vh

(b) 122.4 ft 2 20. (a) h =2A

a + b (b) 11.40 m

2.3

1. V = 1wh = 36.0cm( ) 30.0cm( ) 24.0cm( ) = 25, 900cm 3

2. V = πr 2h = π 2.10in.( )2 7.50in.( ) = 104in 3 3. V =13

πr 2h =13

π 5.40cm( )2 9.30cm( ) = 284cm3

4. V = πr 2h = π11.40cm

2⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

24.00cm( ) = 2450 cm3 5. A = πr 2 = π11.40cm

2⎡ ⎣ ⎢

⎤ ⎦ ⎥

2

= 102.1cm 2



11

6. A = πdh = π 11.40cm( ) 24.00cm( ) = 859.5cm2

7. V = 1w +12

bh⎛ ⎝ ⎜

⎞ ⎠ ⎟ h' = 22.0 ft( ) 10.0 ft( ) +

12

22.0 ft( ) 4.70 ft( )⎡ ⎣ ⎢

⎤ ⎦ ⎥ 37.0 ft( ) = 10,100 ft 3

8. A =a + b

2⎡ ⎣ ⎢

⎤ ⎦ ⎥ h =

3.70 ft + 6.80 ft2

⎡ ⎣ ⎢

⎤ ⎦ ⎥ 19.3 ft( ) = 101 ft 2

9. V = 1wh = 9.00 ft( ) 12.0 ft( ) 8.00 ft( ) = 864 ft 3 10. A = πr 2 = π3.25cm

2⎡ ⎣ ⎢

⎤ ⎦ ⎥

2

= 8.30cm2

11. A =12

bh =12

4.00cm( ) 6.00cm( ) = 12.0cm 2

12. c = a2 + b2 = 4.00cm( )2 + 6.00cm( )2 = 7.21cm

13. A = π r12 − r2

2( )= π3.50cm

2⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

−3.20cm

2⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

= 1.58cm2

14. V =4πr 3

3=

43

π 8.00m( )3 = 2140m3 15. w =A1

=900 m2

25.0m36.0m

16. h =V1w

=192m

8.00m( ) 4.00m( ) = 6.00m 17. V = πr2h = π (2.00cm)2 (4.20cm) = 52.8cm3

18. V = πr 2h = π (3.90cm)2 (8.00cm) = 382cm3

19. d =Cπ

=29.5m

π= 9.39m

20. h =V

πr 2 =1000 m3

π9.39m

2⎛ ⎝ ⎜

⎞ ⎠ ⎟

2 = 14.4m

21. Distance=C no.of rev( )= πd no.of rev( )= π 30.0cm( ) 145( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 137m

22. A = Ch = 29.5m( ) 14.4m( ) 1L5.0m2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 85L 23. h =

Vπr 2 =

500 ,000gal ×1 ft 3

7.50galπ 18.0 ft( )2 = 65.5 ft

24. r =Vπh

=500 , 00gal ×

1 ft 3

7.50galπ 42.0 ft( ) = 22.5 ft 25.

A1

A2=

12.0 ft( ) 15.0 ft( )1.00 ft( ) 3.00 ft( ) =

180 ft 2

3.00 ft2 = 60 panels

26. Atotal = Arec tan lg e − Atrapezoid A = 3.50cm( ) 2.00cm( ) −1.90cm + 2.20cm

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 0.400cm( ) = 6.18cm 2

27. V = Vcylinder + Vcone V = π 1.50m( )2 2.75( ) +13

π 1.50m( )2 2.00m( ) = 24.1m3

28. r =Aπ

=3.05m2

π= 0.985m;yes 29. V = 1wh = 12.0 ft( ) 20.0 ft( ) 0.500 ft( )×

1yd 3

27 ft3 = 4.44yd 3



12

30. 1 =Vwh

=2.00yd 3

4.00 ft( ) 0.333 ft( ) ×27 ft 3

1yd 3 = 40.5 ft

31. V = 1wh − πr 2h = 8.00in .( ) 8.00in.( ) 6.00in.( ) − π 2.50in.( )2 6.00in.( ) = 266in3

32. V = πr 2h − πr 2h = π 25.0cm( )2 60.0cm( )− π 10.0cm( )2 60.0cm( ) = 99, 0 00cm3

Chapter 2 Review Questions 1. c 2. b 3. a 4. (1) To find the volume of liquid storage tanks. (2) To determine the amount of concrete needed for a driveway. 5. As a shorthand way to designate different measured quantities of the same type. 6. Most mistakes are made in problem solving by missing needed information or misinterpreting the information given. 7. Making a sketch helps visualize what is happening in the problem. 8. The basic question. 9. The working equation is found by solving the basic equation for the unknown quantity. 10. Carrying the units through a problem shows whether the answer is the kind expected. 11. Making an estimate of the correct answer shows whether the solution is reasonable. Chapter 2 Review Problems

1. (a) m =F

a (b) a =

F

m 2. h =

v2

2g 3. v f =

2s

t− vi 4. v =

2KE

m

5. b = P – a – c = 36 ft – 12 ft – 6 ft = 18 ft

6. a = 2A

h−

b

2⎛⎝

⎞⎠ = 2

210m2

15.0m−

16.0m

2

⎛⎝⎜

⎞⎠⎟

= 12.0m 7. r =A

π= 2.19m

8. A = ½ b h = ½ (12.2cm) (20.0cm) = 122 cm2 9. h =3V

πr 2=

3(314cm 3 )

π (5.00cm)2= 12.0cm

10. c = a2 + b2 = (41.2mm)2 + (9.80mm)2 = 42.3mm 11. A = 2πrh = 2π (7.20cm)(13.4cm) = 606cm

12. 40.0cm = 2(14.0cm) + 2ω

ω = 6.0cm 13. r =

V

πh=

2100m 3

π17.0m= 6.27m

14. h =2A

b=

2(88.6m2 )

12.3m= 14.4m

15. V = V1 − V2 = (9.0cm ⋅ 6.0cm ⋅12cm) − (6.0cm ⋅ 3.0cm ⋅12cm) = 430cm

16. A = A1 − A2 = (40.0cm ⋅120cm) − (10.0cm ⋅12.0cm) = 4680m2 Chapter 2 Applied Concepts

1. Aproperty − Ahouse = l prop w prop − lhousewhouse = 100 ft( ) 200 ft( )− 35.0 ft( ) 80.0 ft( ) = 17,200 ft 2

$50.00

17200 ft9 ft 2( )1yd 2( )= $0.026 / yd 2 = 2.62 ⊄ /yd2



13

2. V = h × w × l = 8.00 ft( ) 10.0 ft( ) 32.0 ft( ) = 2560 ft 3; 2560 ft 3

20.0 min1min( )60sec( ) = 2.13 ft 3 / s

3. VSolidBeam = lwh = 240 ft( ) 8.00in.( ) 8.00in.( ) = 15400in 3 VIBeam = Vtop + Vvertical + Vbottom = 1.00in.( ) 8.00in.( ) 240 in.( )+ 6.00in.( ) 1.00in( ) 240 in.( )+ 1.00in( ) 8.00in.( ) 240 in.( )= 5280in3 Vsolid

VIBeam=

15400in 3

5280in 3 = 2.91

4. # of balls wide= w2r

=16.8in.

2× 4.00in.= 2.10balls

# of balls high= h2r

=16.8in.

2× 4.00in.= 4.20balls

2 balls x 2 balls x 4 balls = 16 balls

# of balls long=l

2× r=

33.6in.2 × 4.00in.

= 4.20balls

5. (a) Vcylinder = πr 2h = π 1.53m( )2 0.915m( ) = 6.73m3

(b) m = DV = 7750kgm3

⎛ ⎝

⎞ ⎠ 6.73m3( )= 522,000kg

F = m × g = 522, 000kg( ) 9.80ms 2

⎛ ⎝

⎞ ⎠ = 512, 000N



14

Chapter 3

3.1

1. 100 km ×1.0cm50 km

= 2.0cm 2. 125km ×1.0cm50 km

= 2.5cm 3. 140km ×1.0cm50 km

= 2.8cm

4. 260 km ×1.0km50 km

= 5.2cm 5. 315km ×1.0cm50 km

= 6.3cm 6. 187km ×1.0cm50 km

= 3.7cm

7. 8.

9. 10.



15

11. 12.

13. 100 mi ×1

4 in.20 mi .

= 1 14in, 14. 170 mi ×

14in.

20 mi= 2 1

8 in. 15. 210 mi ×1

4 in.20 mi

= 2 58in.

16. 145mi .×1

4 in20 mi

= 11316in 17. 75mi ×

14 in.

20 mi= 15

16in. 18. 160 mi ×1

4 in.20 mi

= 2in.

19. 20.

21.

22. 23. 24.



16

3.2

1. 2. 3.

4. 5. 6.

7. 8. 9. 10.

11. 12



17

13. 14.

15. 16.

17. 18.

19. x= (cos 42.0°)(13.4m)=9.96 m 20. x= (cos 75.5°)(275 mi.) = 68.9 mi

y= (sin 42.0°)(13.4m)= 8.97m y= (sin75.5°)(275 mi) = -266mi



18

21. x= (cos 68.0°)(48.6km)= -18.2 km 22. x= (cos 68.6°)(9780 m) = - 3570 m

y= (sin68.0°)(48.6 km) = -45.1 km y= (sin 68.6°)(9780 m) = 9110 m

23. x= (cos 8.4°)(98.5 km) = 97.4 km 24. x= (cos 15.0°)(48.5 ft) = -46.8 ft

y= (sin 8.4°)(98.5 km) = -14.4 km y= (sin 15.0°)(48.5 ft) = 12.6 ft

25. x= (cos 10.5°)(38.9 m) = 38.2 m 26. x= (cos 15.0°)(478 ft) = -462 ft

y= (sin 10.5°)(38.9 m) = 7.09 m y= (sin 15.0°V)(478 ft) = -124 ft

27. x= (cos 50.0°)(9.60 km) = 6.17 km 28. x= (cos 26.3°0(5430 mi) = -4870 mi

y= (sin 50.0°(9.60 km) = -7.35 km y= (sin 26.3°)(5430 mi) = 2410 mi

29. x= (cos 78.5°)(29.5m) = -5.88 m 30. x= (cos 86.8°)(154 mi) = 8.60 mi

y= (sin 78.5°)(29.5 m) = 28.9 m y= (sin 86.8°)(154 mi) = -154 mi

3.3

1. 61km at 55°north of east 2. 110 km at 56° south of west 3. 1300mi. at 1° west of south

4. 23 mi. at 42° south of west 5. 36 km at 5° east of north 6. 3.1 km at 4° east of south

7. 38 km at 25° north of west 8. 290 km at 3° south of east 9. 1500 mi at 71° north of east

10. 44 mi. at 39° south of west 11. 120 km at 72° south of east 12. 7.4 km at 8° north of east

13. 47 mi at 49° north of east 14. 52 km at 34° north of east

15. 16. 17.

18. 19. 20.



19

21. 22.

23. 24.

25.

26.



20

27.

28. 29.



21

30.

31. R = 8.00m( )2 + 6.00m( )2 = 10.0m 32. R = −15.0 ft( )2 + −45.0 ft( )2 = 47.4 ft

tan A =6.00m8.00m

tan A =45.0 ft15.0 ft

A = 36.9o = Θ A = 71.6o

Θ = 180o + 71.6o = 251.6o



22

33. R = 18.5mi( )2 + −18.5mi( )2 = 26.2mi 34. R = −16.5in( )2 + 10.0in( )2 = 19.3in

tan A =18.5mi18.5mi

tan A =10.0in.16.5in.

A = 45.0o A = 31.2o

Θ = 360o − 45.0o = 315.0o Θ = 180o − 31.2o = 148.8o

35. R = −1.40m( )2 + 9.60m( )2 = 9.70m 36. R = 425km( )2 + −365km( )2 = 560 km

tan A =9.6m

1.40m tan A =

365km425km

A = 81.7o A = 40.7o

Θ = 180o − 81.7o = 98.3o Θ = 360o − 40.7o = 319.3o

37. R = 19.5m( )2 + −49.6m( )2 = 53.3m 38. R = −158km( )2 + 236km( )2 = 284km

tan A =49.6m19.5m

tan A =236km158km

A = 68.5o A = 56.2o

Θ = 360o − 68.5o = 291.5o Θ = 180o − 56.2o = 123.8o

39. R = 14.7mi( )2 + 16.8mi( )2 = 22.3mi . 40. R = −3240 ft( )2 + −1890 ft( )2 = 3750 ft

tan A =16.8mi14.7mi

tan A =1890 ft3240 ft

A = 48.8o = Θ A = 30.3o

Θ = 180 o + 30.3o = 210.3o

41. R = −9.65m( )2 + 4.36m( )2 = 10.6m 42. R = 375km( )2 + −408km( )2 = 554km

tan A =4.36m9.65m

tan A =408km375km

A = 24.3o A = 47.4o

Θ = 180 o − 24.3o = 155.7o Θ = 360o − 47.4o = 312.6o

43. R = Rx

2+ Ry

2= 150m

2

+ −100m2

= 180m

44. R = Rx

2+ Ry

2= −275m 2 + 350m

2

= 445m

45. R= 307 km at 48.4º north of west 46. 61.3 mi at 65.2º south of west

47. 155 km at 58.4º south of west 48. 29.1 km at 12.7º north of west


1. d

2. a



23

3. a

4. Two intersecting lines (axes) determine a plane (the number plane). The origin is the point where the

axes intersect.

5. yes

6. yes

7. Graph each vector placing the initial point of each at the endpoint of the previous one; the resultant is

the vector from the beginning point to the end point of the last vector.

8. Add the x components and then add the y components. These are components of the resultant.

9. No

10. Counterclockwise

11. In the third quadrant, the angle measure must be between 180° and 270°.

12. Complete the right triangle with the legs behind the x- and y- components of the vector. Then find the

lengths of the x and y and determine their signs.

13. Complete the right triangle with the legs being the x- and y- axes.


1. x=(cos 30.0°)(13.0cm) =11.3 cm

y= (sin30.0°)(13.0cm) = 6.50 cm

2. Rx =10.0 cm cos 60.0°= 5.00cm

Ry =10.0 cm sin 60.0° = 8.66 cm

3. Rx = 20.0 cm cos 30.0° = 17.3 cm

Ry = 20.0 cm sin 30.0° = 10.0cm

4. x = (cos 60.0°)(9.0 cm) = -4.50 cm

y = (sin 60.0°)(9.00 cm) = -7.59 cm

5. Rx = 9.00cm cos 40.0° = 6.89cm

Ry = 9.00cm sin 40.0° = 5.79cm

6. Rx = 18.0cm cos 240.0° = −9.00cm

Ry = 18.0cm sin 240.0° = −15.6cm

7. 1.00 cm (3.00 km / 1.00 cm) = 3.00 km North

8. 1.00 cm (3.00 km / 1.00 cm) = 3.00 km South



24

9. x – component y-component

A -13.0 cm 0.00 cm

B 0.00 cm 9.00 cm

C 2.00 cm 0.00 cm

R -11.0 cm 9.00 cm

R = x2 + y 2 = −11.0cm( )2 + 9.00cm( )2 = 14.2cm

14.2cm20.0km( )1.00cm( ) = 284km

Θ = tan−1 9.00

11.0cm= 39.3o west of north


A 14.5 cm 6.76 cm

B 0.00 cm 6.00 cm

C -5.00 cm 0.00 cm

R 9.50 cm 12.8 cm

R = x2 + y 2 = 9.50cm( )2 + 12.8cm( )2 = 15.9cm20.0km1.00cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 318km

tan−1 12.8cm

9.50cm= 53.4onorth of east

11. R = 14.0( )2 + 3.00( )2 = 14.3 12. R = x2 + y 2 = −5.00( )2 + 10.0( )2 = 11.2

13. R = x2 + y 2 = 8.00( )2 + −2.00( )2 = 8.25 14. R = x2 + y 2 = −3.00( )2 + −4.00( )2 = 5.00

15.

x – component y-component

A 3.00 cm 4.00 cm

B 5.00 cm -7.00 cm

C -2.00 cm 1.00 cm

R 6.00 cm -2.00 cm

16.


A 5.00 cm 7.00 cm

B 9.00 cm -3.00 cm

C -5.00 cm 5.00 cm

R 9.00 cm 9.00 cm



25

17.


A -3.00 cm -4.00 cm

B -5.00 cm 7.00 cm

C 2.00 cm -1.00 cm

R -6.00 cm 2.00 cm

18.


A -5.00 cm -7.00 cm

B -9.00 cm 3.00 cm

C 5.00 cm -5.00 cm

R -9.00 cm -9.00 cm

19.


A -6 4

B -3 -8

C 4 0

D 0 -5

R -5 -9

20.


A -2 6

B 5 0

C -8 0

D 2 -4

R -3 2

21. 389 km at 63.6º south of west

22. 344 km at 25.3º south of east

Chapter 3 Applied Concepts

1. ( ) ( )2 22 2 176 378 417R a b ft ft ft= + = + =

( ) ( )2 22 2 176 52.0 184R a b ft ft ft= + = + =



26

2. 800 sin 60.0 693xR m m= =o ; 800 cos 60.0 400xR m m= =o ; 1090x xR R m+ =

3. d = 18.0 m tan 37.0° = 13.6 m

4.


A 0.00 mi 3.50 mi

B 0.707 mi 0.707 mi

C 0.00 mi -1.50 mi

R 0.707 mi 2.71 mi

( ) ( )2 20.707 2.71 2.80R mi mi mi= + =


A 0.00 km 500 km

B 141 km 141 km

R 141 km 641 km

( ) ( )2 2141 641 656R km km km= + =



27

Chapter 4

4.1

1. 50 mih 2. 22 m/s 3. 21.6 m/s 4. 90 km/h 5. 68.3 ft/s

6. v= s/t = 1.00mi30.0s

×3600s

1h= 120 mi

h 7. s = vt = 90 kmh × 3.5h = 320km

8. s = vt = 90 kmh × 1.0s×

1000m1km

×1h

3600s= 25m 9. (a) 55

mih

×5280 ft1mi

×1h

3600s= 81 ft / s

(b) 55mih

×1610m

1mi×

1h3600s

= 25m / s (c) 55mih

×1.61km

1mi= 89km

h

10. (a) 22.0ms

×1km

1000m×

3600s1h

= 79.2 kmh (b) 22.0

ms

×1mi

1610m×

3600s1h

= 49.2 mih

(c) 22.0ms

×3.28 ft

1m= 72.2 ft

s 11. s = vt = (100km / h)(2.75h) = 275km

12. s = vt = (105km / h)(3.85h) = 404km 13. t =s

v=

265km

60.0km / h= 4.42h

14. s = vt = (55.0km / h)(3.65h) = 201km

15. 160km2.0h

= 80 kmh east 16.

10 0km3.0h

= 33kmh north

17. 100 0km8.00h

= 125mih south 18.

31.0mi0.500h

= 62.0 mih west

19. 275km4.50h

= 61.1kmh at30o south of east 20.

426km2.75h

= 155kmh at45o north of west

21. (a) 255mi4.75h

= 53.7 mih (b)

121mi4.75h

= 25.5mih west

22. (a) 120mi4.25h

= 28 mih (b)

45mi4.25h

= 11mih at11oeast of north

23. 370 kmh at90.0o 24. 195km

h at180.0o

25. R = −235km /h( )2 + −45.0km /h( )2 = 239km /h

tan A= 45.0km / h235km /h

A = 10.8o

θ = 180o + 10.8o = 190.8o

26. R = 35.0mi / h( )2 + 185mi / h( )2 = 188mi /h

tan A = 185mi / h35.0mi / h

A = 79.3o = θ



28

27. Plane: 155 mi/h SW (at 225.0 °)

x = (cos 45.0°)(155 mi/h) = -110 mi / h

y= (sin 45.0°)(155 mi/h) = -110 mi/h

Wind: 45.0 mi/h at 0°

x= 45.0 mi/h

y= 0

Sum of Components:

x= -65 mi/h

y= -110 mi/h

R = −65mi / h( )2 + −110 mi / h( )2= 128mi / h

tan A= 110 mi / h65mi /h

A= 59.4°

θ = 180° + 59.4° = 239.4°

28. Plane: 215 km/h SE (at 315.0°)

x= (cos 45.0°)(215 km/h) = 152 km/h

y= (sin 45.0°)(215 km/h) = - 152 km/h

Wind: 75.0 km/h at 270.0°

x= 0

y= -75.0 km/h

Sum of Components:

x= 152 km/h

y= -227 km/h

R = 152km /h( )2 + −227km / h( )2 = 273km /h

tan A= 227km /h152km / h

A= 56.2 °

θ = 360° − 56.2° = 303.8°

29. Plane: 190 kmh at155.0°

x= cos 25.0°( ) 190 km /h( )= −172km /h

y= sin25.0°( ) 190 km / h( )= 80.3km /h

Wind: 45.0 kmh at195.0°

x = cos15.0°( ) 45.0km /h( ) = −43.5km / h

y= sin15.0°( ) 45.0km / h( ) = −11.6km / h



29

Sum of Components:

x= -216 km/h

y= 68.7 km/h

R = −216km / h( )2 + 68.7km /h( )2 = 227km /h

tan A= 68.7km / h216km /h

A= 17.6°

θ = 180° −17.6° = 162.4°

30. Plane: 150 mih at216.0°

x = cos 36.0°( ) 150 mi /h( )= −121mi / h

y = sin36.0°( ) 150 mi / h( )= −88.2mi / h

Wind: 55.0 mi/h at 255.0°

x = cos 75.0°( ) 55.0mi /h( ) = −14.2mi / h

y = sin 75.0°( ) 55.0mi / h( ) = −53.1mi /h

Sum of Components:

x = 135mi / h y = −141.3mi /h

R = −135mi /h( )2 + −141.3mi /h( )2 = 195mi /h

tan A= 141.3mi /h135mi /h

A= 46.3°

θ = 180° + 46.3 = 226.3°

4.2

1. a =Δvt

=15m / s

1.0s= 15m / s 2 2. a =

Δvt

=18m / s

3.0s= 6.0m / s2

3. a =Δvt

=70 ft / s − 60 ft / s

1.0s= 10 ft / s 2 4. a =

Δvt

=65m / s − 45m / s

2.0s= 10 m / s 2

5. a =Δvt

=90 km /h − 25km / h

5.6s×

1000m1km

×1h

3600s= 3.2m / s 2

6. a =Δvt

=50 mi /h −10 mi /h

3.5s×

5280 ft1mi

×1h

3600s= 17 ft / s2

7. a =Δvt

=62.5m / s − 0

10.0s= 6.25m / s2 8. a =

Δvt

=55mi /h − 25mi /h

4.5s×

5280 ft1mi

×1h

3600s= 9.8 ft / s 2

9. a =Δvt

=110 kmh / −10 km /h

135s×

1000m1km

×1h

3600s= 0.206m / s 2



30

10. Δv = at = 30.0 ft / s 2( )3.30s( ) = 99.0 ft / s

11. From # 10: 99.0 ft / s×1mi

5280 ft×

3600s1h

= 67.5mi / h

12. Δv = at = 10.0m / s 2( )20.0s( ) = 200 m / s

13. From # 12: 200 ms

×1km

100m×

3600s1h

= 720 km / h

14. t =Δva

=65.0m / s − 20.0m / s

15.0m / s2 = 3.00s

15. a =Δv

t=

10km / h1h

3600s⎛⎝

⎞⎠

1000m

1km⎛⎝

⎞⎠

5.20s= 0.534m / s2

16. a =Δv

t=

20km / h1h

3600s⎛⎝

⎞⎠

1000m

1km⎛⎝

⎞⎠

4.80s= 1.16m / s2

17. t =Δva

=90.0km /h1.50m / s2

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 16.7s

18. t =Δva

=120 km /h3.50m / s2

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 9.52s

19. a =Δv

t=

3.00m / s

4.20s= 0.714m / s2

20. Δv = at = (3.25m / s2 )(5.80s) = 18.9m / s

21. a =Δvt

=16.0m / s − 2.00m / s

3.50m / s2 = 4.00m / s 2

22. a =Δvt

=−9.00m / s − −2.00m / s( )

2.00s= −3.50m /s 2

23. a =Δvt

=0m / s − 25.0m / s

3.00s= −8.33m / s2 24. a =

Δvt

=0m / s − 25.0m / s

6.00s= −4.17m / s 2

25. a =Δvt

=0m / s − 50.0m / s

3.00s= −16.7m /s 2 26. a =

Δvt

=0m / s − 50.0m / s

6.00s= −8.33m / s2

4.3

1. vavg =v f + vi

2=

6.20m / s + 3.90m / s2

= 5.05m / s

2. vi = v f − aavgt = 16.8m / s − 3.07m / s 2( )4.10s( ) = 4.2m / s

3. s = vi t +12

aavgt 2 = 33.0m /s( ) 3.00s( )+ 1 /2( ) 6.40m / s 2( )3.00s( )2 = 127.8m



31

4. v f = 2aavgs + vi2 = 2 8.41m / s2( )4.81m( ) + 1.24m / s( )2 = 9.08m / s

5. aavg =v f − vi

t=

10.40 ft / s − 4.01 ft / s3.00s

= 2.13 ft / s2

6. t =s

aavg=

35.0m

15.0kmh

×1000m1km

×1h

3600s

= 8.40s

7. vavg =v f + vi

2=

18.0mi / h + 02

= 9.00mi / h

8. t =v f + vi

aavg=

50.0 ft / s − 032.2 ft / s 2 = 1.55s

9. s =1

2(v f + vi )t =

1

2(1.75m / s + 0m / s)(2.50s) = 2.19m

10. s =1

2(v f + vi )t =

1

2(50.0m / s + 0m / s)(9.80s) = 245m

11. a = vit +12

aavgt 2 = 0( )t +12

44.0 ft / s2( )5.00s( )2 = 550 ft

12. t =v f − vi

aavg=

74.0 ft / s − 5.00 ft / s2.00 ft / s 2 = 34.5s

13. aavg =v f − vi

t=

120 km /h − 85km / h( )9.2s

×1h

3600s×

1000m1km

= 1.1m / s2

14. t =2s

aavg=

2 95.0m( )9.80m / s 2 = 4.40s v f = vi + aavgt = 0 + 9.80m / s 2( )4.40s( ) = 43.1m / s

15. a =v f

2 − vi2

2s=

02 − 295km /h( )×1000m

1km×

1h3600s

⎡ ⎣ ⎢

⎤ ⎦ ⎥

2

2 205m( ) = −16.4m / s2

16. v f = 2aavgs + v12 = 2( ) 32.2 ft / s2( ) 43.0 ft( )+ 62.0 ft / s( )2 = 81.3 ft / s

17. aavg =v f − vi

t=

0 − 70 km / h( )12s

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = −1.6m / s 2

18. (a) t =v f − vi

aavg=

90 km / h − 60 km / h( )3.6m / s2 ×

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 2.3s

(b) s =12

v f − vi( )t =12

90 km / h + 60 km /h( ) 2.3s( ) 1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 48m

19. (a) v f = vi + aavgt = 0 + 9.80m / s 2( )2.40s( ) = 23.5m / s

(b) s =12

v f + vi( )t =12

23.5m / s + 0( ) 2.40s( ) = 28.2m



32

20. (a) vi = v f2 − 2aavgs = 02 − 2 32.2 ft / s 2( )−700 0 ft( ) = 671 ft / s

(b) t =v f − vi

aavg=

0 − 671 ft / s( )32.2 ft / s2 = −20.8s

(c) ttotal = tup + tdown = 20.8s + 20.8s = 41.6s

21. (a) s =v f

2 − vi2

2aavg=

0 − −250 m / s( )2

2 9.80m / s 2( ) = 3190m (b) t =v f − vi

aavg=

0 − −250 m / s( )9.80m / s2 = 25.5s

(c) ttotal = tup + tdown = 25.5s + 25.5s = 51.0s

22. (a) v f = vi + aavgt = 30.0 ft / s + 32.2 ft / s2( )3.50s( ) = 143 ft / s

(b) s =v f

2 − vi2

2aavg=

143 ft / s( )2 − 30.0 ft / s( )2

2 32.2 ft / s 2( ) = 304 ft

23. (a) t =v f − vi

aavg=

0 − −10.0m / s( )9.80m / s2 = 1.02s (b) s =

v f2 − vi

2

2aavg=

02 − −10.0m / s( )2

2 9.80m / s2( ) = 5.10m

(c) s = 5.10m + 25.0m = 30.1m

v f = vi2 + 2aavgs = 02 + 2 9.80m / s2( )30.1m( ) = 24.3m / s

(d) tdown =v f − vi

aavg=

24.3m / s − 09.80m / s2 = 2.48s

tup + t down = 1.02s + 2.48s = 3.50s

24. (a) vi = v f2 − 2aavgs = 02 − 2 9.80m / s2( )−15.0m( ) = 17.1m / s

(b) t =v f − vi

aavg=

0 − −17.1m / s( )9.80m / s 2 = 1.74s

(c) s=15.0m + 40.0m = 55.0m

v f = vi2 + 2aavgs = 02 + 2 9.80m / s2( )55.0m( ) = 32.8m / s

(d) tdown =v f − vi

aavg=

32.8m / s − 09.80m / s 2 = 3.35s

t = tup + t down = 1.74s + 3.35s = 5.09s

25. (a) vi = v f2 − 2aavgs = 02 − 2 32.2 ft / s 2( )−30.0 ft( ) = 44.0 ft / s

(b) s = 255.0 ft + 30.0 ft = 285.0 ft; t =2s

aavg=

2 285.0 ft( )32.2 ft / s 2 = 4.21s

(c) v f = vi2 + 2aavgs = 02 + 2 32.2 ft / s 2( )285.0 ft( ) = 135 ft / s



33

26. (a) Hammer Up:

s =v f

2 − vi2

2aavg=

02 − −40.0 ft / s( )2

2 32.2 ft / s 2( ) = −24.8 ft

t =v f − vi

aavg=

0 − −40.0 ft / s( )32.2 ft / s2 = 1.24s

Hammer Down:

s= 275.0 ft + 24.8 ft = 299.8 ft

t =2s

aavg=

2 299.8 ft( )32.2 ft / s 2 = 4.32s

thammer = 1.24s + 4.32s = 5.56s

Wrench:

t = 2s

aavg=

2 275.0 ft( )32.2 ft / s 2 = 4.13s

Δt = 5.56s − 4.13s = 1.43s

(b) v f = vi2 + 2aavgs = 02 + 2 32.2 ft / s 2( )275.0 ft( ) = 133 ft / s

(c) v f = vi2 + 2aavgs = 02 + 2 32.2 ft / s 2( )299.8 ft( ) = 139 ft / s

(d) 1.24 s; from part a

(e) s =12

at 2 =12

32.2 ft / s 2( )1.24s( )2 = 24.8 ft (fell)

Δs = 275.0 ft − 24.8 fft = 250.2 ft (above the ground)

27. Let = time of the second ball

t + 1= time of the first ball

s1 = s2 = vi t +12

at 2

0( )t +12

32.2 ft / s2( ) t + 1( )2 = 40.0 ft / s( )t +12

32.2 ft / s 2( )t 2

t = 2.06 s

28. v f = vi + at = 2.00m / s + 4.00m / s 2( )2.50s( ) = 12.0m / s

29. v f = vi + at = 30.0km / h + 3.50m / s 2( )6.80s( ) 3600s1h

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1km1000m

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 116km /h

30. t =v f − vi

a=

28.0m / s − 0m / s5.50m / s2 = 5.09s 31. t =

v f − vi

a=

3.00m / s − 22.0m / s−2.10m / s2 = 9.05s



34

4.4

1. vx = v cosθ = 35.0m / s cos15.0° = 33.8m / s

vy = v sinθ = 35.0m / s sin15.0° = 9.06m / s

sx = vxt = 33.8 ms t

t =v fy − v iy

ay=

−9.06m / s − 9.06m / s−9.80m / s2 = 1.85s

sx = 33.8m / s( ) 1.85s( ) = 62.5m

2. vx = v cosθ = 35.0m / s cos75.0° = 9.06m / s

vy = v sinθ = 35.0m / s sin 75.0° = 33.8m / s


t =v fy − v iy

ay=

−33.8m / s − 33.8m / s−9.80m / s2 = 6.90s

sx = vxt = 9.06m / s( ) 6.90s( ) = 62.5m

3. vx = v cosθ = 35.0m / s cos35.0° = 28.7m / s

vy = v sinθ = 35.0m / s sin 35.0° = 20.1m / s


t =v fy − v iy

ay=

−20.1m / s − 20.1m / s−9.80m / s 2 = 4.10s

sx = 28.7m / s( ) 4.10s( ) = 118m

4. vx = v cosθ = 35.0m / s cos55.0° = 20.1m / s

vy = v sinθ = 35.0m / s sin 55.0° = 28.7m / s

sx = vxt = 20.1ms t

t =v fy − v iy

ay=

−28.7m / s − 28.7m / s−9.80m / s2 = 5.86s

sx = 20.1m / s( ) 5.86s( ) = 118m

5. vx = v cosθ = 35.0m / s cos45.0° = 24.7m / s

vy = v sinθ = 35.0m / s sin 45.0° = 24.7m / s


t =v fy − v iy

ay=

−24.7m / s − 24.7m / s−9.80m / s2 = 5.04s

sx = 24.7m / s( ) 5.04s( ) = 124m

6. The range increases as the angles approach 45.0°. The angles above and below 45.0° are complimentary

angles and therefore have the same range.



35

7. vx = v cosθ = 67.0 ms cos 62.0° = 31.5m

s

vy = v sinθ = 67ms sin 62.0° = 59.2m

s

sx = vxt = 31.5ms t

t =v fy − v iy

ay=

−59.2m / s − 59.2m / s−9.80m / s2 = 12.1s

sx = 31.5m / s( ) 12.1s( ) = 381m

8. vx = v cosθ = 33.0 fts cos 85.0° = 2.88 ft

s

vy = v sinθ = 33.0 fts sin 85.0° = 32.9 ft

s

sx = vxt = 2.88 fts t

t =v fy − v iy

ay=

−39.2 ft / s − 39.2 ft / s−32.2 ft / s 2 = 2.04s

sx = 2.88 ft / s( ) 2.04s( ) = 5.87 ft

9. vx = v cosθ = 110 ft s cos 25.0° = 99.7 fts

vy = v sinθ = 110 ft s sin 25.0° = 46.5 fts


t =v fy − v iy

ay=

−46.5 ft / s − 46.5 ft / s−32.2 ft / s2 = 2.89s

sx = 99.7 ft / s( ) 2.89s( ) = 288 ft the ball will bounce

10. t =

2sy

g=

2(1.40m)

9.80m / s2= 0.535s

sx = vxt = (0.600m / s)(0.535s) = 0.321m

11. t =

2sy

g=

2(1.50m)

9.80m / s2= 0.553s

sx = vxt = (0.800m / s)(0.553s) = 0.442m


1. c

2. c

3. a

4. b

5. d

6. gravity accelerates all objects at the same rate.

7. Horizontal motion is independent of the vertical pull of gravity

8. Velocity has magnitude and direction, where speed has only magnitude.

9. No; anything speeding up of slowing down has a changing velocity.



36

10. Vectors are necessary to determine the direction of the velocity, acceleration, and other vector

quantities. Examples: navigating planes, boats and spacecrafts.

11. (a) An automobile speeding up of slowing down. (b) Anything being dropped to the ground (c) A

bullet being fired.

12. Acceleration is change in velocity; deceleration is a special case of acceleration where the object is

slowing; acceleration need not be uniform and so may be subject to averaging.

13. 9.80 m / s2 (metric); 32.2 ft / s2 (English)


1. s = vavgt = 17.0mi /h( ) 1.50h( ) = 25.5mi 2. t =3000mi

550 mi / h= 5.45h

3. Plane: 215 km/hat 90.0°

x= 0

y=215 km/h

Wind: 69 km/h at 180.0°

x= -69km/h

y=0

Sum of Components:

x= -69 km/h

y= 215 km/h

R = −69km / h( )2 + 215km /h( )2 = 226km /h

tan A =215km / h69km / h

A= 72.2°

θ = 180° − 72.2° = 107.8°

4. Glider: 25.0 km/h at 320.0°

x = cos 40.0°( ) 25.0km / h( ) = 19.2km / h

y = sin 40.0°( ) 25.0km / h( ) = −16.1km /h

Wind:12.0 km/h at 15.0°

x = cos15.0°( ) 12.0km / h( ) = 11.6km / h

y = sin15.0°( ) 12.0km / h( ) = 3.10km /h

Sum of Components

x= 30.8 km/h

y= -13.0 km/h

R = 30.8km / h( )2 + −13.0km / h( )2 = 33.4km / h



37

tan A =13.0km /h30.8km / h

A= 22.9°

θ = 360° − 22.9° = 337.1°

5. a =Δvt

=8.00 ft / s

2.00s= 4.00 ft / s 2

6. t =v f − vi

a=

150km /h( )×1h

3600s×

1000m1km

− 0

6.0m / s2 = 6.9s

7. vavg =v f + vi

2=

20.0km / h + 10.0km / h2

= 15.0km / h

8. v f = vi + at = 0 + 1.30m / s2( )3.00s( ) = 3.90m / s 9. t =sv

=1.5× 104 km50 0km / h

= 30 h

10. vi = v f − at = 110km /h( )×1h

3600s×

1000m1km

− 0.50m / s 2( )36s( ) = 13m / s

11. vavg =v f + vi

2=

0 + 8.00m / s2

= 4.00m / s

12. a =Δvt

=120km / h ×

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟

13s= 2.6m / s2

13. aavg =v f

2 − vi2

2s=

71.0m / s( )2 − 0m / s( )2

2 1000m( ) = 2.52m / s2

14. v f = 2aavgs + vi2 = 2 3.00m / s2( )535m( )+ 21.0m / s( )2 = 60.4m / s

15. (a) vi = v f2 − 2aavgs = 02−2 9.80m / s 2( )−2150m( ) = 205m / s

(b) t =v f − vi

aavg=

0 − −205m / s( )9.80m / s 2 = 20.9s

(c) t = tup + t down = 20.9s + 20.9s = 41.8s

16. (a) v f = vi + aavg t = 10.0m / s + 9.80m / s 2( )2.75s( ) = 37.0m / s

(b) s = vi t +12

aavgt 2 = 10.0m / s( ) 2.75s( ) +12

9.80m / s 2( )2.75s( )2 = 64.6m

17. vx = v cosθ = 9.43m / s cos 55.0° = 5.41m / s

vy = v sinθ = 9.43m / ssin 55.0° = 7.72m / s

sx = vxt = 5.41ms t

t =v fy − v iy

ay=

−7.72m / s − 7.72m / s−9.8m / s 2 = 1.58s

sx = 5.41m / s( ) 1.58s( ) = 8.55m



38

18. vx = v cosθ = 47.0 ft / s cos25.0° = 42.6 ft / s

vy = v sinθ = 47.0 ft / ssin 25.0° = 19.9 ft / s


t =v fy − v iy

ay=

−19.9 ft / s −19.9 ft / s−32.2 ft / s2 = 1.24s

sx = 42.6 ft / s( ) 1.24s( ) = 52.8 ft The arrow will miss the bull’s eye.


1. (a) v =st

=57.5m15.3s

= 3.76m / s

(b) vPeopleMover = vOverAll − vAmy−Walking = 3.76m / s −1.75m / s = 2.01m / s

2. R = vx2 + vy

2 = 15.7mi / h( )2 + 5.35mi /h( )2 = 16.6mi / h

θ = tan−1 vy

vx= tan−1 5.35mi /h

15.7mi / h= 18.8°

3. d =12

v f + vi( )t =12

29.3 ft / s + 88.0 ft / s( )10.8s = 633 ft

4. vx =sx

t=

7.35mt

t =sy × 2

ay=

2 12m( )9.80m / s 2 = 1.56s

vx =7.35m1.56s

= 4.71m / s

5. vx =sx

t=

13.5mt

t =2sy

ay=

2 1.40m( )9.80m / s 2 = 0.535s

vx =13.5m0.535s

= 25.2m / s



39

Chapter 5

5.2

1. 30.0 N

2. 2.00 N

3. 744 lb.

4. 546 N

5. 252 lb.

6. 126 lb.

7. 230 N

8. 3.37 lb.

9. 40.0 m/s 2

10. 1.71 m/s 2

11. 11.7 16.5 m/s 2

12. 16.5 ft/s 2

13. 0.518 m/s 2

14. 0.393 m/s 2

15. 1.39 ft/s 2

16. 0.195 m/s 2

17. F = ma = 1750kg( ) 3.00m / s 2( )= 5250N

18. a =Fm

=93.0N6.00kg

= 15.5m / s 2

19. F = ma = 120 slugs( )11.0 ft / s2( )= 1320lb

20. F = ma = 25000slugs( ) 28.0 ft / s 2( )= 7.00× 105lb

21. F = ma = 975kg( ) 2.50m / s2( )= 2440N

22. F = ma = 432kg( ) 1.75m / s2( )= 756N

23. a =Δv

t=

25.0km / h − 75.0km / h1000m

1km⎛⎝

⎞⎠

1h

3600s⎛⎝

⎞⎠

8.20s= 1.69m / s2

24. a =Δv

t=

3.00km / h − 15.0km / h1000m

1km⎛⎝

⎞⎠

1h

3600s⎛⎝

⎞⎠

2.70s= 1.23m / s2

25. F = ma = 140kg( ) 41.0m / s 2( )= 5, 740N 26. a =Fm

=300 N

0.750kg= 400 m / s2

27. m =Fa

=90.0N

15.0m / s2 = 6.00kg

28. (a) a =Fm

=150 0 lb

105slugs= 14.3 ft / s2 (b) a =

Fm

=150 0lb

130 slugs= 11.5 ft / s2

29. a =Fm

=6.75× 106 N5.27 × 105 kg

= 12.8m / s 2

30. a =Fm

=8.50N

80.0 + 15.5( )kg= 0.0890m / s2

31. (a) m =Fa

=140 kg m

s2

19.0ms

2 = 7.37kg (b) It is halved. 9.50m / s 2 (c) It is doubled. 38.0m / s 2



40

32. (a) F = ma = 275kg( ) −4.50m / s 2( )= −1240N

(b) The negative sign on the force shows it is directed opposite the scooter velocity.

33. (a) aavg =v f

2 − vi2

2s=

0m / s( )2 − 105kmh

×1h

3600s×

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

2 53.0m( ) = −8.03m / s2

F = 1230kg( ) −8.30m / s2( )= −9880N

The force acts opposite to the direction of the pickup.

(b) Double the force would be required to stop the truck in half the distance.

5.3

1. Ff = μFN = 0.16( ) 2400N( ) = 380N 2. Ff = μFn = 0.40( ) 780lb( ) = 310lb

3. Ff = μFN = 0.23( ) 4700N( ) = 1100N 4. μ =Ff

FN=

850N4700N

= 0.18

5. μ =Ff

FN=

60 lb750lb

= 0.080 6. FN =Ff

μ=

2.5× 106N0.65

= 3.8 × 106 N

7. FN =Ff

μ=

2500lb0.10

= 25, 000lb 8. Ff = μFN = 0.12( ) 3560N( ) = 430N

9. Ff = μFN = 0.13( ) 12,000N( ) = 1600N 10. Ff = μFN = 0.140( ) 14, 000N( )= 1960N

11. Ff = μFN = 0.130( ) 20, 000N( )= 2600N 12. Ff = μFN = 0.080( ) 12, 000N( ) = 960N

13. μ =Ff

FN=

14.0N40.0N

= 0.350 14. Ff = μFN = 0.350( ) 60.0N( ) = 21.0N

15. (a) Ff = μFN = μmg = − 0.500( ) 750 kg( ) 9.80m / s2( )= −3680N

The (-) sign indicates the frictional force acts in the opposite direction to that of the truck.

(b) a =Fm

=−3680N750 kg

= −4.91m / s2

(c) s =v f

2 − vi2

2a=

0m / s( )2 − 30.0m / s( )2

2 −4.91m / s2( ) = 91.6m

16. (a) L arger Ff = μFN( ) (b) Shorter Ff IsGreater( )

5.4

1. 3.0 N; right 2. 180 N; left 3. 15.0 N; left 4. 54 lb; left 5. 4 N; left 6. 163 N; right

7. a = F /m =500 lb −100 lb

100 slugs= 4.00 ft / s 2 8. a = F /m =

220 0N − 450N150 0kg

= 1.17m / s2

9. a = F /m =890 0N − 2230N

13,100kg= 0.509m / s 2 10. a = F /m =

300 lb −100 lb30.0slugs

= 6.67 ft / s 2



41

11. a = F / m =9200N − 2150N

14, 000kg= 0.504m / s2 12. a = F / m =

8750N − 2000N

5000kg= 1.35m / s2

13. 1095N – 975N = 120N 14. a = F / m =1975N − 550N

2000kg= 0.713m / s2

5.5

1. F = mg = 30.0kg( ) 9.80m / s2( )= 294N 2. F = mg = 60.0kg( ) 9.80m / s2( )= 588N

3. F = mg = 10slugs( ) 32.2 ft / s 2( )= 322lb 4. F = mg = 9.00kg( ) 9.80m / s2( )= 88.2N

5. m =Fg

=17.0N

9.80m / s2 = 1.73kg 6. m =Fg

=21.0lb

32.2m / s2 = 0.652slug

7. m =Fg

=12, 0 00N9.80m / s2 = 1220kg 8. m =

Fg

=25, 0 00N9.80m / s2 = 2550kg

9. m =Fg

=6.7× 1012 lb9.80m / s2 = 6.8× 1011kg 10. m =

Fg

=5.5× 106 lb32.2 ft / s 2 = 1.7× 105slugs

11. F = mg = 1150kg( ) 9.80m / s 2( )= 11, 300N 12. F = mg = 81.5slugs( ) 32.2 ft / s2( )= 2620lb

13. m =Fg

=2750lb

32.2 ft / s 2 = 85.4slugs 14. m =F

g=

20, 000N

9.80m / s2= 2040kg

15. m =F

g=

7500N

9.80m / s2= 765kg 16. m =

Fg

=11, 500N9.80m / s2 = 1170kg

17. (a) F = mg = 1350kg( ) 9.80m / s 2( )= 13, 200N (b) F = mg = 1350kg( ) 1.63m / s2( )= 220 0N

18. (a) m =Fg

=115lb

32.2 ft / s 2 = 3.57slugs (b) F = mg = 3.57slugs( ) 5.36 ft / s2( )= 19.1lb

19. (a) m= 65.0kg (b) F= mg= 65.0kg( ) 1.63m / s2( )= 106N

20. Answers vary 21. Answers vary 22. Answers vary 23. Answers vary

24. (a) 65.0kg ×1slug

14.6kg= 4.45slugs (b) F = mg = 4.45slugs( ) 7.85 ft / s2( )= 34.9lb

25. (a) m =Fg

=115lb

32.2 ft / s 2 = 3.57slugs (b) F = mg = 3.57slugs( ) 85.0 ft / s 2( )= 303lb

26. (a) m= 65.0 kg (b) F = mg = 65.0kg( ) 3.72m / s2( )= 242N

27. Answers vary 28. Answers vary

29. m =F

g=

995N

9.80m / s2= 102kg 30. m =

F

g=

210N

9.80m / s2= 21.4kg



42


1. d

2. b

3. b

4. c

5. a

6. (a) A bridge being supported over a river. (b) Isometric exercises (c) A magnetic on a refrigerator.

7. (a)No (b) No

8. A car hit from behind is forced into the car ahead of it.

9. A body in motion continues in motion unless a force acts on it, and a body at rest continues to stay at

rest unless a force acts on it.

10. Acceleration is a change of velocity.

11. Only if they had the same mass.

12. Yes; 3.00 lb=13.35 N

13. More difficult; everything would slide.

14. Weight is a measure of gravitational pull. The moon, having less mass than the earth, exerts less

gravitational pull.

15. For every force applies by object A to B, there is an equal force applied by B to A in the opposite

direction.

16. For every action, there is an equal and opposite reaction.


1. a =Fm

=18.0N6.00kg

= 3.00m / s2 2. m =Fa

=825N

11.0m / s2 = 75.0kg 3. a =Fm

=17.0lb

0.89slug= 19 ft / s 2

4. F = ma = 2400kg( ) 8.0m / s2( )= 19, 000N 5. F = 64.0N 6. μ =Ff

FN=

57N340N

= 0.17

7. FN =Ff

μ=

870N0.23

= 3800N 8. Ff = μFN = 0.20( ) 57N( ) = 11N

9. Fw = mg = 13.0kg( ) 9.80m / s 2( )= 127N 10. Fw = mg = 0.37slug( ) 32.2 ft / s2( )= 12lb

11. F = ma = 50.0kg( ) 4.00m / s2( )= 200N 12. F = ma = 280kg( ) 3.20m / s2( )= 896N

13. μ =Ff

FN

=175N

640N= 0.273 14. Ff = μFN = 0.40( ) 375N( ) = 150N

15. a =F

m=

2100N − 425N

1400kg= 1.20m / s2 16. FN = mg = 375kg( ) 9.80m / s2( )= 3680N

17. m =F

g=

405N

9.80m / s2 = 41.3kg 18. m =F

g=

12.0N

9.80m / s2 = 1.22kg



43


1. (a) a =Fm

=121,000N78, 200kg

= 1.55m / s 2 (b) v f = 2as = 2 1.55m / s2( )500 m( ) = 39.4m / s

(c) v f = 2as = 2 1.55m / s2( )150 0m( ) = 68.2m / s 2. (a) a =Fm

=33.4 × 106 N2.77 × 106 kg

= 12.1m / s2

(b) sy =12

12.1m / s( ) 150 s( )2= 1.36× 105m or136km

3. (a) F = ma = 3.73slugs × 32.2 ft / s2 = 120 lb

(b) F = ma = m ag + ay( )= 3.73slugs 32.2 ft / s 2 + 4.50 ft / s 2( )= 137lb

(c) F = ma = 3.73slugs × 32.2 ft / s2 = 120 lb

(d) F = ma = m ag + ay( )= 3.73slugs 32.2 ft / s 2 + −5.50 ft / s2( )= 99.6lb

4. (a) Ff = μFN = μm × ag = 0.30( ) 242kg( ) −9.80m / s2( )= −714N

(b) Ff = μFN : ma = μmag : a = μag = 0.30( ) −9.80m / s2( )= −2.94m / s 2

(c) t =v f − vi

a=

0m / s 2 − 40.3m / s−2.94m / s2 = 13.7s

5. v f = vi + at = 40.3m / s( )+ −2.92m / s2( )4.55s( ) = 26.9m / s

v f becomes vi for the collision phase

a =v f − vi

t=

0m / s 2 − 26.9m / s0.530s

= −50.8m / s 2

F = ma = 243kg( ) −50.8m / s 2( )= −1.23× 104 N



44

Chapter 6

6.1

1. 80.0 kg m/s 2. 450 kg m/s 3. 765 slug ft/s 4. 3690 kg m/s 5. 9.5× 108kg ms

6. 6.1× 105kg ms

7. m =Fw

a=

1.50 × 105 N( ) 1kg ms

2

1N

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

9.80m / s= 1.53× 104 kg

p = mv = 1.53× 104 kg( )4.50× 104 m / s( )= 6.89 × 108 kg ms

8. m =Fw

a=

3200lb( )1slug ft

s2

1lb

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

32.2 ft / s= 99slugs

p = mv = 99slugs( ) 60 mih

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟

5280 ft1mi

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 8700slug ft

s

9. (a) p = mv = 180 slugs( ) 70.0fts

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 12,600slug ft / s

(b) v =pm

=12600slug ft

s80.0slugs

= 158 ft / s

(c) Fw = mg = 180slugs( ) 32.2 ft / s2( )= 580 0lb; Fw = mg 80.0slugs( ) 32.2 ft / s 2( )= 2580 lb

10. (a) p = mv = 1.0× 10−3 slug( )700 ft / s( )= 0.700slug fts (b) v =

pm

=0.700slug ft

s5.00× 10−4 slug

= 140 0 ft / s

11. (a) p = mv = 2300kg( ) 21.0m / s( ) = 55,200kg ms

(b) v =pm

=55, 200kg m

s1170kg

×1km

1000m×

3600s1h

= 170 km /h

12. (a) p = mv = 0.50kg( ) 6.0m / s( ) = 3.0kg ms (b) p = 0 because v =0 (c) 3.0 km m/s

13. v2 =m1v1

m2=

0.060kg( ) 575m / s( )4.50kg

= 7.67m / s 14. v2 =m1v1

m2=

1.75kg( ) 30 0m / s( )4.500 kg

= 0.117m / s

15. (a) v f = vi2 + 2aavgs = 02 + 2 9.80m / s2( )10.0m( ) = 14.0m / s

(b) p = mv = 125kg( ) 14.0m / s( ) = 1750kg ms



45

16. (a) p = mv = 0.500kg( ) 75.0km / h( ) 1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 10.4kg m

s

(b) p = mv = 0.500kg( ) 85.0km /h + 75.0km /h( )×1000m

1km⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 22.2kg m

s

(c) p = mv = 0.500kg( ) 85.0km /h − 75.0km / h( )×1000m

1km⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 1.39kg m

s

17. (a) t =s

vavg=

0.660m230 m / s

= 0.00287s

(b) F =mv f − mv i

t=

0.0750kg( ) 460 m / s( )− 00.00287s

= 12,0 00N

(c) Ft = 12,0 00N( ) 0.00287s( ) = 34.4kg m s

(d) p = mv = 0.0750kg( ) 460 m / s( )= 34.5kg ms Note: Answer (d) differs slightly from (c) because

of rounding.

18. (a) t =s

vavg=

0.550m263m / s

= 0.00209s


t=

0.0600kg( ) 525m / s( ) − 00.00209s

= 15,100N

(c) Ft = 15,100N( ) 0.00209s( ) = 31.6kg m s

(d) p = mv = 0.0600kg( ) 525m / s( ) = 31.5kg ms

19. (a) 95.0kmh

×1h

3600s×

1000m1km

= 26.4m / s

F =mv f − mv i

t=

1250kg( ) 0( ) − 1250kg( ) 26.4m / s( )4.00s

= −8250N

(b) s =12

v f + vi( )t =12

0 + 26.4m / s( ) 4.00s( ) = 52.8m

20. (a) 90.0kgkmh

×1h

3600s×

1000m1km

= 25.0m / s

25.0kgkmh

×1h

3600s×

1000m1km

= 6.94m / s

F =mv f − mv i

t=

1350kg( ) 6.94m / s( ) − 1350kg( ) 25.0m / s( )4.00s

= 610 0N

(b) s =12

v f + vi( )t =12

6.9m / s + 25.0m / s( ) 4.00s( ) = 63.9m

(c) a =v f − vi

t=

69.4m / s − 25.0m / s4.00s

= −4.52m / s 2

t =v f − vi

a=

0m / s − 25.0m / s

−4.52m / s2= 5.53s



46

21. 35.0km

h×

1h

3600s×

1000m

1km= 9.72m / s

F =mvf − mvi

t=

3000kg( ) 0m / s( ) − 3000kg( ) 9.72m / s( )5.00s

= −5830N

22. 10.0km

h×

1h

3600s×

1000m

1km= 2.77m / s

F =mvf − mvi

t=

5000kg( ) 0m / s( ) − 5000kg( ) 2.77m / s( )6.00s

= −2310N

6.2

1. v2'=m1v1 + m2v2 − m1v1'

m2

=0.500kg( ) 6.00m / s( )+ 0.200kg( ) 0m / s( ) + 0.500kg( ) 2.57m / s( )

0.200kg= 8.58m / s, right

2. (a) v2'=m1v1 + m2v2 − m1v1'

m2=

625g( ) 4.00m / s( ) + 625g( ) 0m / s( ) − 625g( ) 0m / s( )625g

= 4.00m / s , right

(b) same

3.

v2'=m1v1 + m2v2 − m1v1'

m2=

0.600kg( ) 4.00m / s( )+ 1.00kg( ) −5.00m / s( ) − 0.600kg( ) −7.25m / s( )1.00kg

= 1.75m / s

right

4. v2' =m1v1 + m2v2 − m1v1'

m2=

90.0g( ) 3.00m / s( ) + 75.0g( ) −8.00m / s( ) − 90.0g( ) −7.00m / s( )

75.0g= 4.00m / s , left

5. v2' =m1v1 + m2v2 − m1v1'

m2=

98.0kg( ) 1.20m / s( ) + 125.0kg( ) −0.750m / s( ) − 98.0kg( ) −0.986m / s( )

125kg= 0.964m / s , right

6. v2' =m1v1 + m2v2 − m1v1'

m2=

85.0kg( ) −1.30m / s( ) + 75.0kg( ) 0.965m / s( ) − 85.0kg( ) 0.823m / s( )

75.0kg= −1.44m / s , left

7. v' =m1v1 + m2v2

m1 + m2=

2.00 × 104 kg( )6.00m / s( )+ 1.50 × 104 kg( )−4.00m / s( )2.00× 104 kg + 1.50× 104 kg

=1.71 m/s, north



47

8. v' =m1v1 + m2v2

m1 + m2=

2.00 × 104 kg( )6.00m / s( )+ 1.50 × 104 kg( )0m / s( )2.00× 104 kg + 1.50× 104 kg

= 3.43m / s , north

9. v' =m1v1 + m2v2

m1 + m2=

12.0kg( ) 6.00m / s( )+ 4.00kg( ) −3.00m / s( )12.0kg + 4.00kg

= 3.75m / s , right

10. v2 =m1 + v2( )v' −m1v1

m2=

15.0kg + 3.00m / s( ) 1.50m / s( )− 15.0kg( ) 5.00m / s( )3.00kg

= −16.0m / s (to the left)

11. v2 =m1 + v2( )v' −m1v1

m2=

1650kg + 2450kg( ) 3.00m / s( )− 1650kg( ) −12.0m / s( )2450kg

= 13.1m / s

12. v1 =m1 + v2( )v '−m2v2

m1=

16.0g + 4550g( ) 1.20m / s( ) − 4550g( ) 0m / s( )16.0g

= 342m / s

13. (a) v' =m1v1 + m2v2

m1 + m2=

2450kg( ) 12.0m / s( ) + 1650kg( ) −8.00m / s( )2450kg + 1650kg

= 3.95m / s

(b) v' =m1v1 + m2v2

m1 + m2=

2450kg( ) 12.0m / s( ) + 1650kg( ) 8.00m / s( )2450kg + 1650kg

= 10.4m / s

14. (a)

cos 40.0o =

pA'

pA

p

A' = (0.500kg)(0.800m / s) cos 40.0o = 0.306kgm / s

(b)

sin 40.0o =

pB'

pA

p

B' = (0.500kg)(0.800m / s)sin 30.0o = 0.257kgm / s

(c) PA' = mAvA

'

vA' =

0.306kgm / s

0.500kg= 0.612m / s

(d) PB' = mBvB

'

vB' =

0.257kgm / s

0.500kg= 0.514m / s



48

15. (a)

cos 35.0o =

pA'

pB

p

A' = (1000kg)(30.0m / s) cos 35.0o = 2.46x104 kgm / s

(b)

sin 35.0o =

pB'

pA

p

B' = (1000kg)(30.0m / s)sin 35.0o = 1.72x104 kgm / s

(c) PA' = mAvA

'

vA' =

2.46x104 kgm / s

1000kg= 24.6m / s

(d) PB' = mBvB

'

vB' =

1.72x104 kgm / s

1000kg= 17.2m / s

16. p = (1.20x105 kgm / s)2 + (8.50x104 kgm / s)2 = 1.47x105 kgm / s

17. (a)

cos 40.0o =

pA'

pB

p

A' = (950kg)(12.0m / s) cos 40.0o = 8.73x103 kgm / s

(b)

sin 40.0o =

pB'

pA

p

B' = (950kg)(12.0m / s)sin 40.0o = 7.33x103 kgm / s

(c) PA' = mAvA

'

vA' =

8.73x103 kgm / s

950kg= 9.19m / s

(d) PB' = mBvB

'

vB' =

7.33x103 kgm / s

950kg= 9.19m / s



49


1. b

2. d

3. The slow moving has a large mass and a small

velocity while the rifle bullet has a small mass and

a large velocity; the product of either is large.

4. They are the same.

5. The longer the bat (applied force) is on the ball,

the greater the impulse.

6. Total momentum in a system remains constant.

7. Momentum of the escaping gas molecules is

equal to the momentum of the rocket.

8. elastic

9. inelastic

10. They are equal.


1. p = mv = 1475slugs( )57.0mi /h( )×5280 ft

1mi⎛ ⎝ ⎜

⎞ ⎠ ⎟

1h3600s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 1.23× 105 slug ft

s

2. v =pm

=5.50kg m s

27.0kg= 0.204m / s 3. Ft = 125N( ) 2.00 min( ) 60s

1min⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 15, 0 00Ns

4. p = mv = 0.034kg( ) 250m / s( ) = 8.5kg m s 5. v1 =m2v2

m1=

0.00400kg( ) 625m / s( )4.50kg

= 0.556m / s

6. (a) v f = vi2 + 2aavgs = 02 + 2 9.80m / s2( )7.5m( ) = 12m / s

(b) p = mv = 150kg( ) 12m / s( ) = 1800kg m s

7. (a) t =s

vavg=

0.750m1625m / s

= 4.62 × 10−4 s


t=

0.0150kg( ) 3250m / s( )− 04.62× 10−4 s

= 106, 000N

(c) Ft = 106,000N( ) 4.62× 10−4 s( )= 49.0Ns = 49.0kg m s

(d) p = mv = 0.0150kg( ) 3250m / s( ) = 48.8kg m s



50

8. F =m v f − vi( )

Δ t=

1250kg 8.33m / s − 31.9m / s( )3.50s

= −8420N

(a) s =12

v f + vi( )t =12

8.33m / s + 31.9m / s( )3.50s = 70.4m

(b) Δt =m v f − vi( )

F=

1250kg 0m / s − 31.9m / s( )−8420N

= 4.74s

9. v2'=m1v1 + m2v2 − m1v1'

m2=

575g( ) 3.50m / s( ) + 425g( ) 0m / s( ) − 425g( ) 4.03m / s( )575g

= 0.521m / s,right

10. v' =m1v1 + m2v2

m1 + m2=

2.25× 104 kg( )5.50m / s( ) + 3.00× 104 kg( )−1.50m / s( )2.25× 104 kg + 3.00 × 104 kg

= 1.50m / s,east

11. v2'=m1v1 + m2v2 − m1v1'

m2=

0.195kg × 4.50m / s( )+ 0.125kg × −12.0m / s( ) − 0.195kg × −8.40m / s( )0.125kg

= 8.12m / s , right

12. p ' = (9.50x104 kgm / s)2 + (1.05x105 kgm / s)2 = 1.42x105 kgm / s

13. (a) cos 37.0o =

pA'

pA

pA' = (0.35kg)(0.75kg) cos 37.0o = 0.210kgm / s

(b) sin 37.0o =

pB'

pA

pB' = (0.35kg)(0.75kg)sin 37.0o = 0.158kgm / s

(c) pA' = mAvA

'

VA' =

0.210kgm / s

0.35kg= 0.600m / s



51

(d) pB' = mBvB

'

VB' =

0.158kgm / s

0.35kg= 0.451m / s


1. (a) Impulse= m v f − vi( )= 0.123slugs 57.3 ft / s + 11.5 ft / s( ) = 8.46slugs ft s

(b) the outgoing velocity is less, thereby reducing the change in momentum and the impulse.

2. (a) Δpadult = m v f − vi( )= 68.4kg 0m / s − 24.6m / s( ) = −1680Ns

Δpchild = m v f − vi( )= 34.2kg 0m / s − 24.6m / s( ) = −841Ns

(b) Fadult =mΔvΔt

=−1680Ns0.564s

= −2980N

Fchild =mΔv

Δt=

−841Ns0.260s

= −3230N

3. (a) F =m v f − vi( )

Δ t=

70.8kg 0 + 18.5m / s( )0.355s

= 3690N

(b) F =m v f − vi( )

Δ t=

70.8kg 9.75m / s + 18.5m / s( )1.98s

= 1010N

4. (a) vboat =mSallyvSally

mboat=

125lb× 3.50 ft / s65.5lb

= 6.68 ft / s

(b) It is easier to step out of a heavier canoe. The canoe has a greater inertia and does not move

backwards with as large a velocity.

5. (a) v1 =m1 + m2( )v ' −m2v2

m1=

3230kg × −4.31m / s( )− 1510kg × 21.0m / s( )1720kg

= −26.5m / s = −95.4km / h

(The negative sign represents west.)

(b) Yes, the jeep was speeding



52

Chapter 7

7.1

1. 30 N (right) 2. 735 n (left) 3. (a) 400 N (b) 50 N 4. (a) 525 N (b) 75 N

5. x-comp y-comp

F1 0 N 1350 N

F2 925 N 0N

FR 925 N 1350 N

tanα =1350N

925N

α = 55.6° = θ

FR = 925N( )2 + 1350N( )2 = 1650N

6. x-comp y-comp

F1 0 lb -1150 lb

F2 805 lb 0 lb

FR 805 lb -1150 lb

tanα =1150lb

805lb

α = 55.0°

θ = 360° − 55.0° = 305.0°

FR = 805lb( )2 + −115lb( )2 = 1400lb

7. x-comp y-comp

F1 0 N 1000 N

F2 1500 N 0 N

FR 1500 N 1000 N

tanα =1000N

1500N

α = 33.7°

FR = 1500N( )2

+ 1000N( )2

= 1800N



53

8. x-comp y-comp

F1 -100 N 0 N

F2 -50.0N 0 N

F3 0 N 0 N

FR -150 N 175 N

tanα =175N

−150N

α = 49.4°

FR = −150N( )2

+ 175N( )2 = 230N

9. First find the x − and y − components of F2

x-comp y-comp

F1 -1570 lb 0 lb

F2 -523 lb 1740 lb

FR −210 0 lb 1740 lb

tanα =1740lb

2100lb

α = 39.6°

θ = 180° − 39.6° = 140.4°

FR = −2100lb( )2 + 1740lb( )2 = 2730lb

10. x-comp y-comp

F1 -1950N 0 N

F2 -1920 N -3330 N

FR -3870 N -3330 N

tanα =3330N

3870N

α = 40.7°

θ = 180° + 40.7° = 220.7°

FR = −3870N( )2 + −3330N( )2 = 5110N



54

11. x-comp y-comp

F1 2550 N 0 N

F2 -1580 N 2730 N

F3 −350 0N 1140 N

FR -2530 N 3870 N

tanα =3870N

2530N

α = 56.8°

θ = 180° − 56.8° = 123.2°

FR = −2530N( )2 + −3870N( )2 = 4620N

12. x-comp y-comp

F1 2660 lb 0 lb

F2 -1920 lb 1450 lb

F3 -831 lb -2280 lb

FR -91 lb -830 lb

tanα =830lb

91lb

α = 83.7°

θ = 180° + 83.7° = 263.7°

FR = −91lb( )2 + −830lb( )2 = 835lb

13. x-comp y-comp

F1 1150 N 0 N

F2 0 N 875 N

F3 -1260 N -725 N

FR -110 N 150 N

tanα =150N

110N

α = 53.7°

θ = 180° − 53.7° = 126.3° fromF1

FR = −110N( )2 + 150N( )2 = 190N

14. x-comp y-comp

F1 2750 lb 0 lb



55

F2 2380 lb 1375 lb

F3 1375 lb 2380 lb

F4 0 lb 2750 lb

FR 6505 lb 6505 lb

tanα =6505lb

6505lb

α = 45.0°

FR = 6505lb( )2 + 6505lb( )2 = 9200lb

7.2

1. 100 N 2. 100 lb 3. 260 N 4. 500 N 5. 690 0N 6. 720 lb

7. 570 N 8. No 9. Yes 10. 4.5 tons 11. 322N 12. 698N

Note: the sum of the x-components equation is written first; the sum of the y-components is written second.

13. −F1 + 100 N( ) cos 45.0°( ) = 0

70.7N = F1

F2 + − 100 N( ) sin 45.0°( )[ ]= 0

F2 = 70.7N

14. −F1 + 950 N( ) cos 30.0°( ) = 0

823N = F1

−F2 + − 950 N( ) sin 30.0°( )[ ]= 0

475N = F2

15. −F1 cos 30.0°( )+ 500 lb( )= 0

577lb = F1

F1 sin 30.0°( ) + −F2( )= 0

289lb = F2

16. F1 + − 100 0lb( ) cos10.0°( )[ ]= 0

F1 = 985lb

F2 + − 100 0 lb( ) sin10.0°( )[ ]= 0

F2 = 174 lb

17. F2 cos 60.0°( ) + −250 lb( )= 0

F2 = 500 lb

F1 + −F2( ) sin 60.0°( ) = 0

F1 = 433lb



56

18. F1 cos 20.0°( )+ −F2( )= 0

F1 sin 20.0°( ) + −400 N( )= 0

F1 = 1170N

Then F2 = 110 0N

19. −T1 cos 20.0°( ) + T2 cos 20.0°( ) = 0

T2 cos 20.0°( ) = T1 cos 20.0°( )

T2 = T1

T1 sin 20.0°( )+ T2 sin 20.0°( ) + −500 lb( )= 0

2T1 sin 20.0°( ) = 500 lb

T1 = 731lb = T2

20. −T1 cos10.0°( ) + T2 cos10.0°( ) = 0

T2 cos10.0°( ) = T1 cos10.0°( )

T2 = T1

T1 sin10.0°( )+ T2 sin10.0°( ) + −500 lb( )= 0

2T1 sin10.0°( ) = 500 lb

T1 = 1440lb = T2

21. −T1 cos 20.0°( ) + T2 cos 30.0°( ) = 0

T2 =T1 cos 20.0°( )

cos 30.0°

T1 sin 20.0°( )+ T2 sin 30.0°( ) + −500 lb( )= 0

T1 sin 20.0°( )+T1 cos 20.0°( )

cos 30.0°sin 30.0°( ) = 500 lb

T1 = 565lb

T2 = 613lb

22. E cos 25.0°( )+ −T( ) = 0

E cos 25.0°( ) = T

E sin 25.0°( ) + −890 0N( )= 0

E = 21,100N

T = 21,100N( ) cos 25.0°( ) = 19,100N



57

23. E cos 30.0°( )+ −T( ) = 0

E cos 30.0°( ) = T

E sin 30.0°( ) + −150 0N( )= 0

E = C = 300 0lb

T = 300 0 lb( ) cos 30.0°( ) = 260 0lb

24. 20lb + −M cos 45.0°( )[ ]= 0

28lb = M

25. B + −16,200N( ) cos 70.0°( ) = 0

B = 5540N

26. E + −T( ) cos 40.0°( ) = 0

T sin 40.0°( ) + −750 N( )= 0

T = 1170N

E = 1170N( ) cos 40.0°( ) = 896N

27. sum of x-components= 0

E cos 58.0°( )+ −T( ) cos 33.0°( ) = 0

sum of y-components= 0

E sin 58.0°( ) + −T( ) sin 33.0°( ) + −1850lb( ) = 0

0.530E − 0.839T = 0

0.848E − 0.545T = 1850 Note: Solve the above equations simultaneously.

E = 3690lb = C

T = 2330lb



58

28. sum of x-components= 0

E cos 34.7°( )+ −T( ) cos19.6°( ) = 0

sum of y-components= 0

E sin 34.7°( ) + −T( ) sin19.6°( ) + −11,500N( ) = 0

0.822E − 0.942T = 0 0.569E − 0.335T = 11,500

Note: Solve the above questions simultaneously.

E= 41,600 N = C

T= 36,300 N

7.3

1. T = Fst = 16.0 lb( ) 6.00 ft( ) = 96.0lb ft 2. T = Fst= 100 N( ) 0.420 ft( ) = 42.0N m

3. st =TF

=60.0N m30.0N

= 2.00m 4. F =Tst

=35.7lb ft0.0240 ft

= 1490lb

5. F =Tst

=65.4N m35.0cm

= 187N 6. T = Fst = 630 N( ) 0.740m( ) = 466N m

7. F =Tst

=38.0N m0.0237m

= 1.60 × 103 N 8. T = Fst = 56.2 lb( ) 1.50 ft( ) = 84.3lb ft

9. st =TF

=25.0N m

70.0N= 0.357m 10. T = Fst = 112N( ) 0.350m( ) = 3.92N m

11. F =Tst

=175lb ft110 ft

= 159 lb 12. F =Tst

=14.5N m25.0cm

100cm1m

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 58.0N

13. F =Tst

=12.0N m0.0300m

= 400 N 14. F =T

st

=30.0N m

0.29m= 103N

15. F =T

st

=65.0N m

0.30m= 217N 16. F =

Tst

=27.0N m0.300m

= 90.0N

17. (a) F =Tst

=25lb ft1.0 ft

= 25lb (b) It is halved. 18. st =TF

=13N m28N

= 0.46m

19. F =Tst

=40.0N m0.300m

= 133N 20. It is halved. The applied force and the length of the

torque arm is inversely proportional.

21. F =T

st

=60.0N m

0.325m= 171N 22. F =

T

st

=55.0N m

0.325m= 169N



59

7.4

1. F = 100 lb 2. F = 200 N 3. F + 200 N = 700 N;F = 500 N 4. F = 200 N + 150 N = 350 N

5. 900 N = 450 N + F;F = 450 N 6. 650 lb = 100 lb + 250 lb + F;F = 300 lb

7. F + 210 0N = 750 N + 1500 lb + 250 N = 400 N

8. 50.0N + 35.0N + 15.0N = 10.0N + F + 75.0N = 15.0N

9.

Fw( )d1( )= F2( )d2( )

90.0kg( ) 9.80m / s 2( )3.00m( ) = F2 8.00m( )

F2 = 331N

W = mg = 90.0kg( ) 9.80m / s 2( )= 882N

F1 = 882N − 331N = 551N

10.

F2 50.0 ft( )= 500 0lb( ) 20.0 ft( ) + 400 0lb( ) 40.0 ft( )

F2 = 520 0lb

520 0lb + F1 = 400 0lb + 500 0lb

F1 = 380 0lb

11. F2 27.0m( ) = 240 0kg( ) 9.80m / s2( )

6.00m( )+ 150 0kg( ) 9.80m / s2( )10.0m( )

F2 27.0m( ) = 2.88× 105kg m2 / s2

F2 = 1.07× 104 N

Fup = Fdown

Ft + Fc = F1 + F2 or F1 = Ft + Fc − F2

F1 = 2.88 × 105 N −1.07× 104 N

F1 = 2.77 × 105 N

12. F2 2.50m( ) = 165kg( ) 9.80m / s2( )1.00m( )

F2 = 647N

F1 + 647N = 1620N

F1 = 970N



60

13.

20.0kg( ) 9.80m / s 2( )+ 40.0kg( ) 9.80m / s 2( )= Fw

Fw = 588N

588N( )x = 196N( ) 8.00m( )

x = 2.67m 14.

F1 + F2 = 75.0kg( ) 9.80m / s 2( )+ 75.0kg( ) 9.80m / s 2( )+ 21.0kg( ) 9.80m / s2( )= 1680N

so, F1 = F2 = 840N

15.


Select F1 for point of rotation.

F2 12.0m( ) = 75.0kg( ) 9.80m / s2( )3.00m( )+ 21.0kg( ) 9.80m / s2( )6.00m( )+ 90.0kg( ) 9.80m / s2( )9.00m( )

F2 = 948N

F1 = 1820N − 948N = 872N

16.



F2 10.0m( ) = 65.0kg( ) 9.80m / s2( )2.00m( )+ 18.0kg( ) 9.80m / s2( )3.50m( )+ 95.0kg( ) 9.80m / s2( )6.00m( )

F2 = 748N

F1 = 1740N − 748N = 992N



61

17. (1.0m)(76.0kg)(9.80m / s2 ) = F2 (2.22m)

F2 = 335N = ma

m =335N

9.80m / s2 = 34.2kg

(76.0kg)(9.80m / s2 ) − 335N = 410N

F1 = 410N = ma

m =410N

9.80m / s2 = 41.8kg

18. (0.75m)(12.60kg)(9.80m / s2 ) = F2 (2.00m)

F2 = 46.3N = ma

m =46.3N

9.80m / s2 = 4.72kg

(12.60kg)(9.80m / s2 ) − 46.3N = 77.18N

F1 = 77.18N = ma

m =77.18N

9.80m / s2 = 7.88kg

7.5

1. F1 = 22.6 2. Fw = 42.0

3. Select F1 for point of rotation.

F2 15.0 ft( ) = 165lb( ) 7.00 ft( ) + 22.0lb( ) 7.50 ft( )

F2 = 88.0lb

F1 + F2 = Fp + FB

88.0lb + F1 = 165lb + 22.0lb

F1 = 99lb

4 Select F1 for point of rotation.

F1 + F2 = 720N

F2 2.0m( ) = 720N( ) 0.50m( )

F2 = 180N

F1 = 720N −180N = 540N



62

5. Select F1 for point of rotation.

F2 3.30m( ) = 1.50 × 103 N( ) 1.30m( )

F2 = 591N

591N + F1 = 1.50 × 103 N

F1 = 909N

6. F2 10.0 ft( ) = 650 lb( ) 4.00 ft( ) + 75.0lb( ) 5.00 ft( )

F2 = 298lb

298lb + F1 = 650lb + 75.0lb

F1 = 427lb

7. F1 + F2 = 187, 2000N

F2 9.00m( ) = 98, 0 00N( ) 4.00m( ) + 889, 200N( ) 4.50m( )

F2 = 88, 200N

Therefore F1 = 99, 0 00N

8.

200 lb( )12.0 ft( ) = 155lb( ) 9.00 ft( ) + F0( )4.00 ft( ) + 75.0lb( ) 6.00 ft( )

F0 = 139lb

9.

F2 4.40 ft( )= 14.0lb( ) 1.90 ft( ) + 29.0lb( ) 2.20 ft( )+ 125lb( ) 3.40 ft( )

F2 = 117lb

117lb + F1 = 14.0lb + 29.0lb + 125lb

F1 = 51lb



63

10.

F2 5.00m( ) = 735N( ) 2.00m( ) + 294N( ) 2.50m( )

F2 = 441N

F1 + 441N = 735N + 294N

F1 = 588N

11.

F1 21.0m( ) = 1.57 × 105( )10.5m( ) + 3.43× 104( )7.00m( )

F1 = 8.99× 104 N

8.99× 104 N + F2 = 1.57× 105 N + 3.43× 104 N

F2 = 1.01× 105 N

12.

F1 5.00m( ) = 360N( ) 2.50m( )

F1 = 180N

13. F1 4.00m( ) = 315N( ) 1.50m( )

F1 = 118N

F2 + 118N = 315N

F2 = 197N

14. Fx = 350 0kg( ) 9.80m / s2( )

FT = 2.60× 104 kg( )9.80m / s2( )

F1 32.0m( ) = 2.55× 105 N( )16.0m( )+ 3.43× 104 N( )15.0m( )

F1 = 1.44 × 105N

F2 + 1.44 × 105N = 2.55× 105 N + 3.43× 104 N

Therefore F2 = 1.45× 105 N



64

15.

Fx = 295kg( ) 9.80m / s2( )

Fx = 295kg( ) 9.80m / s2( )

F1 4.00m( ) = 441N( ) 2.00m( )+ 2890N( ) 1.00m( )

F1 = 943N

F2 + 943N = 441N + 2890N

F2 = 2390N

16.

F1 + F2 = 325kg( ) 9.80m / s 2( )+ 125kg( ) 9.80m / s 2( )= 4410N


F2 4.00m( ) = 325kg( ) 9.80m / s2( )1.00m( )+ 125kg( ) 9.80m / s2( )2.00m( )

F2 = 1410N

F1 = 4410N −1410N = 300 0N

Note: Since the length of the beam is not given, we need 1.00m, 1.00m, 2.00m as the lengths. Any length

ratios of 1:1:2 will work. Try any other set.

17.

2.00m( )F2 = 1550N( ) 1.00m( )+ 245N( ) 2.00m( )

F2 = 1020N

F1 + 1020N = 1550N + 245N

F1 = 775N

18.

(a) F4 = 255N + 975N + 375N = 1605N

(b) 975N( ) 2.50m( ) + 375N( ) 3.50m( ) = 1605N( ) x( )

x = 2.34 m



65

19.

F6 = 1525N (up)

x = distance from A of F6

375N( ) 1.00m( ) + 1175N( ) 3.00m( )+ 1850 N( ) 4.00m( ) = 625N 7.00m( )+ 1525N( ) x( )

4.54 m =x

20.

F6 = 1375N (down)

x= distance from A’ of F6

500 N( )1.00m( ) + 750 N( ) 2.20m( )+ 2375N( ) 3.95m( )+ 1375N( ) x( ) = 3750 N( ) 4.45m( )

x = 3.75 m from A’ and 4.75 m from A


1. b

2. b

3. a

4. c

5. c

6. a

7. b

8. b

9. No; e.g., bridge

10. They are equal

11. Equilibrium is the condition of a body where

the net force acting on it is zero.



66

12. Toward the center of the earth.

13. They are in equilibrium.

14. It is a diagram showing how forces act on a

body.

15. No, only when the pedals are parallel to the

ground.

16. Even if the vector sum of the opposing forces

is zero, they must also be positioned so there is no

rotation in the system.

17. Choosing a point through which a force acts to

eliminate a variable.

18. (a) stacking bridges

(b) riding a bicycle

(c) lifting any object

(d) hitting a baseball

(e) leaning into the wind

19. No; only if the object is of uniform

composition and shape.

20. The support closer to the bricks.


1. 569 N (left) 2. (a) 500 lb (b) 50 lb

3. x- comp y- comp

F1 0 N 1500 N

F2 -3400 N 0 N

FR -3400 N 1500 N

tan A =1500N3400N

A = 24°

θ = 180° − 24° = 156°

c = −3400N( )2 + 1500N( )2 = 3700N

4. x- comp y- comp

F1 5080 lb 0 lb

F2 2550 lb -3640 lb

FR 7630 lb -3640 lb

tan A =3640lb7630lb

A = 25.5°

θ = 360° − 25.5° = 334.5°

c = 7630lb( )2 + −3640lb( )2 = 8450lb



67

5. x- comp y- comp

F1 54,600 N 0 N

F2 0 N 54,600 N

F3 -38,600 N 38,600 N

FR 16, 0 00 N 93,600 N

tan A =93,200N16,000N

A = 80.3° = θ

c = 16, 0 00N( )2+ 93, 200N( )2 = 94, 600N

6. x- comp y- comp

F1 1250 N 0 N

F2 -313 N 541 N

F3 -1600 N 925 N

FR -660 N 1466 N

tan A =1466N660N

A = 65.8°

θ = 180° − 65.8° = 114.2°

c = −660N( )2 + 1466N( )2 = 1610N

7. 220 N 250 N 8. 6.0 tons

340 N 160 N

180 N 420 N

560 N 830 N

130 0N -830 N

470 N

9. Fx + 375+ 150 = 0

Fx = −525

10. tan A =600 N900 N

11. F1 + − 1650N( ) cos 67.0°( )[ ]= 0

A = 33.7° = θ F1 = 645N

c = 900 N( )2+ 600 N( )2

= 1080N F2 + − 1650N( ) sin 67.0°( )[ ]= 0

F2 = 1520N



68

12. F1 cos 65.0°( )+ −F2( )= 0

F1 sin 65.0°( ) + −4750lb( ) = 0

F1 = 5240lb

F2 = 2210lb

13.

E + −T( ) cos 40.0°( ) = 0

T sin 40.0°( ) + −1150N( ) = 0

T = 1790N

E = 1370N

14.

−T1( ) cos 30.0°( )+ T2 cos 45.0°( ) = 0

T1 sin 30.0°( )+ T2 sin 45.0°( ) + −475lb( ) = 0

−0.866T1 + 0.707T2 = 0

−0.500T1 + 0.707T2 = 475

T1 = 348lb

T2 = 426lb

T3 = 475lb

15.

−T1( ) cos 50.0°( )+ T2 = 0

T1 sin 50.0°( )+ −2200N( ) = 0

T1 = 2900N

T2 = 1900N

T3 = 2200N

16. E cos 30.0°( )+ −T1( ) cos 30.0°( ) = 0

E = T1

E sin 30.0°( ) + T1( ) sin 30.0°( ) + −6250N( ) = 0

2E sin 30.0°( ) = 6250N

E = 6250N = T1 = T2



69

17. T = Fst = 53.0N( ) 0.500m( ) = 26.5N m 18. F =Tst

=81.0lb ft3.00 ft

= 27.0 lb

19. 200 kg( ) 9.80m / s 2( )= 1080N + T 20. 450 lb – 290 lb =160 lb

880 N = T

21. 100.0kg 22. 68.0 cm x 3/4 = 51.0 cm 23. 800 0kg + 320 0kg

2= 560 0kg

24. F1 + F2 = 7.84 × 104 N + 3.14 × 104 N = 11.0 × 104 N

F1 26.0m( ) = 7.84 × 104 N( )13.0m( ) + 3.14 × 104 N( )7.00m( )

F1 = 47, 700N

F2 = 62,300N

25.

392N( ) 0.700m( )+ 21.6N( ) 1.35m( )+ 539N( ) 2.20m( ) = 127N( ) 1.45m( )+ F2 2.70m( )

F2 = 483N

F1 + F2 + 127N = 392N + 21.6N + 539N

F1 + F2 = 826N

F1 = 343N

26. 3475N( ) 2.00m( ) + 1125N( ) 3.00m( ) = F 4.00m( )

F = 2580N


1. (a) τ = Fxd = 894N × 2.57m = 230 0N m

(b) τ = Fxd = 894N × sin 30°( )2.57m = 1150N m

(c) Continually push perpendicular to the door.



70

2. (a) tan−1 5.50 ft33.5 ft

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 9.32° ;

9.32° × 2 = 18.6°

(b) F =Fg

cosθ=

57.5lbcos 9.32°

= 58.3lb

(c) tan−1 5.50 ft5.75 ft

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 43.7°

F =F2

cosθ=

57.5 ftcos 43.7°

= 79.5lb

(d) The angle between the ropes is increased so that Sean and Greg not only pull up but also pull

horizontally.

3. (a) Δτ = τ new −τ old = 25.5N × 0.127m( )− 25.5N × 0.0374m( ) = 2.28N m

(b) F =τd

=25.5N × 0.0374m( )

0.127m= 7.51N

ΔF = Foriginal − Fnew = 25.5N − 7.51N = 18.0N less force

4. (a) F = 85.8N cos 45.0° = 60.7N

τ = F × d = 60.7N × 0.750m = 45.5N m which will not support the torque.

(b) Reduce the angle between the wall and the bracket.

5. (a) F1 + F2 = 350 N + 877N = 1230N

F1 = 1230N − F2

F2 1.53m( ) = 350 N × 2.75m( )+ 877N × 5.50kg( )

F2 = 3780N

F1 = 1230N − 3780N = −2550N (negative sign means that the bracket pulls down)

(b) Between the bracket and the fulcrum.



71

Chapter 8 8.1 1. W = Fs = 10.0N( ) 3.43m( ) = 3.43N m 2. W = Fs = 125N( ) 4875m( ) = 6.09× 105N m 3. W = Fscosθ = 1850N( ) 625m( ) cos 37.5°( ) = 9.17 × 105 N m = 917kJ

4. F =Ws

=697 ft lb976 ft

= 0.714 lb 5. cosθ =WFs

=5.57 × 106 J

25,700N( ) 238m( ) = 0.9106;θ = 24.4°

6. W = Fs = ma( )s = 16.0kg( ) 9.80m / s2( )13.0m( ) = 2040 kg m / s2( ) m( ) = 2040N m = 2040J

7. W = Fs = 900 0kg( )1.80m( ) = 16,200N m = 16,200J 8. W = Fs = 350 lb( ) 2640 ft( ) = 9.2× 105 ft lb 9. W = Fs = 75 4.00kg( ) 9.80m / s2( )[ ]1.50m( ) = 4410 kg m / s2( )m( ) = 4410N m = 4410J

10. s =WF

=4340 ft lb

11( ) 94.0lb( ) = 4.20 ft 11. W = Fs = 450 lb( ) 75 ft( ) = 34,000 ft lb

12. W = Fs = 200 kg( ) 9.80m / s2( )6.50m( ) = 1.27 × 104 kg m / s 2( )m( ) = 1.27× 104 N m = 1.27 × 104 J = 12.7kJ

13. W = Fscosθ = 35.0N( ) 900 m( ) cos 40.0°( ) = 24,100N m = 24,100J = 24.1kJ 14. W = Fscosθ = 35.0lb( ) 5280 ft( ) cos 25.0°( ) = 1.67 × 107 ft lb 15. W = Fscosθ = 375lb( ) 675 ft( ) cos 50.0°( ) = 1.63× 105 ft lb 16. W = Fscosθ = 775N( ) 231m( ) cos 30.0°( ) = 1.55× 105 N m = 1.55× 105J = 155kJ 17. W = Fscosθ = 12, 000N( ) 550m( ) cos 21°( ) = 6.2× 106N m = 6.2 × 106 J = 6.2MJ 18. W = 2 6.2MJ( ) from#17 = 12.4MJ 19. W = Fs = 825N( ) 35.0m( ) = 28, 900N m = 28, 900J = 28.9kJ 20. The work doubles. The work done is directly proportional to the force. 21. (a) W = Fs = 215N( ) 4.20m( ) = 903J Only the vertical distance is considered because gravity acts vertically. (b) W = Fscosθ = 215N( ) 4.20m( ) cos180°( ) = −903J (Force is upward but displacement is downward so θ =180°) 22. W = Fscosθ = 628N( ) 15.0m( ) cos 46.0°( ) = 6540J

23. W = Fs = Fxs =

200N

tan 25o

⎛⎝⎜

⎞⎠⎟

25.0m( ) = 10700J

24. W = Fs = 1000N( ) 1.75m( ) = 1750J 8.2

1. P =Wt

=132N m

7.00s= 18.9N m / s = 18.9J / s = 18.9W 2. W = Pt = 231 ft lb / s( )14.3s( ) = 330 0 ft lb

3. t =WP

=40.0J75.0W

= 0.533N m

N m / s= 0.533s 4. P =

Wt

=55.0J11.0s

= 5.00W

5. P =Wt

=310 J25.0s

= 12.4W 6. (a) W = Fs = 175lb( ) 1320 ft( ) = 2.31× 105 ft lb

(b) P =Wt

=2.31× 105 ft lb

15.0 min×60s

1min

= 257 ft lb / s (c) 257ft lb

s×

1hp550 ft lb / s

= 0.467hp

7. (a) 1.25kW ×1.34hp1kW

= 1.68hp (b) F =Pts

=1.25× 103W( ) 35.0s( )

200 m= 219

Wsm

= 219J / m = 219N



72

8. t =FsP

=475lb( ) 38.0 ft( )

7.00hp×550 ft lb / s

1hp

= 4.69s

9. t =FsP

=360 kg( ) 9.80m / s2( )16.0m( )

950 W= 59.4

kg m / s2( )m( )W

= 59.4N mJ / s

= 59.4N m

N m / s= 59.4s

10. P =Fst

=150 0lb( ) 22.0 ft( )

2.50min= 13, 200 ft lb / min×

1hp33, 000 ft lb /min

= 0.400hp

11. 2.00hp ×0.746kW

1hp= 1.49kW 12. 2200W ×

1kW1000W

×1hp

0.746kW= 2.95hp

13. P =Wt

=Fst

=525kg( ) 9.80m / s2( )30.0m( )

25.0s= 6170W = 6.17kW

14. t =WP

=FsP

=175kg( ) 9.80m / s2( )15.0m( )

450 0W ×1N m / s

1W

= 5.72s

15. P =W

t=

mgs

t=

475kg( ) 9.80m / s2( ) 10.0m( )24.0s

= 1.94kW

16. P =W

t=

mgs

t=

50.0kg( ) 9.80m / s2( ) 15.0m( )12.0s

= 0.613kW

17. (a) P =Wt

=Fst

=75( ) 70 kg( ) 9.80m / s 2( )8.0m( )

60s= 6900W = 6.9kW

(b) 6.9kW ×1hp

0.746kW= 9.2hp

(c) 6.9kW × 135% = 9.3kW

18. P =Wt

=Fst

=750 kg( ) 9.80m / s2( )25.0m( )

60s= 3060N m / s = 3060W = 3.06kW

19. P =Wt

=2W12

t= 4W / t (4 times) 20. P =

Wt

=2.5W13

t= 7.5W / t (7.5times)

21. (a) P =Wt

=Fst

=Pts

=12 × 103W( )60s( )

9.0m= 8.0× 104Ws / m = 8.0× 104 N

m =Fg

=8.0× 104 N

75kg( ) 9.80m / s2( )= 110 passengers

(b) 12kW = 0.55P;P = 22kW

22. F =Pts

=4.00× 103W( )60.0s( )

35.0m= 6860N

m =Fg

=6860N

9.80m / s2 = 700 kg Therefore700 L / min

23. P =Fst

=575N( ) 20.0m( )

10.0s= 1150W = 1.15kW

24. P =W

t=

Fs

t=

875N( ) 21.0m( )11.0s

= 1.67kW

25. P =W

t=

Fs

t=

175N( ) 32.0m( )16.0s

= 0.350kW

26. (a) W = Fs = mgs = 7.50kg( ) 9.80m / s 2( )8.20m( ) = 603J



73

(b) W = Fs = 645N + 7.50kg( ) 9.80m / s2( )8.20m( )[ ]= 5890N m = 5890J

(c) P =Wt

=5890J30.0s

= 196W

8.3 1. PE = mgh = 11.4kg( ) 9.80m / s2( )22.0m( ) = 2460N m

2. PE = mgh = 3.50kg( ) 9.80m / s 2( )15.0m( ) = 515J 3. KE = 1/ 2mv2 = 1 /2( ) 4.70kg( ) 9.60m / s( )2 = 217J

4. h =PEmg

=93.6J

2.30kg( ) 9.80m / s 2( )= 4.15m 5. (a) 55.0mih

×1h

3600s×

5280 ft1mi

= 80.7 ft / s

(b) KE = 1/ 2mv2 = 1 /2( ) 950 slugs( ) 80.7 ft / s( )2 = 3.09 × 106 slug ft / s 2( ) ft( )= 3.09× 106 ft lb

6. KE = 1/ 2mv2 = 1 /2( ) 0.0120kg( ) 415m / s( )2 = 1030J 7. v =2KEm

2 742 ft lb( )7.40slugs

= 14.2 ft / s

8. PE = mgh = 475kg( ) 9.80m / s 2( )17.0m( ) = 79,100J = 79.1kJ

9. (a) PE = mgh = 250 0kg( ) 9.80m / s 2( )12.0m( ) = 2.94 × 105 J = 294kJ

(b) PE = mgh = 250 0kg( ) 9.80m / s 2( )16.0m( ) = 3.92× 105 J = 392kJ

10. h =PEmg

=5.17× 105 ft lb

173slugs( ) 32.2 ft / s 2( )= 92.8 ft 11. v =2KEm

2 1750J( )0.0300kg

= 342m / s

12. PE = mgh = 1.00× 106 ft 3( )62.4 lb / ft 3( )726 ft( ) = 4.53× 1010 ft lb

13. KE = 1/ 2mv2 = 1 /2( ) 250kg( ) 150kmh

×1h

3600s×

1000m1km

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 2.2× 105 J = 220kJ

14. KE = 1/ 2mv2 = 1 /2( ) 250kg( ) 25kms

×1000m

1km⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 7.8× 1010J

15. (a) P =Fst

=250 m3( ) 1000kg

1m3⎛ ⎝ ⎜

⎞ ⎠ ⎟ 9.80m / s 2( )65.0m( )

60s= 2.65× 106W = 2650kW

Note: 1m3 of matter has a mass of 1000kg.

(b) 2650kW ×1.34hp1kW

= 3550hp

(c) PE = mgh = 250 m3( ) 1000kg1m3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 9.80m / s 2( )65.0m( ) = 1.59 × 108 J or159MJ

16. (a) P =Fst

=25.0m3( ) 680kg

1m3⎛ ⎝ ⎜

⎞ ⎠ ⎟ 9.80m / s 2( )10.0m( )

60s= 2.8× 104W = 28kW

(b) PE = mgh = 25.0m3

min⎛

⎝ ⎜ ⎞

⎠ ⎟ 680kg

m3⎛ ⎝ ⎜

⎞ ⎠ ⎟ 9.80m / s 2( )10.0m( ) 10 min( ) = 1.7× 107or17MJ

(c) PE = mgh = 25.0m3

min⎛

⎝ ⎜ ⎞

⎠ ⎟ 680kg1m3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 9.80m / s 2( )5.00m( ) 10 min( ) = 8.3× 106J or8.3MJ

17. KE = 1 /2( )mv2;KE = 1/ 2( )m 2v( )2 = 1/ 2( )m 4v( )2 = 2mv 2 4 times( )

18. v =2KEm

= 2KEm

;v = 22KE( )m

= 2KEm

2times( )



74

19. (a) KE =12

mv2;12

4.20g( ) 1kg1000g

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 965m / s( )2 = 1960J

(b) Work = change in KE = 1960 J

(c) F =Ws

=1960J0.750m

= 2610N

(d) F =Ws

=1960J

0.0150m= 1.31× 105 N

20. (a) PE = mgh = 90.0kg( ) 9.80m / s 2( )45.0m( ) = 3.97× 104 J

(b) PE = mgh = 90.0kg( ) 9.80m / s 2( )40.0m − 85.0m( ) = −3.53× 104 J

21. (a) W = Fs = 630 N( ) 5.00m( ) = 3150J (b) 3150 J (c) The increase in gravitational potential energy is equal to the work done in raising the painter. The work done by the painter is the source of increase in energy.



75

8.4 1. v = 2gh = 2 9.80m / s 2( )2.50m( ) = 7.00m / s 2. v = 2gh = 2 32.2 ft / s 2( )400 0 ft( ) = 508 ft / s

3. v = 2gh = 2 9.80m / s 2( )270 m( ) = 72.7m / s 4. (a) v = 2gh = 2 9.80m / s 2( )25m( ) = 22m / s

(b) v = 2 9.80m / s 2( )125m( ) = 49.5m / s (c) v = 2 9.80m / s 2( )225m( ) = 66.4m / s

(d) v = 2 9.80m / s 2( )325m( ) = 79.8m / s 5. h =v2

2g=

75.0 ft / s( )2

2 32.2 ft / s 2( )= 87.3 ft

6. v = 2gh = 2 9.80m / s 2( )1.75m( ) = 5.86m / s 7. v = 2gh = 2 9.80m / s 2( )125m( ) = 49.5m / s

8. v = 2gh = 2 9.80m / s 2( ) 5.25m( ) = 10.1m / s

9. v = 2gh = 2 9.80m / s2( ) 3.60m( ) = 8.40m / s 10. (a) KE i ; + PE i = KEf + PEf 0 + PEi = KE f + 0 0 + mgh = KE f + 0

KEf = 15.0kg( ) 9.80m / s 2( )8.00m( ) = 1180J

(b) v =2 KE( )

m=

2 1180J( )15.0kg

= 12.5m / s

11. v f = vi2 + 2 as = 75.0 ft /s( )2 + 2 32.2 ft /s 2( )475 ft( ) = 190 ft / s

12. h =v2

2g=

95 ft / s( )2

2 32.2 ft / s 2( )= 140 ft / s

13. t s v KE h PE Total 0.000 0.00 0.00 0 300.0 11,760 11,760 1.000 4.90 9.80 190 295.1 11,570 11,760 2.000 19.6 19.60 770 280.4 10,990 11,76o 3.000 44.1 29.40 1730 255.9 10,030 11,760 4.000 78.4 39.20 3070 221.6 8690 11,760 5.000 122.5 49.00 480 0 177.5 6960 11,760 6.000 176.4 58.80 6910 123.6 4850 11,760 7.000 240.1 68.60 9410 59.9 2350 11,760 7.800 300.0 76.68 11,760 0.00 0.00 11,760

14. t v KE h PE Total 6.00 98.0 960 0 0 0 960 0 1.00 88.2 7780 93 1820 960 0 2.00 78.4 6150 176 3450 960 0 5.00 49.0 240 0 367 720 0 960 0 10.00 00.0 0 490 960 0 960 0 12.00 19.6 380 470 9220 960 0 15.00 49.0 240 0 368 720 0 960 0 20.00 98.0 960 0 0 0 960 0



76

Chapter 8 Review Questions 1. c 2. a 3. a 4. d 5. c 6. No 7. No

8. J=kg ms 2 × m

9. No 10. No 11. Yes

12. By measuring the force applied, distance traveled and time taken. 13. It possesses the ability to do work (e.g., turn a wheel) in falling to the lower level. 14. (a) elevator counterweights (b) roller coaster

15. Yes, KE=12

mv 2

16. At its lowest point. 17. At its highest point. 18. No 19. Yes 20. The bolt has accelerated to a higher velocity.

Chapter 8 Review Problems 1. 3.6 × 106 2. 0 (s = 0)

3. W = Fs = mgs; s = Wmg

=1000 J

10.0kg( ) 9.80m / s 2( )= 10.2m

4. W = Fs = 40.0kg( ) 9.80m / s2( )250 0m( )= 9.80 × 105 J or 980 kJ

5. (a) P =Wt

=9.80× 105J

10.0h ×3600s

h

= 27.2W (b) 27.2W ×1hp

746W= 0.0365hp

6. h =PEmg

=10.0J

10.0kg( ) 9.80m / s 2( )= 0.102mor10.2cm 7. h =PEmg

=20.0 ft lb10.0 lb

= 2.00 ft

8. v =2 KE( )

m=

2 1.00J( )1.00kg

= 1.41m / s 9. v =2 KE( )

m=

2 1.00J( )10.0N( ) 9.80m / s2( ) = 1.40m / s

10. KE = 1 /2mv 2 =12

300 0lb32.2 ft / s 2

⎛

⎝ ⎜ ⎞

⎠ ⎟ 55mih

×1h

3600s×

5280 ft1mi

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 303,000 ft lb

11. PE = mgh = 80.0kg( ) 9.80m / s2( )3.00m( ) = 2350J

12. KE = 1 /2( )mv 2 =12

0.020kg( ) 550m / s 2( )= 30 00J

13. PE = mgh = 85.0kg( ) 9.80m / s2( )2.00m( ) = 1670J

14. (a) W = FS = 300 N( ) 10.0m( ) = 300 0J (b) W = Fscosθ = 200 N( )10.0m( ) cos 20.0°( ) = 1880J

15. v = 2gh = 2 9.80m / s 2( )50.0m( ) = 31.3m / s Chapter 8 Applied Concepts

1. (a) P =Wt

=10.0kg( )9.80m / s 2( )2.75m( )

60.0s= 4.49W

(b) Mass of the water, distance the substance is carried, and the time to move the water.



77

2. (a) E = mgh = 355kg( ) 9.80m / s 2( )60.0m( ) = 2.09× 105 J

(b) " A" v =2Em

=2 2.09 × 105 J( )

355kg= 34.3m / s

“ B” E= U + K

2.09 × 105 J = 355kg( ) 9.80m / s 2( )20.0m( ) +12

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 355kg( )v 2

v = 28.0 m/s “C” E = U + K

2.09 × 105 J = 355kg( ) 9.80m / s 2( )30.0m( ) +12

355kg( )v2

v = 24.3 m/s “D” E = U + K

2.09 × 105 J = 355kg( ) 9.80m / s 2( )30.0m( ) +12

355kg( )v2

v = 19.8 m/s (c) The higher the elevation, the less the velocity.

3. (a) K =12

mv 2 =12

22, 500kg( ) 65.3m / s( )2 = 4.80 × 107 (The same as the jet’s original kinetic energy.)

(b) F =Wd

=4.80× 107J

115m= 4.17× 105 N

(c) The force needed to stop the jet is less because the time is more.

4. (a) P =Wt

=mgh

t=

3.79× 106kg( )9.80m / s2( )142m( )1.00s

= 5.27× 109W (This is enough energy to power all

of Paraguay.) (b) P × 2 = 1.050× 1010W (c) The higher the dam, the more potential energy changes to kinetic energy. 5. (a) U = mgh = 1250kg( ) 9.80m / s 2( )12.7m( ) = 1.56× 105 J

(b) Utop = Kbottom = 1.56 × 105 J

(c) F =Wd

=1.56× 105J

0.437m= 3.57× 105 N

(d) When the wrecking ball is at its lowest point, all of its energy has been converted to kinetic energy, thus striking the wall with the greatest velocity.



78

Chapter 9

9.1

1. (a) (6.5 )(2 / ) 40.8rev rad rev radπ = (b) (6.5 )(360 / ) 2340rev rev° = °

2. (a) (2880 )(1 / 360 ) 8.00rev rev° ° = (b) (2880 )(2 / 360 ) 50.3rad radπ° ° =

3. (a) (25 )(1 / 2 ) 12.5rad rev rad revπ π = (b) (25 )(360 / 2 ) 4500rad radπ π° = °

4. (a) (12.0 )(2 / ) 75.4rev rad rev radπ = (b) (12.0 )(360 / ) 4320rev rev° = °

5. 525

1543.42 min

revrpm

t

θω = = = 6.

7360199 /

37.0

revrev s

t s

θω = = =

7. 4.00 2

8.38 /3.00 1

rev radrad s

t s rev

θ πω = = × = 8.

32565.0

5.00 min

revrpm

t

θω = = =

9. 6370

354 /18.0

revrev s

t s

θω = = = 10.

6.25 27.78 /

5.05 1

rev radrad s

t s rev

θ πω = = × =

11. 1 60

675 64502 min

rad rev srpm

s radπ× × = 12.

2 1min285 29.8 /

min 60

rev radrad s

rev s

π× × =

13. 2 1min

136 14.2 /min 60

rev radrad s

rev s

π× × = 14.

1 6088.4 844

2 min

rad rev srpm

s radπ× × =

15. 60

11.0 660min

rev srpm

s× = 16.

1min180 3.0 /

min 60

revrev s

s× =

17. (a) 1

6.67 /0.150

revrev s

t s

θω = = = (b)

606.67 400

min

rev srpm

s× =

(c) 2

6.67 41.9 /rev rad

rad ss rev

π× =

18. (a) 50.0

15.4 /3.25

revrev s

t s

θω = = = (b)

6015.4 924

min

rev srpm

s× =

(c) 2

15.4 96.8 /rev rad

rad ss rev

π× =

19. (a) 41 609.52 10 min 0.0571

1050 min

rev st s

rpm

θ

ω−= = = × × =

(b) 1min

(1050 )(5.00 ) 87.560

t rpm s revs

θ ω= = =

20. (a) 2

0.17536.0 /

radt s

rad s

θ π

ω= = = (b)

1(36.0 / )(8.00 ) 45.8

2

revt rad s s rev

radθ ω

π= = =

21. (7.00 / )(1.20 ) 8.40t rad s s radθ ω= = =

22. (4.00 / )(13.0 ) 52.0t rev s s revθ ω= = =



79

23. 2

(1.50 )(5.0 ) 0.131360

rads r m m

πθ= = ° =

°

24. (5.00 )(4 ) 62.8s r mi rad miθ π= = =

25. 47.0 2 1min

0.270 1.33 /min 60

rev radv r m m s

rev s

πω= = × × × =

26. 2 1min

275 0.150 4.32 /min 60

rev radv r m m s

rev s

πω= = × × × =

27. (a) 2 1min

655 68.6 /min 60

rev radrad s

rev s

π× × =

(b) 60

(68.6 / )(3.00 min) 12, 300min

st rad s radθ ω= = =

(c) (25.0 )(2 ) 157s r cm rad cmθ π= = =

(d) (0.250 )(12, 300 ) 3080s r m rad mθ= = =

(e) (68.6 / )(0.250 ) 17.2 /v r rad s m m sω= = =

28. (a) 2 1min

1150 120 /min 60

rev radrad s

rev s

π× × =

(b) (120 / )(4.00 ) 480t rad s s radθ ω= = =

(c) (120 / )(2.00 ) 240 /v r rad s m m sω= = =

(d) (120 / )(1.00 ) 120 /v r rad s m m sω= = =

29 (a)

1 100060.0

3600 150.5 /

0.330

km h mv h s km

rad sr m

ω

× ×

= = =

(b) (50.5 / )(30.0 ) 1520t rad s s radθ ω= = =

(c) (0.330 )(1520 ) 502s r m rad mθ= = =

d) 502m

30. (a) 2 1min

0.105 /1min 60

radrad s

t s

θ πω = = × =

(b) 32 11.75 10 /

1 3600

rad hrad s

t h s

θ πω −= = × = ×

(c) 42 11.45 10 /

12 3600

rad hrad s

t h s

θ πω −= = × = ×

31. 2 2

15.0 . 188 . /rev rad

v r in in ss rev

πω= = × × = or 15.7 ft/s



80

32. 30.0 /

20.0 / ;1.50

v ft srad s

r ftω = = =

40.628

20.0 /

radt s

rad s

θ π

ω= = =

33. (a) 1 2

6400 1680 /24

rev radv r km km h

h rev

πω= = × × =

(b) cos (6400 )(cos 30.0 ) 5540Hr r km kmθ= = ° =

25540 1450 /

24

radv r km km h

h

πω= = × =

(c) cos (6400 )(cos 45.0 ) 4530Mr r km kmθ= = ° =

24530 1190 /

24

radv r km km h

h

πω= = × =

(d) cos (6400 )(cos 60.0 ) 3200Ar r km kmθ= = ° =

23200 838 /

24

radv r km km h

h

πω= = × =

34. 2

28.0 21.51.86 /

3.50f

avg

i

rad rad

s s rad st s

ω ωα

−−= = =

Δ

35. 2

15.4 8.501.33 /

5.20f

avg

i

rad rad

s s rad st s

ω ωα

−−= = =

Δ

36. (a) 37.7

15.1 /2.50avg

radrad s

t s

θω = = =

(b) 2

2 2

2 2 37.712.1 /

(2.50 )avg

radrad s

t s

θα

×= = =

(c) 2 2 15.1 / 30.2 /f avg rad s rad sω ω= × = × =

(30.2 / )(20.0 ) 604 /v r rad s cm cm sω= = =

(d) 2604 / 0 /242 /

2.50f iv vv cm s cm s

a cm st t s

−Δ −= = = =

Δ Δ

37. (a) 20 / 28.8 /3.60 /

8.00f i

avg

w w rad s rad srad s

t sα

− −= = = −

Δ

(b) (28.8 / )(15.0 ) 432 /v r rad s cm cm sω= = =

(c) 1 1 1

( ) (28.8 / 0 / )8.00 115 ( ) 18.32 2 2f i

revw w t rad s rad s s rad rev

radθ

π= + = + = =



81

38. (a) 2( ) 250 /

16.7 /15.0

f i

avg

w w rad srad s

t sα

−= = =

Δ

(b) (250 / )(20.0 ) 5000 / 50.0 /v r rad s cm cm s m sω= = = =

(c) 1 1 1

( ) (250 / 0 / )15.0 1875 ( ) 2982 2 2f i

revw w t rad s rad s s rad rev

radθ

π= + = + = =

9.3

1. 2 2(64.0 )(34.0 / )

435017.0

mv kg m sF N

r m= = = 2.

2 2(11.3 )(3.00 / )31.4

3.24

mv slug ft sF lb

r ft= = =

3.22

(2500 )(72.0 )79.4

(47.6 / )

Fr lb ftm slugs

v ft s= = = 4.

2 2(67.0 )(0.780 / )0.0694

587

mv kg m sr m

F N= =

5.(602 )(3.20 )

5.53 /63.0

Fr N mv m s

m kg= = = 6.

2 2(37.5 )(17.0 / )2890

3.75

mv kg m sF N

r m= = =

7.( )( )22 100 1.20 /

1.9275.0

kg m smvr m

F N= = = 8.

(80.0 )(17.5 )5.71 /

43.0

Fr N mv m s

m kg= = =

9. 2 2

3(117 )(49.3 / )3.60 10

79.0

mv slugs ft sF lb

r ft= = = ×

10.2 2(7.12 )(2.98 / )

23.22.72

mv kg m sF N

r m= = = 11.

2 2(800 )(15.0 / )28.8

6250

mv kg m sr m

F N= =

12. 1

175 5.43 ;32.2

sluglb slugs

lb× =

1 1min 528014.0 20.5 /

60 min 60 1

mi h ftft s

h s mi=⎛ ⎞⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

( )( )22 5.43 20.5 /

13417.0

slugs ft smvr ft

F lb= = =

13. 1 1min 1000

80.0 22.2 / ;60 min 60 1

km h mm s

h s km× × × =

( )( )22 1650 22.2 /

5420150

kg m smvF N

r m= = =

14. 1 1min 1000

95 / 26.4 / ;60 min 60 1

h mkm h m s

s km× × × =

( )( )22 510 26.4 /8900

40

kg m smvF N

r m= = =

15. ( )( )22 3.2 3.5 /

192.1

kg m smvF N

r m= = =

16. F =

mv2

r=

1500kg( )16.7m / s( )2

35.0m= 11,900N



82

17. F =

mv2

r=

750kg( ) 8.33m / s( )2

40.0m= 1300N

18. (a) ( )( )22

4215 62.5 /1.58 10

53.0

slugs ft smvF lb

r ft= = = × (b) It is quadrupled. (c) 46.32 10 lb×

19.35 /

1400.25 /

v m sr m

rad sω= = =

( )( )22 225 35 /2000

140

kg m smvF N

r m= = =

20. F =

mv 2

r=

55,000 kg( )17.2 m / s( )2

38.0 m= 4.29x105 N

21. F =

mv 2

r=

10,000 kg( )9.72 m / s( )2

27.5m= 3.44x104 N

9.4

1. ( )1min 2555 125 7260 /

min 60 1rev rad

P T lb ft ft lb ss rev

πω= = × × =

⎛ ⎞⎜ ⎟⎝ ⎠

2. ( )( )39.4 6.70 / 264P T N m s Wω= = =

3. ( )2 1min 1264 / min 372 18.7

1 60 550

rad hpP T rev ft lb hp

ft lbrev ss

πω= = × × =

⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎜ ⎟⎝ ⎠

4. ( )( )45.0 / 650 29.3P T s N m kWω= = = 5. 8950

18604.80 /

P WT N m

sω= = =

6. 650

1.2 /540

P Ws

T N mω = = =

7. ( ) 1min 2 1400 4500 343

min 60 1 550

rev rad hpP T ft lb hp

ft lbs revs

πω= = × × =

⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟⎝ ⎠

8.175 /

4.241min 2

394min 60 1

P ft lb sT ft lb

rev rad

s rev

πω= = =

× × 9.

6494.95 /

131

P Ws

T N mω = = =



83

10.

550 /0.500

11.64

1min 21600

min 60 1

ft lb shp

P hpT ft lb

rev rad

s rev

πω

×

= = =× ×

11. ( )( )8.3 / 550 4600P T rad s N m Wω= = = 12. 33.0 /

5.5 /6.0

P N m ss

T N mω = = =

13. (a) ( ) 1min 2 1524 3000 299

min 60 1 550 /


s rev ft lb s

πω= = × × =

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(b) ( ) 1min 2 1524 6000 599

min 60 1 550 /


s rev ft lb s

πω= = × × =

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

14. 650

5.0 /130

P Ws

T Nmω = = =

15.

10000.500

12.65

1min 21800

min 60 1

WkW

P kWT N m

rev rad

s rev

πω

×

= = =× ×

16. ( ) 1min 2750 4500 353

min 60 1

rev radP T N m kW

s rev

πω= = × × =

⎛ ⎞⎜ ⎟⎝ ⎠

17. ( ) ( )2 0.45175 150 37 / min

min 1 2trev rad m

P T Fs N kJrev

πω ω= = = =

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

18. P = ωT = 209.3

rad

sec

⎛⎝⎜

⎞⎠⎟

1200 Nm( )= 251kW

19. P = ωT = 157

rad

sec

⎛⎝⎜

⎞⎠⎟

1600 Nm( )= 251kW

20. ( ) 51min 21250 5000 6.54 10 654

min 60 1

rev radP T N m W or kW

s rev

πω= = × × = ×

⎛ ⎞⎜ ⎟⎝ ⎠

21. 1000

6.67 /150

P Ws

T N mω = = = 22.

10000.750

136

2000 1min 2

10 min 60 1

WkW

P kWT N m

rev rad

s rev

πω

×

= = =× ×

23. tT Fs P Tω= =

36 2

175 0.43 28006.0 1t

rev radP Fs N m W

s rev

πω= = × × × =



84

24. 25.0 2

250 0.570 1720 / 172013.0t

rev radP T Fs N m Nm s W

s rev

πω ω= = = × × × = =

25.

10001.50

17.17 7.17 7.17

10, 000 1min 2

5.00 min 60 1

WkW

P NmkWT Ws s Nm

rev rad ss rev

πω

×

= = = = =× ×

⎛ ⎞⎜ ⎟⎝ ⎠

26. P = ωT = 2.75

rad

sec

⎛⎝⎜

⎞⎠⎟

45.5 Nm( )= 125W

27. P = ωT = 2.25

rad

sec

⎛⎝⎜

⎞⎠⎟

52.5 Nm( )= 118W

9.6

1. ( )( )156 16

5248

rpmNTn rpm

t= = = 2.

( )( )225 24150

36

rpmntn rpm

T= = =

3. ( )( )72.0 18

2454.0

rpmNTt teeth

n rpm= = = 4.

( )( )55.5 6424

148

rpmntT teeth

N= = =

5. ( )( )276 36

20748

rpmntN rpm

T= = = 6.

( )( )144 16192

12

rpmNTn rpm

t= = =

7.( )( )85.0 36

42.572

rpmNTn rpm

t= = = 8.

( )( )1250 541500

45

rpmNTn rpm

t= = =

9. ( )( )

( )250 30

75100

rpmNTt teeth

n rpm= = = 10.

( )( )( )

154 4028

220

rpmNTt teeth

n rpm= = =

11. ( )( )4.2 15

631

ntT teeth

N= = = 12.

( )( )10 1550

3

ntT teeth

N= = =

13. ( )( )162 72

14481

rpmNTn rpm

t= = = 14.

( )( )1600 60200

480

rpmNTt teeth

n rpm= = =

15.( )( )

( )3 20

601

ntT teeth

N= = = 16.

( )( )1500 602250

40

rpmNTn rpm

t= = =

17. 1 60

2.030 1min

rev srpm

s=

⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

; ( )( )0.50 36

92.0

rpmNTt teeth

n rpm= = =

18.( )( )44 22

11 ;88

rpmntN rpm

T= = = ( )( )11 10.0 min 110rpm rev=



21. Clockwise




85

23. Clockwise

24. Clockwise

25. Clockwise

26. Clockwise



29.

( )( )( )( )( )

1 2

1 2

1680 60 151050

30 48

rpmNTTn rpm

t t= = =

30.( )( )( )

( )( )1 2

1 2

738 30 20205

45 48

rpmnt tN rpm

TT= = =

31. ( )( )( )

( )( )160 45 48

57630 20

rpmn rpm= =

32.( )( )( )( )

( )( )( )240 20 30 45

180030 10 12

rpmn rpm= =

33. ( )( )( )( )

( )( )( )370 60 48 45

148048 45 15

rpmn rpm= =

34.

( )( )( )( )( )

1 21

2

1850 60 1537

1500 30

rpmNTTt teeth

nt rpm= = =

35. ( )( )( )

( )( )1

780 30 2040

45 260

rpmt teeth

rpm= =

36.( )( )( )

( )( )1

160 45 4820

30 576

rpmt teeth

rpm= =

37. ( )( )( )( )

( )( )( )1

250 20 30 4520

30 10 1125

rpmt teeth

rpm= =

38. ( )( )( )( )

( )( )( )1

370 60 48 4540

48 45 555

rpmt teeth

rpm= =

39. B is reversed in all problems.

40. The number of teeth on gear D is doubled.

9.7

1. 2250

2. 243

3. 3600

4. 4.50

5. 147

6.

( )( )6.50 . 1650413

26.0 .

in rpmD Nn rpm

d in

⋅= = =

7.

( )( )25.0 12062.5

48.0

cm rpmD Nn rpm

d cm

⋅= = =

8.

( )( )36.0 60060.0

360

cm rpmD Nd cm

d rpm

⋅= = =

9. ( )( )15.0 . 675

22.5 .450

in rpmD Nd in

n rpm

⋅= = =

10.

( )( )10.0 12080.0

15.0

cm rpmD Nn rpm

d cm

⋅= = =

11. Clockwise

12. Clockwise


14. Clockwise


16. ( )( )1200 12.0

96.0150

rpm cmd cm

rpm= =



86


1. d 2. a 3. b 4. Curvilinear motion is motion along a curved path; rotational motion occurs when the body itself is spinning. 5. Radians and revolutions 6. A radian is an angle with its vertex at the center of a circle whose sides cut off an arc on the circle equal to its radius. 7. Angular displacement is the angle through which any point on a rotating body moves. It can be measured in radians and revolutions. 8. linear velocity = angular velocity × radius 9. They are alike except for the substitution of θ for s, ω for v, and α for a.

10. Law of Conservation of Angular Momentum. 11. Yes 12. No; it tends to cause a body to continue in a straight line (tangent to the curve). 13. The number of teeth on driver × number of revolutions of driver = number of teeth on driven gear × number of revolutions of driven gear. 14. An idler changes the directions of rotations of the driver gear. 15. Opposite 16. Since the gears are connected, they both rotate together. 17. Diameters are used in similar equations rather than teeth. 18. The small pulley. 19. The belt is one continuous piece of material.


1. (a) ( )( )13 2 / 26 81.7rev rad rev rad or radπ π= (b) 360

13 4680revrev

°× = °

2. 25

1.7 /45

radrad s

t s

θ πω = = =

3. 65 2 1min

0.130 0.885 /min 60

rev radv r m m s

rev s

πω= = × × × =

4. ( )2

21000 1

0.300 90.03600

7.5025.0

km m hkg

mv h km sF N

r m

× ×

= = =

⎛ ⎞⎜ ⎟⎝ ⎠

5. 2

;mv

Fr

= ( )( )12.0 0.600

2.75 /0.950

N mFrv m s

m kg= = =

6. ( )( ) min1.20 45.0 0.105 5.67 /P T lb ft rpm ft lb s

rev sω= = =

⎛ ⎞⎜ ⎟⎝ ⎠

7. 300

6.00 /50.0

P Wrad s

T N mω = = =

8. ( )( )15.0 2.00

6.005.00

rpm cmndN rpm

D cm= = =

9. ( )( )30 60.0

2090.0

rpmT Nt teeth

n rpm

⋅= = =



87

10. ( )( )13 115

57.526

rpmT Nn rpm

t

⋅= = =

11. Yes

12. ( )( )14.0 75.0

10510.0

cm rpmD Nn rpm

d cm

⋅= = =

13.( )( )5.00 100

2.00250

cm rpmd nD cm

N rpm

⋅= = =


15.( )( )( )

( )( )200 45 48

72020 30

rpmn rpm= =

16. ( )( )( )

( )( )800 30 20

32300 50

rpmt teeth

rpm= =


1. (a) ( )( )22 9.80 / 6.25 11.1 /Fr mg r

v m s m m sm m

= = = =

11.1 /

1.78 17.06.25

v m s radrpm

r m sω = = = =

(b) ( )( )24 9.80 / 6.25 15.7 /Fr mg r

v m s m m sm m

= = = =

15.7 /2.51 24.0

6.25

v m s radrpm

r m sω = = = =

(c) No, the mass of the person is unrelated to the rotational speed.

2. (a) ( )( )( )210.0 9.80 / 2.45 240m g d kg m s m N mτ = = =

(b) 2

2

240405

0.593

T NmI kgm

rad

sα

= = =

(c) Increase the radius of the waterwheel or make the waterwheel a ring and not a solid disk. More mass farther away from the axis increases the rotational inertia.



88

3. (a) " " 28.0 /At B v m s=

2v

F mr

=

( )2

228.0 /78.4 /

10.0

m svma m m s

r m= = =

2

2

78.4 /8.00 '

9.80 /

a m sg s

g m s= =

" " 19.8 /At D v m s= 2v

F mr

=

( )22219.8 /

19.6 /20.0

m svma m m s

r m= = = ;

2

2

19.6 /2.00 '

9.80 /

a m sg s

g m s= =

(b) “B” is not safe, the “B” loop should have a larger radius to reduce the centripetal force. “D” is safe.

4. (a) 2

fv

F mr

=

2v

mg mr

μ =

( )( )( )22.00 32.2 / 117 86.8 / 59.2 /v gr ft s ft ft s mi hμ= = = =

(b) 2

f

vF m

r=

2v

mg mr

μ =

( )( )( )22.00 32.2 / 234 123 / 83.9 /v gr ft s ft ft s mi hμ= = = =

(c) As the radius increases, the less the centripetal force and the greater the maximum velocity.

5. (a) ( )( )( ) 472 3000 2

2.26 10 / 41.160.0 1

ft lb rpm radTP lb ft s hp

t s rev

πω= = = × =

(b) ( )

( ) ( )

42.26 10 / 60.043.2

5000 2

lb ft s sPT lb ft

rpm radω π

×= = =

(c) The lower the gear, the lower the speed and the higher the torque. More torque is required to move up steep hills.



89

Chapter 10 10.2 1. 14.8 2. 273 3. 36.3 4. 22.6 5. 52.4 6. 4.00 7. 2.39 8. 17.0 9. 48.8 10. 3.08 11. 1.55 12. 208 13. 2.27 14.16.9

15. (a) ( )( )275 8.00

11002.00

E ER

R

lb ftF dF lb

d ft= = = (b)

11004.00

275R

E

F lbMA

F lb= = =

16. (a) ( )( )2100 9.80 / (1.00 )

3922.50

R RE

E

kg m s mF dF N

d m= = = (b)

9802.50

392R

E

F NMA

F N= = =

17. (a) ( )( )45.0 6.00

1.50180

E ER

R

lb ftF dft

lbd

F= = = (b)

6.004.00

1.50E

R

ftMA

ft

dd

= = =

18. (a) ( )325 (3.00 )

6501.50

R RE

E

N mF dF N

d m= = = (b)

3250.500

650R

E

F NMA

F N= = =

10.3 1. 14.0 2. 0.363 3. 271 4. 22.6 5. 524 6. 17.0 7. 48.8 8. 1.55 9. 20.4 10. 2.27

11. FE =FR RR

rE

=1000N( ) 0.136m( )

0.750m=181N

12. FE

FR RR

rE

=(975N )(0.120m)

0.620m= 189N

13. (a) ( )( )73 1.50

438.250

.R RE

E

lb ftFF lb

R

ftrΟ

= = =Ο

(b) MA =rE

rR=

1.50 ft0.250 ft

= 6.00

14. (a) FE =FR r R

rE=

20.Οcm( )150 0N( )70.0cm

= 429 N (b) MA =rE

rR=

70.0cm20.0cm

= 3.50

(c) FR =FE dE

dR=

575N( )70.0cm( )20.0cm

= 2010 N

15. (a) dE =FR dR

FE=

142N( )10.0cm( )475N

= 2.99cm (b) MA =rR

rE=

10.0m2.99 m

= 3.34

16. (a) FE =FR rR

rE=

4610 N( )0.040m( )0.48m( ) = 380 N (b) MA =

rE

rR=

0.48 m( )0.040m( )=

121

(c) ( )( )

( )( )4610 0.040 640

640 ; 3200.60 0.48 2

output output

input

input

F S N m NF N N

efficiency S m= = = =

10.4 1. 1 2. 3 3. 3

4. 4 5. 6 6. 5

7. 2 8. 7



90

#’s 9 – 14 are represented by the following diagram. (9. MA = 4 & 12. MA = 7)

15. 2

16. (a) FE =FR

MA=

250 lb2

=125 lb (b) dE = MA dR = 2( ) 15.0 ft( )= 30.0 ft

17. (a) FR = MA FE = 4( ) 97.0 N( )= 388 N (b) dE = MA dR = 4( ) 20.5m( )=82.0m 18. (a) Effort Force Effort Distance

FE =FR

MA=

400 lb4.00

= 10 0 lb dE = MA dR = 4.00( ) 30.0 ft( )=120 ft

(b) FR = MA FE = 4.00( ) 65.0 N( )= 260 N dR =dE

MA=

13.0m4.00

= 3.25m

19. MA = FR

FE

3 < 3.67 No

20. (a) FE =FR

MA=

1950N6.00

= 325N

(b) dE = Ma dR = 6.00( ) 3.00m( )= 18.0m 21. Mechanical Advantage (MA)

# Pulleys 1 2 3 4 5 6 7 8 Fixed Movable

1 0

1 1

2 1

2 2

3 2

3 3

4 3

4 4

Fixed Movable

0 1

1 1

1 2

2 2

2 3

3 3

3 4

22. No; for 10 pulleys, the maximum number of strands holding the resistance is 10, which is also the

maximum mechanical advantage.

10.5

1. 12.8 2. 0.905 3. 21.2 4. 8120

5. 36.3 6. 0.378 7. 4.62 8. 12.6

9. 5.61 10. 233

11. (a) MA =lengthheight

=2.75 m0.75 m

= 3.67 (b) FE =FR

MA=

727N4.00

=182N

(c) MA =FR

FE=

815N200 N

= 4.08; 4 < 4.08 therfore, no



91

12. (a) length = FRheight

FE=

538 lb( ) 5.50 ft( )140 lb

= 21.1 ft (b) MA =FR

FE=

538 lb140 lb

= 3.84

(c) FE =FR height

length=

257 lb( )5.50 ft( )21.1 ft

= 67.0 lb

13. MA =length

height=

3.00 m

1.00 m= 3.00

14. MA =length

height=

2.75 m

0.75 m= 3.67

15. (a) FE =FRheight

length=

255N( ) 1.20 m( )3.80m

= 80.5N (b) MA =FR

FE=

325N75.0 N

= 4.33

16. (a) FE =FR

MAincline=

480 lb4.0

= 120 lb (b) MAincline =Lincline

hincline=

12 ft3.0 ft

= 4.0

17. FE =FRheight

length=

5500N( ) 0.75 m( )2.30 N

= 1790N

18. FE =FRheight

length=

255N( ) 1.20 m( )3.80m

= 80.5N

10.6 1. 2.30 2. 20,800 3. 14.2 4. 8.35

5. 2.28 6. 30.5 7. 29.3 8. 1.39

9. 35.2 10. 4.34

11. (a) FE =FR pitch

2πr=

3650 lb( )1/8 in.( )2π 15.0 in.( ) = 4.84 lb (b) MA =

FR

FE=

3650 lb4.84 lb

= 754

12. (a) pitch =2πrMA

=2π 34.5cm( )

97.0= 2.23cm (b) FR = MA FE = 97.0( ) 405N( )= 39, 300N

13. (a) MA =2πrpitch

=2π 0.750 in.( )

0.125 in.= 37.7 (b) FR = MA FE = 37.7( ) 15.0 lb( )= 566 lb

(c) FR =FE 2πrpitch

=15.0 lb( )2π 0.250 in.( )

0.125 in.= 188 lb

14. (a) pitch =2πrMA

=2π 60.0cm( )

78.0= 4.83cm (b) FR = MA FE = 78.0( ) 430 N( )= 33,500 N

10.8

1. MAtotal = MA plane( )MApulley( )=21.0m7.00m

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5( ) = 15.0

2. FE =FR

MA=

9790N15.0

= 653N



92

3. 50.0

5.0010.0plane

ftMA

ft= =

MAwh =rE

rR=

1.00 ft0.500 ft

= 2.00

MAtotal = 5.00( ) 2.00( ) 4.00( ) = 40.0 4. FR = MA FE = 40.0( ) 300 lb( )= 12, 0 00 lb

5. FE =FR

MA=

300 0 lb40.0

= 75.0 lb

6. MA =lengthheight

=8.00 m2.00m

= 4.00; MAtotal = 4.00( ) 5.0( ) = 20

7. FE =FR

MA=

250 0 N20.0

=125N

8. MA =lengthheight

=12.0m0.500m

= 24.0; MA =rE

rR=

40.0cm12.0cm

= 3.33

MAtotal = 24.0( ) 3.33( ) 4( ) = 320 9. FR = MA FE = 320 ( ) 450 N( )= 1.44× 105 N

10. FE =FR

MA=

250 0kg( ) 9.80 m / s2( )320

= 76.6 N

Chapter 10 Review Questions 1. d 2. a 3. b 4. d 5. b 6. b 7. (a) bicycle (b) auto transmission (c) high-speed drill 8. resistance force 9. effort force x effort distance = resistance force x resistance distance 10. Mechanical advantage 11. efficiency 12. No 13. the fulcrum

14. lengthof effort arm

lengthof resis tance arm= MA of lever

15. First class

16. FR × dR = FE × dE 17. The opposite end of the resistance force with

the effort force between. 18. Resistance force x resistance radius =

effort force x effort radius 19. No; it depends on the radii 20. A fixed pulley does not move. It is

suspended by its center axle. A movable pulley is free to move and is suspended by the strand around the groove.

21. No

22. MA =lengthof planeheight of plane

23. The distance a screw advances into the wood in one revolution. 24. It is greater because the handle of the jackscrew can be longer than the radius of the screwdriver.


1. MA =FR

FE=

250 N125N

= 2.00

2. 1150N1575N

= 73.0%



93

3. MA =dE

dR=

2.75m0.720m

= 3.82

4. (a) F =540N( )2.00m( )

2.00 m= 540N (b) F =

540N( )2.00m( )3.00 m

= 360N

5. (a) rR =FE ⋅ rE

FR=

125N( )17.0cm( )325N

= 6.54 cm (b) MA =rE

rR=

17.0cm6.54 cm

= 2.60

6. 12 7. FR = MA( ) FE( )= 5( ) 135N( )= 675N

8. (a) FE =FR heightlength

=875N( )1.50 m( )

4.50 m= 292N

(b) MA =rE

rR=

4.50 m1.50m

= 3.00

9. height =FE length

FR=

230 lb( )10.0 ft( )100 0 lb

= 2.30 ft

10. FR =FE ⋅ 2πrpitch

=29.0 N( )2π 1.80cm( )

0.0200cm= 16,400 N

11. pitchFE ⋅ 2πr

FR=

13.5N( ) 2π 120mm( )945N

=1.08mm

12. MA =2πrpitch

=2π 36.0cm( )

1.50cm=151

13. (a) IMA =gear radius

wheel radius=

4.00cm35.6cm

= 0.112

(b) Efficiency = 95% ; Therefore 95% (0.112) = 0.106 (c) Fr = MA Fe = 0.106( )155N( )=16.4 N 14. (a) The mechanical advantage is doubled. (b) rE = IMA ⋅ rR = 0.112( ) 14.0cm( )=1.57cm

15. MA =FR

FE=

225N129 N

=1.74

16. Efficiency =workin

workoutand FRdR = FE dE( )efficiency( )

Therefore, dE =FR dR

FE eff .=

1250 N( )13.0cm( )225N( )88.7%( ) = 81.4 cm

17. 5 18. 4

19. (a) MA =32.0cm8.00cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟

16.0m4.00 m

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 4( ) = 64.0 (b) FR = MA( )FE = 64.0( ) 75N( )= 4800N

20. MA =FR

FE=

270N45N

= 6.0


1. (a) dE =FR dR

FE=

5.85× 1025N( )3.84 × 108 m( )858 N

= 2.62× 1031 m

(b) MA =dE

dR=

2.62× 10 31 m3.84 × 108 m

= 6.82 × 1022 This was clearly a theoretical statement by Archimedes.



94

2. (a) MA =dE

dR=

0.255 m2.13 m

= 0.120 (b) dR =dE

MA=

0.305m0.120

= 2.54 m

(c) v =dt

=2.45m0.554 s

= 4.58 m / s (d) Increases the speed of the resistance end. The tip of the

dE = MA dR = (0.188)(87.5 in.) = 10.4 in. fishing pole needs a high speed to cast the lure.

3. (a) MA =2πrpitch

=2π 7.75 in.

6.80 in.= 7.16 (b) FE =

FR

MA=

23.4 lb7.16

= 3.27 lb (c) The MA would increase.

4. (a) MA =dE

dR=

1.65in.14.0 in.

= 0.118 (b) FR = MA FE = 0.118( ) 54.5 lb( )= 6.42 lb

(c) C = 2πr = 2π 14.0 in.( )= 87.9 in. (d) An MA less than one is advantageous for speed. The bike must be in a high gear and moving quickly.

5. (a) MA =dE

dR=

1.87m0.568m

= 3.29 (b) MA =rE

rR=

0.265m0.075m

= 3.53

(c) Increase length of the lever arm and/or the radius of the wheel.



95

Chapter 11 11.1 1. (a) Fg = mg = 84.3kg( ) 9.80 m / s2( )=826 N

(b) Fg = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg284.3kg( )1.99× 10 30 kg( )

1.50× 1011 m( )2 = 0.497N

2. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg25.97× 1024 kg( )1.99× 1030 kg( )

1.50 × 1011 m( )2 = 3.52× 1022 N

3. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg23.30× 10 23 kg( )1.99× 1030 kg( )

5.97× 1010 m( )2 =1.23× 1022 N

4. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg21.90× 10 27 kg( )1.99× 1030 kg( )

7.78 × 1011( )2 = 4.17 × 1023 N

5. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg21.5× 1022 kg( )1.99 × 1030 kg( )

5.91× 1012 m( )2 = 5.70× 1016 N

6. Jupiter is much more massive than the earth despite its distance.

7. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg23.80× 104 kg( )5.97× 10 24 kg( )6.38 × 106 m + 3.22 × 105 m( )2 = 3.37× 10 5 N

8. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg23.80× 104 kg( )5.97 × 1024 kg( )

6.38× 106 m + 2( ) 3.22 × 105 m( )[ ]2 = 3.07 × 105 N

9. FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg29.11× 10−31 kg( )1.67 × 10−27 kg( )

5.30× 10 −9 m( )2 = 3.61× 10−51N

10. (a) FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg24240kg( ) 5.97 × 1024 kg( )

6.38 × 106 m( )2 = 4.16 × 104 N

(b) FG = Gm1 m2

r 2 = 6.67 × 10−11 Nm 2

kg24240kg( ) 7.35× 1022 kg( )

1.74 × 106 m( )2 = 6.87 × 103 N

11.3

1. v =G m

r=

6.67× 10−11 Nm2

kg2

⎛

⎝ ⎜ ⎞

⎠ ⎟ 5.97× 1024 kg( )3.84 × 108 m

= v =1020ms

2. Τ = 2πr 3

G m= 2π

3.84 × 108 m( )3

6.67× 10−11 Nm2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

= 2.37× 106 s = 27.4 d

3. v =G m

r=

6.67 × 10−11 Nm 2

kg2 1.99 × 1030 kg( )5.79× 1010 m

= 4.79 × 104 ms



96

4. v =G m

r=

6.67 × 10−11 Nm 2

kg2 1.99 × 1030 kg

1.50 × 1011 m= 2.97 × 104 m

s

5. v =G m

r=

6.67 × 10−11 Nm 2

kg2 1.99 × 1030 kg

1.43× 1012 m= v = 9.63× 103 m

s

6. v =

G mr

=6.67 × 10−11 Nm 2

kg2 1.99 × 1030 kg

2.87 × 1012 m= v = 6.80× 103 m

s

7. Τ = 2πr 3

G m= 2π

5.79× 1010 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1.99× 1030 kg( )

= 7.59 × 106 s = 0.241 yr

8. Τ = 2πr 3

G m= 2π

1.50× 1011 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1.99× 1030 kg( )

= 3.17 × 107 s =1.00 yr

9. Τ = 2πr 3

G m= 2π

1.43× 1012 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1.99× 1030 kg( )

= 9.32 × 108 s = 29.6 yr

10. Τ = 2πr 3

G m= 2π

2.87× 1012 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1.99× 1030 kg( )

= 2.65× 109 s = 84.0 yr

Chapter 11 Review Questions 1. d 2. b 3. a 4. c 5. a 6. The earth’s mass gives it a greater gravitational force. 7. The increased radius would cause you to weigh 1/4 of your original weight. 8. The increased mass would cause you to weigh twice your original weight.

9. The satellite’s horizontal velocity enables it to continually miss the earth as it falls. 10. With no horizontal velocity, the apple struck the ground. The moon’s horizontal velocity enables it to maintain its orbit. 11. The force at the perigee would be greater than that at the apogee. 12. The period of the orbit would be greater. 13. No, the mass in the period equation is the mass of the object being orbited.


1. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 20.150kg( )0.150kg( )

0.250 m( )2 = 2.40× 10−11 N

2. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 265.0kg( )65.0kg( )

1.00 m( )2 = 2.82× 10−7 N



97

3. Fg = m g = 65.0kg( ) 9.80ms 2

⎛ ⎝

⎞ ⎠ = 640 N =140 lbs

4. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 265.0kg( )1.90 × 1027 kg( )

7.15× 107 m( )2 =1.61× 103 N = 362 lb

5. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 265.0kg( )1.50 × 1022 kg( )

1.15× 106 m( )2 = 49.2N = 11.1 lb

6. Τ = 2πr 3

G m= 2π

1.92 × 108 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

= 8.37 × 105 s = 9.69 d

7. Τ = 2πr 3

G m= 2π

7.68× 108 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

= 6.70× 106 s = 77.5d

8. Τ = 2πr 3

G m= 2π

1.54 × 109 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

=1.90× 107 s = 220 d

9. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 285.0 kg( )5.97× 1024 kg( )

6.38 × 106 m( )2 = 832N

10. FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 285.0 kg( ) 5.97 × 1024 kg( )

6.38 × 106 m + 3.62 × 105 m[ ]2 = 745 N


1. (a) FG = Gm1 m

2

r 2

m a = Gm1 m2

r 3 =6.67× 10−11 Nm2

kg 2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 6.42× 1023 kg( )

3.40 × 106 m( )2 = 3.70ms

(b) FG −Mars = m ag = 85.0kg( ) 3.70ms 2

⎛ ⎝

⎞ ⎠ = 315N

FG −Earth = m g = 85.0kg( ) 9.80ms 2

⎛ ⎝

⎞ ⎠ = 833N

(c) The astronaut’s muscles would waken and have a difficult time supporting the returning astronaut on the earth.



98

2. (a) FG −Moon = Gm1 m2

r 2

ma = Gm1 m2

r 2 =6.67 × 10−11 Nm 2

kg 2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 7.35× 1022 kg( )

1.74 × 106 m( )2 = 1.62ms 2

aMoon = Gmearth

r 2

r =G mEarth

amoon=

6.67× 10−11 Nm2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

1.62ms 2

=1.57 × 107 m

(b) 1.57× 107 m6.38 × 106 m

= 2.46Earth Radii.

3. (a) ( ) ( )( )224 11 24

2 23 7

2 2

3 8.64 10 6.67 10 5.97 104.22 10

4 4

Nms kgG m kg

r mπ π

−× × ×Τ

= = = ×

from the center of the earth OR 3.58 × 107 m from the surface of the earth.

(b) v =G m

r=

6.67× 10−11 Nm2

kg 2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5.97× 1024 kg( )

4.22× 107 m= 3.07 × 103 m

s

4. (a) v =G m

r=

6.67× 10−11 Nm2

kg 2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 7.35× 1022 kg( )

1.74 × 106 m + 1.50 × 105 m( ) =1.61× 103 ms = 5.80× 103 km

h

(b) Τ = 2πr 3

G m= 2π

1.74 × 106 m +1.50× 105 m( )3

6.67 × 10−11 Nm 2

kg2⎛ ⎝ ⎜

⎞ ⎠ ⎟ 7.35× 1022 kg( )

= 7370 s= 2.05hr

(c) Increasing the altitude would decrease the velocity and increase the period.

5. (a) FG = Gm1 m

2

r 2 = 6.67× 10−11 Nm 2

kg 265.7 kg( ) 5.97 × 1024 kg( )

6.38 × 106 m + 4.27 × 105 m( )2 = 565N

(b) The astronauts only feel weightlessness because they are in continuous free fall.



99

Chapter 12 12.2

1. S =FA

=1.41×105N

2.50 m( )Ο.80 m( ) = 71kPa

S =FA

=1.41× 105 N

2.50 m( )Ο.45m( )=130 kPa

S =FA

=1.41× 105 N

0.80 m( )Ο.45m( )= 390 kPa

2. S =FA

=975kg( )9.80m / s 2( )

2.50m( )0.200 m( ) =19.1kPa

S =FA

=975kg( )9.80m / s 2( )

2.50m( )0.300 m( ) =12.7 kPa

S =FA

=975kg( )9.80m / s 2( )0.200 m( )0.300m( ) =159kPa

3. (a) k =

FΔl

=54.0 lb24.0 in.

= 2.25 lb / in.

Δl =

Fk

=104 lb

2.25 lb / in.= 46.2 in.

(b) F = k Δl( ) = 2.25 lb / in.( )9.00 in.( )= 20.3 lb

4. (a) k =

FΔl

=17.0 N

0.650cm= 26.2N /cm

F = k Δl( ) = 26.2 N / cm( )1.87cm( )= 49.0N

(b) Δl =

Fk

=21.3N

26.2N /cm= 0.813cm

5. k =

FΔl

=36.0N0.180m

= 200 N /cm

6. Δl =

Fk

=5.00 N

0.250N /cm= 20.0cm

7. (a) k =

FΔl

=1.30×105 N

5.90×10−3 cm= 2.20× 107 N /cm

Δl =

Fk

=5.50× 105 N

2.20× 107 N /cm= 0.0250cm

(b) F = k Δl( ) = 2.20×107 N /cm( )0.0710 cm( )= 1.56× 106 N

8. k =

FΔl

=30, 0 00 lb

0.00234 in.=1.28× 107 lb / in.

Δl =

Fk

=125, 000 lb

1.28× 107 lb / in.= 9.77×10−3 in.

9. F = k Δl( ) = 1.28×107 lb / in .( )0.0279 in.( )= 3.57× 105 lb



100

10. (a) k =

FΔl

=5.00 N40.0cm

= 0.125N /cm Δl =

Fk

=15.0 N

0.125 N /cm= 120 cm

(b) F = k Δl( ) = 0.125N /cm( )60.0cm( )= 7.50 N

11. S =FA

=12.0 00N

π 0.00800m( )2 = 5.97× 107 N /m2 = 5.97 ×107 Pa = 59.7 MPa

12. S =FA

=6.80×106 N

0.250 m( )2=1.09 × 108 Pa =109 MPa

13. (a)

(b) 180N (c) 9.0cm (d) 25 N/cm

14. Δl =

Fk

=15.0 N

96.0 N /cm= 0.280 m −x

x = 0.124m original length = 0.280 m – 0.124 m = 0.156 m



101

15.

( )( ) ( )

( )( ) ( )( )( )( ) ( )( )

1 2

2

2

1

1 1 1

2 2 2

26.0

26.0 2.00 6.00

8.67

17.3

17.3 0.970 / 16.8

8.67 1.45 / 12.6

F F N

N m F m

F N

F N

F k N N cm cm

F k N N cm cm

+ =

=

=

=

Δ = = =

Δ = = =

l

l

16. (a) S =FA

=725N

0.0206m2 = 3.52× 104 N /m2 = 35.2kPa

(b) The stress is doubled because it is inversely proportional to the area applied, which is halved. 17. 489 N. The breaking force is dependent on the diameter of the wire and not on its length. 18. Fb = tensile strength( ) area( )= 1.00×105 N /cm 2( )2.50×10−3 cm2( )= 250 N

19. W = Favg s =ks2

2=

1.25N /cm( )11.5cm( )2

2=82.7Ncm or 0.827 Nm = 0.827 J

12.5

1. Dm =mV

=215g

75.0cm 3 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

×1kg

1000g= 2870 kg /m3

2. ( )( )( )

33

3

1810 1728 .31.0 /

155.9 . 71.1 . 25.4 .w

w lb inD lb ft

V ftin in in= = × =

3. w = Dw V = 30.0 lb / ft 3( )55.9 in.( )71.1 in.( )25.4 in.( )× 1 ft3

1728 in.=1750 lb

4. V =m

Dm=

1350g13,600 kg / m3 ×

1kg1000g

×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 99.3cm 3

5. V =m

Dm=

1350g240 kg /m3 ×

1kg1000 g

×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 5600cm3

6. V =m

Dm=

1350 g1.25kg /m3 ×

1kg1000 g

= 1.08 m3

7. ( )( )( )

33

3

302 1728 .1210 /

19.00 . 8.00 . 6.00 .w

w lb inD lb ft

V ftin in in= = × =

8. w = Dw V = 555 lb / ft 3( )1.40 in.( )2 9.00 in.( )×1 ft 3

1728 in.=17.8 lb

9. Dm =mV

=6.24 kg

12.0cm( )2 16.0cm( )×

100cm1m

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 2710kg / m3

10. 3

3

0.907 8 154.1 /

1.00 1 0.134w

w lb pt galD lb ft

V pt gal ft= = × × =

11. 3

3

3

106 1 100684 /

155 1000 1m

m g kg cmD kg m

V cm g m= = × × =

⎛ ⎞⎜ ⎟⎝ ⎠



102

12. w = Dw V = 42.0 lb / ft 3( )1.00gal( )0.134 ft 3

1gal= 5.63 lb

13. V =m

Dm=

3045kg870kg / m3 = 3.5m3

14. V =w

Dw=

573 lb

62.4 lb / ft 3= 9.18 ft 3

15. Dw =wV

=0.694 lb20.4 in 3 ×

1728 in 3

1 ft 3 = 58.8 lb / ft 3

16. Dw =wV

=0.00301 lb108 in 3 ×

1728 in 3

1 ft 3 = 0.0482 lb / ft 3

17. V =m

Dm=

3.00 kg

2.02kg / m3=1.49m3

18. 2650kgm3 ×

2.21 lb1kg

×1m

3.28 ft⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

=166 lb / ft3

19. Dm =mV

=9.76 kg

18.0cm( )24.0cm( )8.00cm( )×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 2820 kg /m3

20. (a) m = DmV = 100 0 kg / m3( )1.00m3( )=100 0 kg

(b) m = DmV = 680kg /m3( )1.00 m3( )= 680 kg

(c) m = DmV = 8890 kg / m3( )1.00m3( )=8890 kg

(d) m = DmV = 790 kg / m 3( ) 1250L *1m 3

1000L

⎛⎝⎜

⎞⎠⎟

= 990 kg

(e) m = DmV = 1.29 kg / m3( )1.00m3( )=1.29 kg

21. (a) V =m

Dm=

100 0 kg100 0kg / m3 =1.00 m3 ×

1000L1m3 = 100 0 L

(b) V =m

Dm=

100 0 kg680 kg /m3 =1.5m3 ×

1000L1m3 = 1500L

(c) V =m

Dm=

100 0kg13,600 kg / m3 = 0.0735 m3 ×

1000 L1m3 = 73.5L

22. 8890kgm3 ×

1m100cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

×1000g1kg

= 8.89 g /cm3

23. m = DmV = 680 kg / m 3( ) 1250L *1m 3

1000L

⎛⎝⎜

⎞⎠⎟

= 850 kg

24. m = DmV = 790 kg / m 3( ) 1250L *1m 3

1000L

⎛⎝⎜

⎞⎠⎟

= 990 kg



103

25. spgr =Dmaterial

Dwater=

917 kg /m3

100 0 kg /m3 = 0.917

26. 2.3

27. 7.8

28. 0.00129

29. 0.68

30. 0.24

31. Floats

32. Sinks

33. Floats

34. Floats

35. Dm =mV

=m

43

πr 3=

1.67×10−27 kg

43

π 4.1× 10−16 m( )3= 5.8× 1018 kg /m3

spgr =Dmaterial

Dwater=

5.8× 1018 kg / m3

100 0 kg /m3 = 5.8× 1015

36. Dm =mV

=0.315kg0.275m3 = 1.15kg /m3

37. Dm =mV

=0.500 kg

215cm 3 ×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3 = 2330kg / m3

Chapter 12 Review Questions 1. a, c, e

2. a, b, c, d, e

3. d

4. d

5. d

6. b

7. b

8. e

9. Mass density refers to the mass per unit volume while weight density refers to weight per unit volume.

10. Yes; no

11. Capillary action refers to the effect of surface tension of liquids that causes the level of liquid in small diameter tubes to be higher or lower than that of the liquid in a large diameter tube due to the adhesive force between the liquid and the tube.

12. Adhesion refers to the attractive force between different molecules; cohesion refers to the attractive force between similar molecules.

13. Surface tension of water allows the base of many insects’ legs to be supported by the surface of the water, allowing them to “walk” on the surface of a pond.

14. 1.673× 10−27 kg9.109× 10−31 kg

=1800

15. pressure

16. Stress is directly proportional to strain as long as the elastic limit is not exceeded

17. kPa

18. The specific gravity of an object can be found by dividing the density of the object by the density of water. The density can be found by determining the mass of an object on a scale and dividing by its volume.

19. mass density

20. viscosity

21. elastic limit

22. solid, liquid, gas



104

23. An atom consists of one nucleus and its surrounding electrons. A molecule consists of two or more atoms.

24. A proton has a positive charge; a neutron has a neutral charge.

25. Tension, compression, shear, bending, and twisting.

26. The hygrometer measures the density of the battery electrolyte. The density is related to the amount of sulfuric acid in the electrolyte and therefore the charge of the battery. Temperature does affect the measurement.


1. k =

FΔl

=32.5N

0.470cm= 69.1N /cm

F = k Δl( ) = 69.1N /cm( )2.39cm( )=165N

2. Δl =

Fk

=7.33N

0.298N /cm= 24.6cm

3. k =

FΔl

=42,100 lb0.0258 in.

=1.63× 106 lb / in.

Δl =

Fk

=51,700 lb

1.63× 106 lb / in.= 0.0317 in.

4. S =FA

=7.95× 106 N

16.0cm( )2×

100cm1m

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 3.11× 108 N /m2 = 3.11× 108 Pa = 311MPa

5. Dw =wV

=425 lb

7.00 in.( )6.50 in.( )8.00 in.( )×1728 in.3

1 ft3 = 2020 lb / ft 3

6. F = mg = DmVg = 2700kg / m3( )π 1.95cm( )2 4.25cm( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

9.80 m / s2( )= 1.3N

7. DmmV

=8.36kg

9.34 cm( )10.0cm( )2×

100cm1m

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 8950kg /m3

8. Dw =wV

=602 lb

27.7 in.( )36.3 in.( )12.4 in.( )×1728 in.3

1 ft 3 = 83.4 lb / ft 3

9. V =m

Dm=

759g13,600 kg / m3 ×

1kg1000g

×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 55.8cm3

10. V =m

Dm=

1970 g0.0899kg / m3 ×

1kg1000g

= 21.9m3

11. m = DmV = 1.43kg /m3( )1510m3( )= 2160kg

12. w = DwV = 62.4 lb / ft 3( )951 ft 3( )= 59, 300 lb

13. Dw =wV

=0.982 lb

4.27 in.( )3.87 in.( )5.44 in.( )×1728 in.3

1 ft3 =18.9 lb / ft 3

14. Dw =wV

=3.67 lb2.00 qt

×4 qt1gal

×1gal

0.134 ft 3 = 54.8 lb / ft 3

15. V =w

Dw=

4.65 lb39.8 lb / ft3 = 0.117 ft 3

16. spgr =Dmaterial

Dwater=

694 kg /m3

100 0 kg /m3 = 0.694



105

17. w = DwV = 135 lb / ft 3( )4.30 gal( )×0.134 ft3

1gal= 77.8 lb

18. m = DmV = 19, 300kg / m3( )4.00cm( )6.00cm( )20.00cm( )× 1m100cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 9.26 kg

19. 3

3

3

225 1 1002960 /

75.9 1000 1m

m g kg cmD kg m

V cm g m= = × × =

⎛ ⎞⎜ ⎟⎝ ⎠

20. sink (volume of 1 g of water is 1cm3 ) Chapter 12 Applied Concepts

1. (a) P =FA

=1650kg( ) 9.80 m / s2( ) 1

4( )0.0826 m( )0.159m( ) = 3.08× 105 N

m2

(b) P =FA

=1650kg( ) 9.80 m / s2( ) 1

4( )0.152m( )0.210 m( ) =1.27× 105 N

m2

(c) High speed driving and the high pressure tire together increase the heat energy in the tire. This could eventually break down the tire, resulting in a blowout. 2. (a) A shearing strain causes the warping of the lines. (b) An automobile exerts a tremendous amount of force on the pavement to begin its acceleration. On the open road, the vehicle exerts much less force on the pavement to continue its motion.

3. V =Fg

D=

235 lb63.2 lb / ft 3 = 3.72 ft 3

4. V = A ⋅ l = πr 2l = π 1.43m( )2 11.3m = 72.6m3 (a) m = DV = 100 0kg /m3( )72.6 m3( )= 7.26× 104 kg

(b) ( )( )3 3 4870 / 72.6 6.3 10m DV kg m m kg= = = ×

(c) m = DV = 680 kg / m3( )72.6m3( )= 4.9× 104 kg

5. (a) k =

FΔl

=134 lb

0.752 in.=178 lb / in.

(b) F = k Δl = 178 lb / in.( )1.13 in.( )= 201 lb (c) At some point the spring will not be able to compress any farther.



106

Chapter 13 13.1

1. P = hDw = 50.0 ft( )62.4 lb / ft 3( )×1 ft 2

144 in 2 = 21.7 lb / in 2

2. 2 2

3 2

20.0 / 14446.2

62.4 / 1w

lb in inh ft

D lb ft ft

ρ= = × =

3. Dw =Ph

=0.400 lb / in 2

42.0 in.×

1728 in 3

1 ft 3 = 16.5 lb / ft 3

4. (a) Ft = AhDw = π 1.30m( )2 0.750 m( ) 980 0 N /m3( )= 3.90× 104 N

(b) Fs =12

AhDw =12

2π 1.30 m( )0.750m( )[ ]0.750 m( ) 980 0 N /m3( )= 22, 500N

5. P = hDw + 40.0 lb / in2 = 35.0 ft( )62.4 lb / ft 3( )×1 ft 2

144 in 2 + 40.0 lb / in 2 = 55.2 lb / in 2

6. (a) h =Ft

ADw=

165 lb5.00 in.( )9.00 in.( ) 0.490 lb / in3( )= 7.48 in.

(b) Fs =12

AhDw =12

9.00 in.( ) 7.48 in.( )2 0.490 lb / in 3( )= 123 lb

7. P = hDw = 25.0m( ) 980 0 N /m3( )= 2.45× 105 N /m2 = 245kPa

8. h =P

Dw=

115kPa980 0N / m3 =

1.15× 105 N / m2

980 0N / m3 = 11.7 m

9. h =P

Dw

=95.0 kPa

9800 N / m 3=

9.50 × 104 N / m2

9800 N / m 3= 9.69 m

10. DW =P

h=

180 kPa

30.0m=

1.80 × 105 N / m 2

30.0 m= 6000N / m3

11. P = hDw = hDmg

Dm =Phg

=1.78 × 105 N / m2

24.0 m( )9.80m / s 2( )=1.78× 105 kg m / s 2( )/ m2

24.0 m( )9.80m / s 2( ) = 757kg / m3

12. (a) Ft = AhDw = π 23.0m( )2 15.0 m( ) 680kg / m3( )9.80 m / s2( )= 1.7 × 108 N

(b) Fs =12

AhDw =12

2π 23.0 m( )15.0m( )[ ]15.0 m( ) 680kg / m3( )9.80m / s 2( )= 1.1× 108 N

13. P = hDw + 50.0 lb / in2

P = 18.0 ft( )62.4 lb / ft3( )×1 ft2

144 in 2 + 50.0 lb / in 2 = 57.8 lb / in2

14. P = hDw + 325kPa P = 8.00 m( ) 980 0 N /m3( )+ 325kPa = 78.4 kPa + 325kPa = 403kPa

15. P = hDw + 25 lb / in2

P = 30( )16.0 ft( )62.4 lb / ft 3( )×1 ft2

144 in 2 + 25 lb / in 2 = 233 lb / in2

16. P = hDw = 3550 m( )980 0N / m3( )= 3.48× 107 N /m2 = 34,800 kPa



107

17. Ft = AhDw = π 12.0m( )2 50.0 m( ) 980 0 N /m3( )= 2.22× 108 N

18. Fs =12

AhDw =12

2π 12.0 m( )50.0m( )[ ]50.0 m( ) 980 0 N /m3( )= 9.24 × 108 N

19 P = hDw = 50.0m( ) 980 0 N /m3( )= 490 kPa

20. P = 490 kPa + 32.5 m( ) 9800 N / m 3( )= 809 kPa

21. P = 490 kPa + 32.5 m − 8.20 m( ) 9800 N / m 3( )= 728 kPa

22. h =P

Dw=

94.0 lb / in 2

451 lb / ft3 ×144 in 2

1 ft 2 = 30.0 ft

13.2

1. F1 =A1F2

A2=

3.00 in 2( )15.0 lb( )0.750 in 2 = 60.0 lb

2. FE =FR

MA=

2400N25

= 96N

3. MA =FR

FE=

8250 N375N

= 22.0

4. FE =FR

MA=

990 lb18

= 55 lb

5. MA =FR

FE=

1320N55.0 N

= 24.0

6. A2 =A1F2

F1=

8.00cm 2( )360 0N( )25.0 N

= 1150cm2

7. FE =FR

MA=

12000N250

= 48N

8. A2 =F2A1

F1=

865 lb( )4.00 in 2( )10.0 lb

= 346 in 2

9. FR = MA FE = 420( ) 55.0N( )= 23, 000N

10. FR = MA FE = 450( ) 60.0N( )= 27,000 N

11. (a) F1 =F2 A1

A2=

5250 N( )π 2.00cm( )2

π 12.0cm( )2= 146N

(b) P =FA

=5250 N

π 12.0cm( )2 = 11.6 N /cm2

P =FA

=146N

π 2.00cm( )2 = 11.6 N /cm2

(c) MA =A2

A1=

π 12.0cm( )2

π 2.00cm( )2= 36.0

12. A2 =F2A1

F1=

3650N( )8.00cm2( )25.0 N

= 1170cm 2



108

13. (a) F2 =F1A2

A1=

75.0 N( )π 40.0cm( )2

π 5.00cm( )2 = 480 0N

(b) P =FA

=480 0 N

π 40.0cm( )2 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 9550N / m2 = 9.55kPa

(c) P =FA

=75.0N

π 5.00cm( )2 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 9550N / m2 = 9.55kPa

(d) ` MA =FR

FE=

480 0 N75.0N

= 64.0

(e) The list will exert twice the force on the large piston; the MA and the pressure will double. (f) The list will exert four times the force on the large piston; the MA and the pressure will be four

times.

14. F2 =F1A2

A1=

20.0 N( ) 50.0cm 2( )25.0cm2 = 40.0N

15. Since the area of the piston is proportional to its square of its diameter, the force is increased by a factor of 4 to 160 N .

16. 2

2 11 2

2

(1600 )(3.75 )80.0

75.0

F A N cmF N

A cm= = =

17. F1 =F2 A1

A2=

25.0cm2( )1.20 × 104 N( )115 cm2 = 2610N

18. (a) F1 =F2 A1

A2=

π 5.00cm( )2 1.33× 104 N( )π 15.0cm( ) 2 = 1480N

19. (b) P =FA

=1480 N

π 5.00cm( )2 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 1.88 × 105 Pa = 188 kPa

19. A2 =A1F2

F1

=(12.0cm2 )(9800N )

75.0= 1570cm2

20. F2 =A2F1

A1

=(1)(19.600N )

150= 131N

13.3

1. P = 815kPa ×14.7 lb / in 2

101.32kPa= 118 lb / in 2

2. P = 64.3 lb / in 2 ×101.32kPa14.7 lb / in 2 = 443kPa

3. P = 42.5 lb / in 2 ×101.32kPa14.7 lb / in 2 = 293kPa

4. P = 215 kPa ×14.7 lb / in 2

101.32 kPa= 31.2 kPa



109

5. (a) P = 3atm× 101.32kPa / 1atm = 303.96 kPa

(b) P = 2atm × 101.32 kPa /1atm = 202.64 kPa

(c) P = 6atm × 14.7 lb / in2 = 88.2 lb / in2

(d) P = 5atm× 101.32kPa = 506.6 kPa

(e) P = 1/ 3 atm× 101.32 kPa = 33.77 kPa

(f) P = 1/ 4 atm× 101.32 kPa = 25.33 kPa

6. (a) P = 516mm ×101.32kPa

760 mm= 68.8 kPa (b) P = 516mm ×

14.7 lb / in 2

760 mm= 9.98 lb / in 2

7. Pabs = Pga + Patm = 485kPa + 101.32kPa = 586 kPa


9. Pga = Pabs − Patm = 45.0 lb / in2 −14.7 lb / in 2 = 30.3 lb / in 2

10. Pga = Pabs − Patm = 425kPa −101.32kPa = 324 kPa


12. Pabs = Pga + Patm = 362 lb / in 2 + 14.7 lb / in 2 = 377 lb / in 2

13. Pga = Pabs − Patm = 1275kPa −101.32 kPa = 1174 kPa

14. Pga = Pabs − Patm = 218 lb / in 2 − 14.7 lb / in2 = 203 lb / in 2

15. Pabs = Pga + Patmand 3.00 × 105 Pa = 300 kPa

Pabs = 300 kPa + 101.32 kPa = 401 kPa

16. P =FA

=125N

23.0cm2 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 54, 300N / m2 = 54.3kPa

Pabs = Pga + Patm = 54.3kPa + 101.32 kPa = 155.6 kPa

13.4 1. F = 81.0 lb− 68.0 lb = 13.0 lb

2. F = 67.0N − 62.0 N = 5.0N

3. F = 25.7N − 21.8= 3.9N

4. F = 455N − 437N = 18N

5. F = DwV = 62.4 lb / ft 3( )1.21 ft 3( )= 75.5 lb

6. F = DwV = 980 0 N /m3( )16.8m3( )= 1.65×105 N

7. F = DwV = 980 0 N /m3( )327cm 3( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 3.20N



110

8. F = DwV = 980 0 N /m3( )657cm 3( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 6.44 N

9. F = DwV = 49.4 lb / ft 3( )2.12 ft 3( )= 105 lb

10. F = DwV = 680kg /m3( )9.80 m / s2( )515cm 3( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 3.4 N

11. mg − DwV = 75.0 kg( ) 9.80m / s 2( )− 980 0N / m3( )3.10× 104 cm 3( )1m3 /106 cm3( )= 431N

12. 125 lb− 0.800 ft3( )62.4 lb / ft 3( )= 75 lb

13. (a) V = 30.0 ft( )85.0 ft( )12.0 ft( )= 3.06× 104 ft 3

(b) F = DwV −160 tons =62.4 lb / ft 3( )3.06× 104 ft3( )

2000 lb /T−160 T = 795 tons

14. (a) V = 12.0 m( )30.0m( )5.00m( )= 180 0m3

(b) F = DwV − 3.55× 106 N = 980 0N / m3( )180 0 m3( )− 3.55× 106 N = 1.41× 107 N

15. 7.50 × 104 N = m 9.80m / s 2( )

m = 7650kg

V =m

Dm=

7650kg100 0kg / m3 = 7.65m3

16. Fbuoyant = DmgV

V =600 N

1000 kg /m3( )9.80 m / s2( )= 0.0612m3

17. T + Fbuoyant = Fw where Fw is the air weight of thecamera

T = Fw − Fbuoyant = Fw = DmVg

= 1250N − 1000kg / m3( )0.0830m3( )9.80m / s 2( )= 440 N

13.5

1. Q = vA = 5.00m / s( )π 0.0195m( )2 ×1000L1m3 ×

60 s1min

= 358L / min

2. (a) Q = vA = 5.00m / s( )π 0.0750m( )2 ×1000 L1m3 ×

60s1min

= 5.30× 103 L /min

(b) 5.30 × 103 L /min( )30.0min( )= 159,000 L



111

3. (a) ( )

( )

2 34 650 / min4 1min 1; 0.096 9.6

4 1.5 / 60 1000

Lv d Q mQ vA d m or cm

v m s s L

π

π π= = = = × × =

(b) v =QA

=650 L /minπ 0.100 m( )2

×1min60s

×1m3

1000 L= 0.34 m / s

4. (a) Q = vA = vπr 2 = 3.96 m / s( )π 0.060 m( )2 103 L1m3

⎛⎝⎜

⎞⎠⎟

60s1min

⎛⎝⎜

⎞⎠⎟

= 2, 690L / min

(b) 0.226 m3 / min×1min60 s

×1000L1m3 = 3.77 L / s

5. (a) v =QA

=50.0 gal / minπ 3.00 in.( )2 ×

0.134 ft3

1gal×

144 in 2

1 ft 2 = 34.1 ft /min

(b) v =QA

=50.0 gal / minπ 1.50 in.( )2

×0.134 ft3

1gal×

144 in 2

1 ft 2 = 136 ft /min

6. from #3

( )( )

34 36.0 / min4 0.134 12 .0.784 9.40 .

10.0 / min 1 1

galQ ft ind ft in

v ft gal ftπ π= = × = × =

7. ( )( )

34 2620 / min4 1min 10.167 16.7

32.0 /16.0 60 1000

LQ md m cm

v m s s Lπ π= = × × = =

8. 30.0 s. The volume rate of flow is constant.

9. Q = vA = vπr2 = 4.20 m / s( )π 0.25 m( )2=

103 L

1m 3

⎛⎝⎜

⎞⎠⎟

60s

1min⎛⎝

⎞⎠ = 49, 500L / min

10. Q = vA = vπr2 = 3.96 m / s( )π 0.060 m( )2 103 L

1m 3

⎛⎝⎜

⎞⎠⎟

60s

1min⎛⎝

⎞⎠ = 2, 690L / min

Chapter 13 Review Questions 1. c 2. b

3. b 4. c

5. a 6. kPa

7. Pressure is the force applied per unit area.

8. Fs =12

AhDw or one-half the force exerted by the water on the bottom of the tank.

9. External pressure applied to a fluid is transmitted to all inner surfaces of the liquid’s container.

10. A ship floats because it is lighter than an equal volume of water.

11. A spinning baseball creates a different air velocity on one side than on the other. This creates a higher pressure on one side that causes the ball to “curve.”

12. The top side of the wing is curved more than the bottom so that the velocity of air rushing past the top must be larger than that going past the bottom. This creates a lower pressure area on the top of the wing.

13. In streamline flow, all particles of the fluid passing a given point will follow the same path. In turbulent flow, the particles passing a given point may follow different paths from that point.

14. A balloon filled with a gas with lower density than air such as helium, rises.



112

15. Absolute pressure is the actual air pressure given by the gauge reading plus normal atmospheric pressure. Gauge pressure is the pressure measured relative to atmospheric pressure. Gauge pressure is used in an automobile tire gauge.

16. The pressures are identical. The forces are different. 17. This force depends upon the horizontal surface area and the average pressure on that surface. The

average pressure depends upon the density and the height of the liquid. 18. The pressure at the bottom is greater than at the top. 19. Smaller; the total force exerted on the brake pads must be larger than the force applied to the brake

pedal. 20. Yes, but only if the fluid to be drawn through the straw is in an airtight sealed container. Chapter 13 Review Problems 1. P = Dwh = 980 0 N /m3( )3.24 m( )= 31.8kPa

2. h =P

Dw=

197 lb / in2

62.4 lb / ft3 ×144 in 2

1 ft 2 = 455 ft

3. h =P

Dw=

297, 000N / m2

980 0 N /m3 = 30.3m

4. Ft = Dw hA = 62.4 lb / ft 3( )3.56 ft( )8.67 ft( )4.83 ft( )= 930 0 lb

5. P = Dwh = 980 0 N /m3( )20.0 m( )= 196kPa

6. Ft = Dw hA = 980 0N / m3( )55.0m( )π 7.53m( )2 = 9.60× 107 N

7. Fs =12

DwhA =12

980 0N / m3( )55.0m( )2π 7.53m( )55.0m( )= 7.01× 108 N

8. Fs =12

DwhA =12

980 0N / m3( )1.25m( )1.55m( )2.95m( )= 2.80 × 104 N

9. P = Dwh + 252N /cm 2 = 980 0 N /m3( )9.00m( )+ 252N /cm 2 ×100cm

1m⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 2.61× 106 N /m2

= 2610kPa

10. P = Dwh + 26 lb / in 2 = 62.4 lb / ft 3( )24( )16.0 ft( )×1 ft 2

144 in 2 + 26 lb / in 2 = 192 lb / in 2

11. P = Dwh = 980 0 N /m3( )3150 ft( )×1m

3.28 ft= 9410kPa

12. (a) F1 =A1F2

A2=

0.564 in2( )650 lb( )4.75in 2 = 77.2 lb (b) MA =

FR

FE=

650 lb77.2 lb

= 8.42

13. FE =FR

MA=

11500N324

= 35.5N

14. (a) F1 =A1F2

A2=

π 0.543cm( )2 4350 N( )π 3.53cm( )2

=103N

(b) P =FA

=4350N

π 3.53cm( )2= 111N /cm 2 (c) MA =

FR

FE=

4350N103N

= 42.2

15. Pabs = Pga + Patm = 202 kPa + 101.32 kPa = 303kPa 16. Pga = Pabs − Patm = 655kPa −101.32kPa = 554 kPa 17. Pga = Pabs − Patm = 314 lb / in2 −14.7 lb / in 2 = 299 lb / in 2 18. F = 55.4 N − 52.1N = 3.3N



113

19. F = DwV = 980 0 N /m3( )643cm3( )×1m

100cm⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 6.30 N

20. F = DwV = 790kg /m3( )9.80 m / s2( )314cm 3( ) 1m100cm

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= 2.4 N

21. (a) V = lwh = 87.5 ft( )22.3 ft( )13.3 ft( )= 2.60× 104 ft 3

(b) F = DwV −157 tons = 62.4 lb / ft3( )2.60× 104 ft 3( )×1 ton

2000 lb− 157 tons = 654 tons

22. Q = vA = 4.43m / s( )π 0.0150 m( )2 ×1000L1m3 ×

60 s1min

= 188L / min

23. (a) Q = vA = 4.53m / s( )π 0.0650 m( )2 ×1000L1m3 ×

60 s1min

= 3610L / min

(b) V = Qt = 3610L / min( )25.0 min( )= 90, 300L

24. (a) Fs =12

AhDw

433N =12

2π 0.913m( )1.75m( )[ ]1.75m( )Dw

Dw = 49.3N /m3 (b) Alcohol


1. (a) r =v

π h=

758 m3

π 14.5m= 4.09 m

( ) ( ) 22 14.5 2 4.09 372A c r m m mπ π= = = =l l

(b) Fs =12

A h Dw =12

372m2( )14.5m( )1025kg / m3( )= 2.76 × 106 m

(c) The bands should be spaced closer together because of the increased pressure at the bottom of the tank.

2. (a) P1 =FA1

=45.5 lb

π 0.570 in.( )2= 44.6 lb / in.2

(b) F = Ps A2 = 44.6 lb / in.2( )π 1.75 in.( )2 = 429 N on each piston or brake drum.

3. (a) V = l w h = 1.25m( )1.25m( )0.450 m( )= 0.703m3 (b) m = DmV = 2, 300kg / m3( )0.703m3( )= 1600kg 9.80m / s 2( )=16,000 N

(c) m = DmV = 1025kg / m3( )0.703m3( )= 720 kg 9.80 m / s2( )= 7060N (d) F = 16,000 N − 7060 N = 8900N 4. (a) 1 1 2 2v A v A=

( )( )7 2

1 12 7 2

2

25 / 3.14 1050.3 /

1.56 10

mi h miv Av mi h

A mi

−

−

×= = =

×

(b) The space between the buildings acts as a venturi tube by limiting the area that the air can pass through.



114

5. (a) Q = v1 A1 = 47.3 in. / s( )π 0.250 in( )2 = 9.28 in.3 / s

v2 =QA 2

=9.28 in.3 / s

15 π 3.13×10−2 in.( )2 = 201in. / s = 16.8 ft / s

(b) Limit the number of holes or make the holes smaller. Limiting the area for the water to come out causes it to travel faster.



115

Chapter 14 14.1 1. 25°C 2. 45°C 3. 125°C 4. 59°F 5. 293°F 6. 95°F 7. −12.2°C 8. −6.7°C

9. 203°F 10. −45.6°C 11. 298K 12. 415°R 13. 133° C 14. 348 K 15. 223 K 16. -98°C

17. 5727°C 18. 2741°F 19. –38.9°C 20. 2552°F 21. 29°C 22. 41°F 23. -321°F 24. -117°C 25. 120 0

26. 2010 27. 941 28. 1324 29. 1175 30. 482 31. 590

14.2 1. 23 cal 2. 1.79 kcal 3. 1.21 × 106 ft lb 4. 1.61 × 107 J or 16.1MJ

5. 3.21 × 106 J or 3.21 MJ 6. 35,800 J or 35.8 kJ 7. 4450 Btu 8. 182 kcal

9. 625kcal ×4190 J1kcal

= 2.62 MJ

10. 20 0 g ×150cal

28g×

4190J1kcal

= 4.5 MJ

11. 1.15 × 104 cal / g( ) 1000 g( ) ×4.19 J

1cal= 4.82x107 J

12. 1.60 × 104 cal / g( ) 500 g( ) ×4.19 J

1cal= 3.35x107 J

13. 38, 000kcal ×4190J

1kcal= 1.59x108 J

14. 1.25 × 104 Btu / lb( )2000 lb( )×778 ft lb

1Btu= 1.95 × 1010 ft lb

15. (a) 1.10 × 105 cal /g( )2.50g( ) 0.24( ) ×4.19 J1cal

= 2.77× 105 J / s

(b) 2.77 × 105 J / s = 2.77 × 105 W = 277 kW 16. 0.25( )x = 100 0 ft / lb

x = 400 0 ft lb ×1 Btu

778 ft lb= 5.14 Btu

17. 50, 000kcal ×4190J

1kcal= 2.10x108 J

18. 38, 000kcal ×4190J

1kcal= 1.59x108 J



116

14.3

1. R =LK

=0.125in.

0.50 Btu / ft °Fh×

1 ft12 in .

= 0.021 ft 2 °F h /Btu

2. R =LK

=4.0 in .

0.48 Btu / ft °Fh×

1 ft12 in .

= 0.69 ft2 °F h / Btu

3. R =LK

=0.50 in .

0.092 Btu / ft °Fh×

1 ft12 in .

= 0.45 ft 2 °F h /Btu

4. K =LR

=0.25in.

1.6 ft 2 °F h /Btu×

1 ft12 in.

= 0.013Btu / ft °F h( )

5. R =LK

=0.50 in.

0.024 Btu / ft °F h×

1 ft12 in.

= 1.7 ft 2 °F h / Btu

6.Q =KAt T2 −T1( )

L=

At T2 −T1( )L /K

=At T2 −T1( )

R=

4 20 ft( )100 ft( )24 h( )55°F − 20 °F( )11 ft 2 °F h / Btu

= 6.1×105 Btu

7.Q =KA T2 −T1( )

L=

0.50Btu / (ft °F h )( )15 ft 2( )30.0 days( )24 h / day( )25°F( )

0.20 in.×1 ft

12 in.

= 8.1×106 Btu

8.Q =KA T2 −T1( )

L=

45J / (sm°C)( )0.45m( )0.75m( )30.0days( )24 h / day( ) 95°C( )

0.25cm ×1m

100cm

=3600 s

1h= 1.5×1012 J

9.Q =KAt T2 − T1( )

L=

45J / (sm°C)( )π 0.0025m( )2 75s( )100 °C − (−25°C)( )0.85m

= 11J

10.( ) ( ) ( )( ) ( )( )( )22

2 1380 /( ) 150 1 /100 15 min 60 / min 985

10.10

100

J sm C cm m cm s CKAt T TQ

mL cmcm

π° °−= =

×

= 5.1×109 J

11.Q =KAt T2 − T1( )

L=

0.023Btu / ( ft°F h )( )30.0 in.( )58.0 in.( )1 ft2 /144 in 2( )24 h( )34°F( )

2.0 in. ×1 ft

12 in.

= 1400 Btu

12. Q =KAt T2 − T1( )

L=

0.023Btu / ( ft°F h )( )30.0 in.( )58.0 in.( )1 ft2 /144 in 2( )30days( )24 h /day( )82°F( )

2.0 in. ×1 ft

12 in.

= 9.8 × 104 Btu

13.Q =0.039 J / (sm°C( )0.76m( )1.550 m( )24 h( )18°C( )

0.050m×

3600s1h

= 1.4 × 106 J or14 MJ



117

14.Q =0.039 J / (sm°C( )0.76m( )1.550 m( )30 days( )24 h / day( )39°C( )

0.050 m×

3600 s1h

= 9.3 × 107 J or 93MJ

15. Q =0.75 J / (sm°C( )2π 0.070 m( ) 0.18 m( ) 45 s( ) 15.3°C( )

0.00600 m= 6800 J

14.4 1. Q = cwΔT = 0.115Btu / (lb°F)( )3.00 lb( )500 °F( )= 173Btu

2. Q = cmΔT = (0.092 kcal / (kg °C ))(155kg)(170°C) = 2400kcal

3. Q = cwΔT = (1.00 Btu / (lb °F))(19.0 lb)(200 °F) = 380 0 Btu

4. Q = cmΔT = (1.00cal /g °C)(250 g)(17.0°C) = 4250cal

5. Q = cmΔT = (2100J / (kg °C)(5.00 kg)(20 °C) = 2.1× 105 J or 210kJ

6. (0.48 /( )(5.00 )(40 ) 96Q cw T Btu lb F lb F Btu= Δ = ° ° = =

7. Q = cmΔT = (0.22cal / (g°C)(79.0 g)(16°C) = 280cal

8. Q = cmΔT = (390J / (kg °C)(750kg)(125°C) = 3.7× 107 J or 37 MJ

9. Q = cmΔT = (481J / (kg °C)(1.25kg)(50.0°C) = 3.01× 104 J or 30.1kJ

10. Q = cmΔT = (0.22 kcal / (kg °C )(0.850kg)(115°C) = 22 kcal

11. Q = cmΔT = (1.00 kcal / (kg °C ))(0.800 kg)(80.0°C) = 64.0 kcal

12. Q = cmΔT = (130J / (kg °C))(475kg)(245°C) = 1.5× 107 J or15MJ

13. Q = cwΔT = (0.092 Btu / (lb °F)) (1200 lb)(450 °F −100 °F) = 39,000 Btu

14. Q = cwΔT = (0.22 Btu / (lb °F)) (500 lb)(650 °F − 75°F) = 6.3× 104 Btu

15. Q = cmΔT = (0.092 kcal / (kg °C )) (1250 kg)(275°C − 25°C) = 29,000 kcal

16. Q = cmΔT = (4190J / (kg °C))(1.85kg) (80.0°C −10.0°C) = 5.43× 105 J or 543kJ

17. Q = cmΔT = (4190J / (kg °C)) (750 kg)(75.0°C −15.0°C) = 1.89× 108 J or189 MJ

18. Q = cmΔT = (0.115kcal / (kg °C)) (750 kg)(300 °C − 75°C) = 1.94 × 104 kcal

19. Q = cmΔT = (481J / (kg °C)) (125kg)(1425°C − 82°C ) = 8.07× 107 J or 80.7 MJ

20. Q =boiler = cmΔT = (0.115kcal / (kg °C)) (525kg)(100.0°C − 40.0°C) = 3620 kcal

Qwater = cmΔT = (1.00kcal / (kg °C )) (315kg)(100.0°C − 40.0°C) = 18, 900kcal

Q =Qboiler + Qwater

0.75=

3620 kcal +18,900 kcal0.75

= 3.00× 104 kcal

21. Q = cmΔT ; 3.36×105 J = (390J /kg °C)) (5.00 kg)(T1 −80.0°C);T1 = 252°C

22. Q = cmΔT = (481J / (kg °C))(1658kg)(13°C) = 1.0× 107 J

23. Q = cmΔT = (481J / kg°C )(0.400kg)(325°C − 295°C) = 5770 J



118

24. (390 / )(0.0600 )(80.0 20.0 ) 1400Q cm T J kg C kg C C J= Δ = ° ° − ° =

25. (a) ΔT =Qcm

=8.45×105 J

(4190J /kg °C)(20.0 kg)= 10.1°C

(b) m = DV = (0.800g /cm3)1000cm3

1L⎛

⎝ ⎜ ⎞

⎠ ⎟ (20.0L) =1.60× 104 g = 16.0 kg

and ΔT =Qcm

=8.45× 105 J

(2400J /kg °C) (16.0 kg)= 22°C

(c) Water is the better coolant since its temperature increase is only about half that of alcohol in

absorbing the same amount of heat. 14.5

1. Tl =

cg wg (T f −Tg )c l wl

= + T f =(1.00 Btu / (lb°F)) (11.0 lb)(84.0°F − 75.0°F)

(0.115Btu / (lb°F)) (2.50 lb)+ 84.0°F = 428°F

2.(1.00 /( ))(5.00 )(200 ) (1.00 /( ))(7.00 )(65.0 )

(1.00 /( ))(5.00 ) (1.00 /( ))(7.00 )g g g

f

g g

c w T c w T Btu lb F lb F Btu lb F lb FT

c w c w Btu lb F lb Btu lb F lb

+ ° ° + ° °= =

+ ° + °l l l

l l

= 121°F

3. c l =

cgmg (T f −Tg )m l (Tl −T f )

=(1.00cal / (g °C))(100 g)(20 °C −10 °C )

(250 g )(99°C − 20 °C )= 0.051cal / (g °C)

4. mg =

c l m l (Tl −T f )cg (T f − Tg)

=(1.00cal / (g °C))(800 g )(90 °C − 50 °C)

1.00cal / (g °C )) (50 °C − 20 °C)= 1100 g

5.

T f =c l wl Tl + cg wgTg

c l wl + cg wg= (0.22 Btu / (lb °F))(159 lb)(500 °F) + (1.00 Btu / (lb °F))(400 lb)(60 °F)

(0.22 Btu / (lb°F))(159 lb) + (1.00Btu / (lb °F))(400 lb)= 95°F

6. T f =

c l wl Tl + cg wgTg

c l wl + cg wg

=(0.115Btu / (lb °F)) (42.0 lb)(670 °F) + (1.00Btu / (lb °F)) (100 lb)(75.0°F)

(0.115Btu / (lb°F)) (42.0 lb) + (1.00Btu / (lb °F)) (100 lb)= 102°F

7. T f =

c l m l Tl + cgmgTg

c l m l + cg mg=

(1.00cal / (g °C )) (500 g ) (95.0°C) + (0.092cal / (g °C)) (1250g)(20.0°C )(1.00cal / (g °C )) (500 g ) + (0.092cal / (g°C)) (1250g )

= 81°C



119

8. T f =

c l m l Tl + c l ml Tl + cg mgTg + cg mgTg

c l ml + c l m l + cgmg + cg mg

=(0.092cal / (g °C )) (500 g) (200 °C) + (0.115cal / (g°C)) (300 g) (150 °C) +

(0.092cal / (g °C)) (500 g) + (0.115cal / (g °C)) (300 g) +

(1.00cal / (g °C )) (900 g) (20.0°C) + (0.22cal / (g °C)) (150 g )(20.0°C)

(1.00cal / (g °C)) (900 g ) + (0.22cal / (g °C )) (150 g )= 33°F

9. c l =

(cgmg + cg mg )(T f −Tg )(m l )(Tl −T f )

=(0.092cal / (g °C)) (153g ) + (1.00cal / (g°C)) (275g)[ ](22.7°C −16.2°C)

(236g )(99.6°C − 22.7°C)= 0.104cal / (g °C )

10. c l =

(cgmg + cg mg )(T f −Tg )(m l )(Tl −T f )

=(920 J / (kg °C)) (0.132 kg) + (4190J / (kg °C ) (0.285kg)[ ](18.6°C −12.6°C)

(0.215kg)(99.1°C −18.6°C)= 456 J / (kg °C)

11. Tl =

(cg mg + cgmg )(T f −Tg )c l m l

+ T f

=(1.00cal / (g °C)) (325g ) + (0.092cal / (g°C))(165g)[ ](31.0°C −18.0°C )

(0.031cal / (g°C))(560 g)+ 31°C = 286°C

12.Q = cmΔT = (481J / kg°C)(0.565kg)(100.0°C − 20.0°C ) = 2.17 × 104 J

13. m =Q

cΔT=

2.17× 104 J(4190J /kg°C)(20.0°C)

= 0.259kg

14.With a much higher specific heat, an equal mass of water will hold approximately 8.7 times

4190 J / kg°C481J /kg°C

⎛ ⎝ ⎜

⎞ ⎠ ⎟ as much heat as the same mass of steel.

14.6 1. l = αl ΔT = (9.5× 10−6 / °F)(200.0 ft)(200.0°F − 40.0°F) = 0.30 ft 2. l = αl ΔT = (2.6× 10−5 / °C)(50.0 m)(130.0°C −15.0°C ) = 0.15m 3. l = αl ΔT = (1.7× 10−5 /°C)(300.0 m)(34°C −14°C ) = 0.10 m 4. l = αl ΔT = (1.3× 10−5 /°C)(8.25m)(45°C + 60 °C) = 0.011m 5. ′ l = l (1+αΔT ) = 200.0 m 1+ (1.3× 10−5 / °C)(40 °C )[ ]= 200.10 m

6. 5

0.0030150

(1.9 10 / )(1.020 )

mT C

C mα −

ΔΔ = = = °

× °

l

l

7. Δl = αl ΔT = (6.0 × 10−6 /°F)(500.0 ft)(180 °F) = 0.54 ft

8.

′ T = T −Δlαl

= 40.0°C −0.003cm

(2.3× 10−5 /°C)(10.003cm)= 27°C

9. ′ d = d (1+ αΔT ) = 0.750 in. 1+ (6.5× 10−6 / °F)(330 °F)[ ]= 0.752 in.



120

10. 5

0.01120.0 20.0 28

(1.9 10 / )(12.000 )

cmT C C C

C cmα − °

Δ= ° − = ° − = − °

×

l

l

5

0.01120.0 20.0 74

(1.7 10 / )(12.000 )

cmT C C C

C cmα − °

Δ= ° + = ° + = °

×

l

l

11. Δ A = 2α A ΔT = 2(1.9× 10−5 / °C)(482cm2 )(100 °C ) = 1.8cm 2 12. ΔV = 3α V ΔT = 3(1.9× 10−5 / °C)(4820cm3 )(315.0°C ) = 87cm 3 13. ′ A = A(1+ 2α ΔT ) = 88.40cm2 1 + 2(2.3× 10−5 /°C )(140 °C )[ ]= 88.97cm2

14. ′ A = A(1+ 2α ΔT ) = 127.20 in 2 1+ 2(6.5× 10−6 / °F)(150 °F)[ ]= 127.45 in 2

15. ′ V = V(1+ 3α ΔT ) = 60.00cm3 1 + 3(9.0 × 10−6 / °C )(64°C)[ ]= 60.10cm3

16. ′ d = d (1+ α ΔT ) = 6.500cm 1 + (1.9× 10−5 /°C)(154°C )[ ]= 6.519cm

A = πd2 /4 = π (6.519cm)2 / 4 = 33.38cm2 17. Δl = αl ΔT = (1.3× 10−5 / °C )(15.000m)(25.0°C ) = 4.9 × 10−3 m = 4.9 mm

18. (a) Δl = α l Δ T = (6.5 × 10−6 / °F)(60, 000 ft )(160 F)(12 in. / ft ) = 0.749 in. (b) Δl = (6.5× 10−6 /°F)(120.00 ft)(55°F)(12 in. / ft) = 0.51 in.

19. α =

Δll ΔT

=0.00711m

(13.00m)(50.5°C )= 1.08 × 10−5 /°C

20. ΔV = 3α V ΔT = 3(6.0 × 10−6 /°F)(8.00 × 104 ft 3)(130 °C) = 190 ft 3 21. ΔV = 3α V ΔT = 3(9.0 × 10−6 /°C)(7238cm3 )(75.0°C) = 15cm 3 22. Δl = αl ΔT = (1.1× 10−5 / °C )(4.250m)(45.0°C ) = 0.0021m Therefore new height = 4.252 m 23. V = πr 2h = π (0.03000 m)2 (1.200 m) = 0.3393m3 ΔV = 3α V ΔT = 3(9.0 × 10−6 /°C)(0.3393m3)(43.0°C) = 3.9 × 10−4 m3 V = 0.3393m3 + 3.9× 10−4 m3 = 0.3397 m3

24. α =

Δll ΔT

=3.40 × 10−3 m

(2.60m)(72.0°C )= 1.82 × 10−5 /°C

14.7 1. ′ V = V(1+ β ΔT ) = 625L 1+ (1.24 × 10−3 /°C)(36°C )[ ]= 653L

2. ΔV = Vβ ΔT = (157 in 3)(1.0× 10−4 / °F)(120 °F) = 1.9 in3 3. ′ V = V(1+ β ΔT ) = 11.7m3 (1+ (9.6× 10−4 / °C)(45°C )[ ]= 12.2 m3

4. ΔV = V β ΔT = 35L(1.49× 10−3 /°C)(10 °C) = 0.52L

5. ′ V = V(1− β ΔT ) − 3780 ft 3 1− (1.17× 10−4 / °F)(58°F)[ ]= 3754 ft 3

6. ΔV = V β ΔT = (1200 L)(9.6× 10−4 / °C )(36°C) = 41L 7. ΔV = V β ΔT = (215cm 3)(1.8 × 10−4 /°C)(15°C ) = 0.58cm 3 8. ΔV = V β ΔT = (2000 ft 3)(6.62 × 10−4 / °F)(21°F) = 28 ft3

9. Δ$ = ($0.40 / L)V β Δ T = ($0.75 / L)(34, 000 L)(9.6 × 10−4 / °C)(17°C) = $416.16

10. ΔV = V β Δ T = (500 L)(9.6 × 10−4 / C°)(30.0°C − 4.0°C) = 12.5L



121

11. ΔV = V β ΔT = 20 mL = (1.49 × 10−3 / °C)(180 mL)(98°C −Tο ) Tο = 23°C

12. ΔV = V β Δ T = (1200L)(9.6 × 10−4 / C°)(25.0°C − 3.0°C) = 25.3 L

13. ΔV = V β Δ T = (500 L)(9.6 × 10−4 / C°)(30.0°C − 4.0°C) = 12.5L 14.8

1. Q = L f m = (80 cal /g )(14.0g ) − 1100cal 2. w =QL f

=635Btu

144 Btu / lb= 4.41 lb

3. Q = wLv = (11.0 lb)(970 Btu / lb) = 10,700 Btu

4. m =QLv

=1520cal

540 cal /g= 2.81g

5. Q = L f m = (80 cal /g )(320 g) − 26,000cal

6. Q = Lvm = (540 cal / g)(3250g ) = 1.76 × 106 cal

7. Q = L f m = (335kJ /kg)(20.0 kg) = 6.70 × 103 kJ or 6.70 × 106 J

8. Q = L f m = (80 kcal /kg)(20.0 kg) = 1600 kcal

9. Q = Lvm = (2.26× 106 J / kg)(1.50kg) = 3.39 × 106 J or 3.39 MJ

10. m =QLv

=5.00MJ

2.26MJ / kg= 2.21kg ×

1L1kg

= 2.21L

11. Q = L f m + cm ΔT = (144 Btu / lb)(33.0 lb) + (1.00 Btu / lb°F)(33.0 lb)(40.0°F) = 6070Btu

12. Q = L f m + cm ΔT = (144 Btu / lb)(20.0 lb) + (1.00 Btu / lb°F)(20.0 lb)(48°F) = 3840Btu

13. Q = L f m + cm ΔT + Lvm + cm ΔT = (144 Btu / lb)(9.00 lb) + (1.00Btu / lb °F)(9.00 lb)(180 °F)

+(970 Btu / lb)(9.00 lb) + (0.048Btu / lb °F)(9.00 lb)(20 °F) = 11,650 Btu

14. Q = Lvm + cm ΔT +L f m + cm ΔT = (540 cal / g)(200 g ) + (1.00cal /g °C)(200 g)(100°C ) +

(80 cal /g )(200 g) + (0.51cal / g°C)(200 g)(12°C ) = 1.45× 105 cal

15. Q = L f m + cm ΔT = (80 kcal / kg)(50.0kg) + (1.00 kcal / kg°C )(50.0kg)(20 °C) = 50 00kcal

16. Q = L f m + cm ΔT = (3.35× 105 J /kg)(15.0 kg) + (4190 J / kg °C)(15.0kg)(75°C ) = 9.7 × 106 J

or 9.7 MJ

17. Q = csmΔTs + Lvm + cw mΔTw = (20 00 J / kg °C )(1.25kg)(15°C) + (2.26 MJ /kg)(1.25kg) +

(4190J /kg °C)(1.25kg)(50 °C) = 3.12× 106 J or 3.12MJ

18. (540 / )(5.00 ) (0.48 / )(5.00 )(45 ) 2810vQ L m cm T kcal kg kg kcal kg C kg C kcal= + Δ = + ° ° =

19. Q = cmΔT + L f m + cmΔT + Lvm + cmΔT

= (0.51cal /g °C)(625g )(24.0°C) + (80 cal /g)(625g) + (1.00cal / g°C)(625g)(100 °C ) +

(540 cal / g)(625g) + (0.48cal / g°C)(625g)(32.0°C) = 4.67× 105 cal



122

20. Q = cm ΔT + Lvm + cm ΔT + L f m + cm ΔT

= (0.48 Btu / lb°F)(5.65 lb)(24.0°F) + (970 Btu / lb)(5.65 lb) + (1.00 Btu / lb°F) (5.65 lb)(180.0°F) +

(144 Btu / lb)(5.65 lb) + (0.51Btu / lb °F)(5.65 lb)(20.0°F) = 7430Btu

21. m =Wg

=143N

9.80 m / s2 = 14.6kg

ΔT = 78.5°C − 65.0°C =13.5°C

Q = mc ΔT + Lvm = (14.6kg)(0.58 kcal /kg °C)(13.5°C ) + (204 kcal /kg)(14.6 kg) = 3090 kcal

22. Q = mL f + cmΔT + mLv + cmΔT = (620 g)(2.82cal / g) +

(0.033cal /g °C)(620 g)(357°C + 38.9°C ) + (620 g)(65.0cal / g) = 5.01× 104 cal

Chapter 14 Review Questions 1. a, b, d

2. b, d

3. a, b, d

4. a, b, d

5. d

6. c

7. The total heat gained by warm objects is gained by cold objects.

8. 778 ft lb of work is equivalent to 1 Btu.

9. Rankine scale.

10. Kelvin scale.

11. Each Celsius degree is 1.8 times the Fahrenheit degree. The freezing point of water is 0°C and 32°F. The boiling point of water is 100°C and 212°F.

12. Heat is an amount of energy. Temperature is a measure of the average kinetic energy of atoms and molecules in matter.

13. Burning fuel in a furnace; burning fuel in an engine. Conversion of heat into electricity in a steam driven generator.

14. Heat generated (a) in a solid by a drill bit. (b) by rubbing two objects together, and (c) in automobile brakes. All are due to friction.

15. Light clothing should be worn because it absorbs less heat.

16. It increases; the metal block increases in size, as does the hole.

17. (a) 4°C (b) Most other liquids have their highest density at the temperature at which the change of state to a solid occurs. (c) Water below 4°C is less dense than water at 4°C and rises. Freezing therefore occurs at the top of a body of water.

18. 10 kg of ice at 10°C; the difference is the heat of fusion, which must be added to the ice to turn it to water.

19. The large amount of heat released when steam changes to water can cause severe burns.

20. Because water expands as it solidifies.

21. Conduction is the transfer of kinetic energy from atoms or molecules in a warmer location through nearby atoms or molecules to atoms or molecules in a colder section if the material. Convection is the transfer of kinetic energy from one region to another via the motion of warmer atoms or molecules from one region to another. Radiation is the transfer of energy through the emission and absorption of electromagnetic radiation.



123

22. Coolants boil at a higher temperature at high pressure. Therefore the engine coolant can be at a higher temperature and transfer more heat from the engine to the atmosphere because of the greater temperature difference.

23. Heat is extracted from the outside air in winter to vaporize the refrigerant. The heat is given up by the condensing fluid to warm the inside air. In summer, heat is extracted from the inside air to vaporize the refrigerant. The condensing fluid then gives up this heat to the outside air.

Chapter 14 Review Problems 1. 344 = TC + 273; TC = 71°C

2. TK = TC + 273 = 24 + 273 = 297 K

3. TF =95

TC + 32° =95

(5110°) + 32° = 9230°F 4. TC =59

(TF − 32°) =59

(635°− 32°) = 335°C

5. 43.0 J ×1cal4.19 J

= 10.3cal

6. 6530J ×1kcal4190 J

= 1.56 kcal

7. 435Btu778 ft lb

1Btu= 3.38× 105 ft lb

8. Q =KAt (T2 −T1)

L=

(0.50 Btu / ft 2Fh)(33 ft 2)(4.10h)(26°F)

0.15 in.×1 ft

12 in.

= 1.4× 105 Btu

9. Q =KAt (T2 −T1)

L=

(0.039 J / sm°C)(0.800 m)(1.44 m)(25.0 days)(24 h / day)(22°C + 14°C)0.040m

= 2.4 × 104 J or 24 kJ

10. Q = cwΔT = (0.115Btu / lb °F)(835 lb)(455°F − 20.0°F) = 4.18 × 104 Btu

11. Q = cmΔT = (0.22 kcal /kg °C)(148 kg)(485°C − 21.5°C) = 1.5× 104 kcal

12. Q = cmΔT + ′ c ′ m ΔT = (0.115kcal / kg °C )(161kg)(91.1°C) +

(1.00 kcal / kg °C )(115kg)(91.1°C) = 1.22× 104 kcal

13. ( ) (1.00 / )(8.35 )(50.2 )

98.2 1300(0.092 / )(3.80 )

g g f g

f

c w T T Btu lb F lb FT T F F

c w Btu lb F lb

− ° °= + = + ° = °

°l

l l

14. c l =

cgmg (T f −Tg )m l (Tl − Tg )

=(1.00cal /g °C )(111g )(32.5°C −15.0°C)

(355g)(48.0°C − 32.5°C)= 0.353cal /g°C

15. ΔT =

ΔlαL

=0.0734cm

(1.9× 10−5 / °C )(45.2cm)= 85°C

16. ′ d = d (1+α ΔT ) = (12.500 m) 1+ (1.3× 10−5 / °C)(154°C − 5°C )[ ]= 12.524 m



124

17. ′ d = d (1+α ΔT ) = (7.50 mm) 1 + (2.3× 10−5 /°C)(89°C − 22°C)[ ]= 7.51mm

A =πd2

4=

π (7.51mm)2

4= 44.3mm2

18. ΔV = 3αVΔT = 3(1.3× 10−5 / °C )(4π / 3)(1.54cm)3 (84.5°C − 35.0°C ) = 0.0295cm3

19. ΔV = β VΔT = (1.49 × 10−3 /°C)(44.8L)(75.5°C − 37.0°C ) = 2.57 L

20. ΔV = β VΔT = (6.62 × 10−4 / °F)(3450 ft 3)(87.0°C − 33.0°C) = 123 ft 3

21. Q = mLv = (21.5kg)(540 kcal / kg) = 11,600 kcal

22. Q = mLv = (8.35 lb)(144 Btu / lb) = 120 0 Btu

23. Q = csmΔTs + mLv + cw mΔTw + mL f + cimΔTi = (0.48 kcal / kg°C)(4.56kg)(25°C)

+(4.56kg)(540 kcal /kg) + (1.00 kcal / kg°C)(4.56kg)(100 °C) +

(4.56 kg)(80 kcal / kg) + (0.51kcal / kg°C)(4.56 kg)(44.5°C ) = 3440kcal

24. Q = mL f + cmΔT + mLv = (0.336kg)(1.04 × 105 J /kg) +

(2400J /kg °C)(0.336 kg)(195.5°C ) + (0.336 kg)(8.54 × 105 J / kg) = 4.80× 105 J


1. ( ) 2

42 1

0.035 (0.206 )(3600 )(25.3 0.00 )( )1.25 10

0.0525

J m s C CKAt T T s m CQ J

L m

° − °− °= = = ×

2. 3 5(1000 / )(24.5 )(33.8 )(0.127 ) 1.05 10m V kg m m m m kgρ= = = ×

Q = c m ΔT = 1.00 kcalkg °C

⎛ ⎝

⎞ ⎠ (1.05× 105 kg)(4.30°C) = 4.52× 105 kcal

3. (a) Q = L f m = (80 kcal /kg)(1.05× 105 kg) = 8.4 × 106 kcal (b) 8.4× 106 kcal + 4.52× 105 kcal =8.9× 106 kcal

(c) Q = cmΔT = 0.51kcalkg °C)

⎛ ⎝

⎞ ⎠ (1.05× 105 kg)(3.43°C ) =1.8 × 105 kcal

4. (a) Steel contracts less.

(b) Steel Δl = αl ΔT = 6.5× 10−6

°F⎛ ⎝

⎞ ⎠ (50.0 ft)(49.5°F) = 0.0161 ft = 0.193in.

Aluminum Δl = αl ΔT = 1.3× 10−5

°F⎛ ⎝

⎞ ⎠ (50.0 ft)(49.5°F) = 0.0322 ft = 0.386 in.

5. (a) ΔV = βVΔT = 5.33× 10−4

°F⎛ ⎝

⎞ ⎠ (2.50 gal)(54.8°F) = 0.073gal

ΔV = 2.50 gal − 0.073gal = 2.43gal

(b) (0.073gal )$1.97

gal⎛ ⎝ ⎜

⎞ ⎠ ⎟ = $ 0.14



125

Chapter 15 15.1 1. 288 K 2. 259 K 3. 44°C

4. −38°C 5. 532°R 6. 405°R

7. 90 °F 8. −85°F

9. V =′ V T′ T

=(225cm 3)(315K)

275K= 258cm3

10. ′ V =V ′ T T

=(60.3 in 3)(455°R)

615°R= 44.6 in3

11. T =V ′ T

′ V =

(200 ft 3)(95+ 460 )°R250 ft 3 = 444°R or −16°F

12. ′T =′V T

V=

(25.2 L)(51+ 273)K

19.7 L= 414 K or141°C

13. ′ V =V ′ T T

=(325m3)(94 + 273) K

(41+ 273) K= 380 m3

14. ′ V =V ′ T T

=(275in 3)(95+ 460 )°R

(35+ 460 )°R= 308 in 3

15. ′ V =V ′ T T

=(1575L)(15+ 273) K

(45+ 273) K= 1430 L

16. ′ T =′ V T

V=

(600 ft 3)(70 + 460 )°R120 0 ft 3 = 265°R = − 195°F

17. ′ T =′ V T

V=

(10, 250cm3)(25.6 + 273)K14,300cm3 = 214 K = − 59°C

18. ′ T =′ V T

V=

(1530cm 3)(18.0 + 273)K1270cm3 = 351K = 78°C 19. ′ V =

V ′ T T

=(34.5L)(12.0 + 273)K

(49.0 + 273)K= 30.5L

20. ′ T =′ V T

V=

(18.5 ft 3)(11.0 + 460 )°R28.7 ft3 = 304°R = −156°F

21. TC = 5 /9(40.0 − 32) = 4.4°C = 277.4 K ′ T =′ V T

V=

(36.0L)(277.4 K)26.0L

= 384 K

Therefore ΔT = 107K 22. (a) temperature is doubled (b) temperature is tripled (c) volume is doubled (d) they are directly

proportional

23. T =V ′ T

′ V =

(38.0L)(383K)90.0L

= 162 K = −111°C



126

24. ′ V =V ′ T T

=(3.00 L)(285K )

277 K= 3.09 L 25. ′ V =

V ′ T T

=(147m3 )(305K )

292K= 154 L

26. V ' =V ′T

′T=

(500 L)(31.0°C + 273K )

17.0°C + 273K( )= 524 L

27. V ' =V ′T

′T=

(2000L)(9°C + 273K )

21°C + 273K( )= 1920 L

28. V ' =′V ′T′T

=(1400L)(3°C + 273K )

19°C + 273K( )= 1320 L

29. V ' =′V T′T

=(250 L)(35°C + 273K )

5.0°C + 273K( )= 277 L

30. V ' =V ′T

′T=

(500 L)(31.0°C + 273K )

17.0°C + 273K( )= 524 L

15.2

1. V =′ V ′ P P

=(315cm 3)(85.0kPa)

101kPa= 265cm3

2. ′ P =VP

′ V =

(450 L)(750 kPa)700 L

= 482kPa

3. P =′ V ′ P V

=(139 m3)(41.0 kPa)

76.0m3 = 75.0 kPa

4. ′ V =VP

′ P =

(439 in 3)(47.1 psi)38.7 psi

= 534 in3

5. ′ D =D ′ P P

=(1.80kg /m3 )(125kPa)

108 kPa= 2.08kg / m3

6. ′ P =′ D P

D=

(1.85kg / m3)(87.0kPa)1.65kg /m3 = 97.5 kPa

7. 3

3(0.231 / )(51.0 )0.180 /

65.3

D P lb ft psiD lb ft

P psi

′= = =

′

8. ′ V =VP

′ P =

(140 0 in 3)(22.5 psi)18.0 psi

= 1750 in3

9. ′ P =VP

′ V =

(185m3 )(110.0kPa)225m3 = 90.4 kPa

10. ′ V =VP

′ P =

(65.0 L)(185.0kPa)95.0 kPa

= 127 L

11. ′ D =D ′ P P

=(3.75kg / m3)(725kPa)

815kPa= 3.34 kg / m3

12. ′ P =′ D P

D=

(1.45kg / m3)(101kPa)1.75kg / m3 = 83.7kPa

13. ′ P =VP

′ V =

(750 m3 )(601kPa)500 m3 = 902 kPa (absolute) = 801kPa (gauge)



127

14. ′ V =VP

′ P =

(20.0 ft 3)(29.7 psi)34.7 psi

= 17.1 ft 3

15. ′ V =VP

′ P =

(13.0 ft 3)(94.7 psi)64.7 psi

= 19.0 ft3

16. ′ D =D ′ P P

=(6.35kg / m3)(556kPa)

786kPa= 4.49kg / m3

17. ′ P =′ D P

D=

(30.6oz / ft 3)(279.7 psi)48.5oz / ft3 = 176 psi (absolute) = 162 psi (gauge)

18. ′V =VP

′P=

(2.30 m 3 )(137kPa)

141kPa= 2.23m 3

19. ′V =VP

′P=

(2.60 m 3 )(121kPa)

127kPa= 2.48 m 3

20. ′ V =VP

′ P =

(4.5m3 )(147 kPa)114 kPa

= 5.8 m3

21. (a) ′V =VP

′P=

(75.0 in.3 )(1)

2= 37.5 in.3 (b) ′V =

VP′P

=(75.0in.3 )(1)

3= 25.0in.3

(c) ′ V =VP

′ P =

(75.0 in 3)(1)1 /2

= 150 in 3

22. (a) ′ P =VP

′ V =

(1)(300 kPa)2

= 150 kPa (b) ′ P =VP

′ V =

(1)(300 kPa)3

= 100 kPa

(c) ′P =VP

′V=

(1)(300kPa)

1 / 2= 600 kPa

23. ′ V =V ′ T T

=(58.0 L)(341K )

306 K= 64.6 L

′ P =VP

′ V =

(58.0L)(120 kPa)64.6 L

= 108kPa

′ D =D ′ P P

=(4.85kg / m3)(108kPa)

120 kPa= 4.37kg / m3

24. ′ P =VP

′ V =

(42.0L)(320 kPa)51.0L

= 264 kPa

25. ′ V =VP

′ P =

(2.00 L)(33.0kPa)57.0 kPa

= 1.16L

26. (a) Less (b) P =′ V ′ P V

=(2.75cm 3)(48.3kPa)

3.25cm3 = 40.9kPa

27. D =mV

=1.31kg3.00 m3 = 0.437 kg / m3

′ D =D ′ P P

=(0.437kg /m3 )(97.4 kPa)

121kPa= 0.352 kg /m3

28. (a) Increase

(b) ′ D =D ′ P P

=(1.40kg /m3 )(700 kPa)

314 kPa= 3.12kg / m3



128

29. (a) Lower

(b) ′ P =′ D P

D=

(1.35kg / m3)(13.3kPa)1.45kg / m3 = 12.4 kPa

15.3 VPT

=′ V ′ P ′ T

1. V (825 psi)

575°R=

(1550 in 3)(615 psi)525°R

2. (500 in 3)P

500 °R=

(800 in 3)(800 psi)450 °R

V = 1270 in 3 P = 1420 psi

3. (900 m3)(105kPa)

300 K=

′ V (165 kPa)265K

4. (18.0 m3)(112kPa)

T=

(15.0m3 )(135kPa)235K

V’ = 506 m3 T = 234 K = -39 °C

5. (532m3 )(135kPa)

360 K=

(379 m3)(123kPa)′ T

6. (600 in 3)(1500 psi)

525°R=

′ V (1200 psi)550 °R

T = 234 K = -39 °C V’ = 786 in.3

7. (800 m3 )(235kPa)

303K=

(1200 m3 )(215kPa)′ T

8. (1400 L)(135kPa)

327K=

(800 L)(275kPa)′ T

T’ = 416 K = 143 °C T’ = 381 K = 108 °C

9. 200 0 psi500 °R

=′ P

550 °R

10. 200 0 psi500 °R

=′ P

430 °R

11. (5.00 L)(101.32kPa)

273K=

(2.50L) ′ P 673K

′ P = 220 0 psi ′ P = 1720 psi ′ P = 500 kPa (absolute) = 399kPa (gauge)



129

12. (a) (5.00 L)(101.32kPa)

273K=

(2.50L)(202.64 kPa)′ T

(b) (1)(1)273K

=(2)(3)

′ T

′ T = 273K = 0°C ′ T = 1638K = 1365°C

OR (2)(1)273K

=(1)(2)

′ T

′ T = 273K = 0°C

13. (a) (2.00 m3)(350 kPa) = (1.00 m3) ′ P (b) (1)(1)291K

=(2)(2)

′ T

′ P = 700 kPa ′ T = 1164 K = 891°C

(c) (2.00 m3)(2)

291K=

′ V (1)261K

(d) (2.00 m3)(350 kPa)

291K=

′ V (130 0kPa)304 K

′ V = 3.59m3 ′ V = 0.563m3

14. (320 0 mL)(223kPa)

300 K=

′ V (213kPa)338K

15. (7.85L)(101.32 kPa)

297 K=

′ V (60.5kPa)279 K

′ V = 3770 mL ′ V = 12.3L Chapter 15 Review Questions 1. c 2. a

3. c 4. c

5. b 6. b

7. Standard pressure is 101.32 kPa or 14.7 lb/in 2 and standard temperature is 0°C or 32°F . 8. A temperature increase tends to cause a volume increase. A pressure increase tends to cause a volume

decrease. If both temperature and pressure are increased, the volume change is given by the combined Charles’ and Boyle’s Laws.

9. The temperature will increase. 10. Heating a gas increases the kinetic energy of the gas molecules. This causes an increase in pressure

and volume. 11. When gas is compressed, work must be done on it to decrease its volume. This work is transferred

into increased kinetic energy of the molecules of the gas, causing a temperature increase. 12. When the volume is increased, the number of gas molecules striking the surface of the container per

unit area decreases. Thus the pressure exerted by the gas molecules on the container decreases. Chapter 15 Review Problems

1. ′ V =V ′ T T

=(13.5 ft 3)(88.6 + 460 )°R

35.8 + 460 )°R= 14.9 ft 3

2. ′ V =V ′ T T

=(3.45m3 )(98.5+ 273)K

(18.5+ 273)K= 4.40m3

3. ′ T =′ V T

V=

(132 ft 3)(54.5+ 460 )°R115 ft 3 = 591°R = 131°F



130

4. ′ T =′ V T

V=

(44.2L)(38.5 + 273) K45.3L

= 304 K = 30.9°C

5. ′ T =′ V T

V=

(156cm 3)(12.4 + 273) K145cm3 = 307K = 34.1°C

6. ′ P =VP

′ V =

(1)(276 kPa)2

= 138kPa

7. ′ V =VP

′ P =

(35.0 ft 3)(17.5+ 14.7) psi(32.4 + 14.7) psi

= 23.9 ft3

8. ′ D =D ′ P P

=(6.35kg / m3)(355+ 101)kPa

(685 + 101) kPa= 3.68 kg /m3

9. ′ V =VP ′ T

′ P T=

(435in 3 )(1340 + 14.7) psi (45+ 460 )°R(1150+ 14.7) psi (75+ 460 )°R

= 478 in 3

10. ′ T =′ V ′ P TVP

=(1330 m3)(197kPa)(25+ 273) K

(775m3)(344 kPa)= 293K = 19.9°C

11. (a) ′ P =P ′ T T

=(1950 psi )(98 + 460 )°R

(38 + 460 )°R= 2180 psi

(b) ′ T =′ P TP

=(1870 psi )(38 + 460 )°R

1950 psi= 478°R = 18°F

12. ′ P =VP ′ T

′ V T=

(4.50L)(101.32 kPa)(546 K)(2.25L)(273K)

= 405kPa (abs) = 304 kPa (gauge)

13. ′ P =VP ′ T T ′ V

=(5.35L)(101.32kPa)(45.5 + 273) K

(273K)(0.5)(5.35L)= 236 kPa (abs) = 135kPa (gauge)

14. ′ V =PV ′ T T ′ P

(400 0Pa)(1120 L)(77 + 273) K(45 + 273) K (600 0Pa)

= 822 L

15. ′ D =D ′ P P

=(2.50kg /m3 )(525kPa)

(327 + 101) kPa= 3.07kg / m3

Chapter 15 Applied Concepts 1. (a) The temperature of the helium begins to decrease, and the volume of the balloon decreases.

(b) ′ V =V ′ T T

=(1.85 ft 3)(471°R)

526°R= 1.66 ft 3

(c) ′ V =V ′ T T

=(1.85 ft 3)(561°R)

526°R= 1.97 ft 3

(d) Over inflate the balloon in winter; under inflate the balloon in summer.



131

2. (a) ′ P =P ′ T T

=

32 lbin 2

⎛ ⎝

⎞ ⎠ (605°R)

50 0°R= 39 lb / in 2

(b) P =′ P T′ T

=

32 lbin 2

⎛ ⎝

⎞ ⎠ (50 0°R)

605°R= 26 lb / in 2

3. (a) V = πr 2l = π (0.020m)2(0.150m) = 1.88 × 10−4 m3 (b) V = πr 2Δl = π(0.02m)2(0.11m) = 1.38× 10−4 m3

′ P =VP

′ V =

(1.88× 10−4 m3)(101kPa)1.38× 10−4 m3 = 138kPa

(c) ′ P =VP ′ T

′ V T=

(1.88× 10−4 m3)(101kPa)(293K )(1.38× 10−4 m3)(273K)

= 148kPa

4. (a) ′ V =VP

′ P =

(0.030 m3)(22.5× 103 kPa)1.01× 102 kPa

= 6.68 m3

(b) ′ V =VP ′ T

′ P T=

(0.03m3 )(22.5× 103 kPa)(244.5K)(0.285kPa)(293K)

= 1980m3

5. (a) ′ V =VP ′ T

′ P T=

(2.50 L)(1.35× 105 Pa)(245K)(285Pa)(296 K)

= 980L

(b) No, the balloon should not be fully inflated. If it were, the balloon would probably burst on the way up.



132

Chapter 16 16.2 1. T = 1/ f = 1/ 500 Hz = 2.00 × 10−3 = 2.00ms

2. f = 1/T = 1/ 0.550 s = 1.82

3. v = fλ = (400 Hz)(2.00 m) = 8.00 m / s

4. (a) f = v /λ =3.00 × 108 m / s5.00 × 10−7 m

= 6.00 × 1014 Hz

(b) 14 151 / 1/(6.00 10 ) 1.67 10T f Hz s−= = × = ×

5. v = fλ = (3.50 / s)(0.550 m) = 1.93m / s

6. λ = v / f =1.50 m / s0.650 / s

= 2.31m

7. λ = v / f =6.10 m / s

7.50 s= 0.813m

8. (a) T = 3.50 s / 20 = 0.175s (b) f = 1/T = 1/ 0.175s = 5.71Hz

9. f = 30 /2.50s = 12.0 Hz

10. λ = c / f =3.00 × 108 m / s

50.0 × 106 / s= 6.00m

11. f = v / λ =3.00 × 108 m / s

0.25 m= 1.20 × 109 Hz

12. λ = v / f =3.00 × 108 m / s

1014 Hz= 3 × 10−6 m

13. λ = v / f =3.00 × 108 m / s

1014 Hz= 3 × 10−6 m

14. f = 1/T = 1/ 0.0125s = 80.0 / s = 80.0Hz

λ = c / f =2.68 × 106 m / s

80.0 / s= 3.35 × 104 m

15. f =63

8.3 × 10−6 min ×60 s

1min

= 1.27 × 105 / s

λ = v / f =662 m / s

1.27 × 105 /s= 5.21 × 10−3 m

16. (a) v = λf = (3.00 × 10−9 m)(3.00 × 1018 / s) = 9.00 × 109 m / s

(b) T =1f

=1

3.00 × 1018 / s= 3.33 × 10−17 s



133

17. v = λf = (0.750 m)(2.75 × 1010 / s) = 2.06 × 1010 m / s

16.4 1. v = 331m / s + (0.61(m / s) / °C)T = 331m / s + (0.61(m / s) / °C)(10 °C) = 337m / s

2. v = 331m / s + (0.61(m / s) / °C)T = 331m / s + (0.61(m / s) / °C)(35°C) = 352 m / s

3. v = 331m / s + (0.61(m / s) / °C)T = 331m / s + (0.61(m / s) / °C)(−23°C) = 317m / s

4. t =sv

=21.0m

337 m / s= 0.0623s

5. t =sv

=7500 m

150 0m / s= 5.00 s

6. t =sv

=2(15km)

1.50km / s= 20 s

7. s = vt = (150 0 m / s)(1.76s) = 2640m

8. vair = 331m / s + (0.61(m / s) / °C)T = 331m / s + (0.61(m / s) / °C)(23°C) = 345m / s

s =4.00s

1vair

−1

vwater

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=4.00 s

1345m / s

−1

150 0 m / s⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 1790m

9. ′ f = fv

v − vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 550 Hz

345m / s345m / s − 40 m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 622 Hz

10. 345 /

450 409345 / 35 /s

v m sf f Hz Hz

v v m s m s′ = = =

+ +

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

11. ′ f = fv

v − vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 480 Hz

1090 ft / s1090 ft / s − 58.7 ft / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 507 Hz

12. ′ f = fv

v + vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 480 Hz

1090 ft / s1090 ft / s + 58.7 ft / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 455Hz

13. v = 331m /s + 0.61m /s°C

⎛ ⎝ ⎜

⎞ ⎠ ⎟ (−6.0°C ) = 327m / s

′ f = fv

v + vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 660 Hz

327m / s327m / s + 3.61m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 653Hz

14. f = ′ f v − vs

v⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 850 Hz

150 0m / s − 5.00 m / s150 0s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 847Hz

15. t =s

v=

30.0 m

345 m / s= 0.0870 s

16. s = vt = (470 0 m / s)(0.136s) = 639m



134

17. (a) ′ f = fv

v − vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = (605Hz)

345m / s345m / s − 44.7m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 695Hz

(b) 345 /

(605 ) 522345 / 55.0 /s

v m sf f Hz Hz

v v m s m s′ = = =

+ +

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

18. t =s

v=

72.0 m

345 m / s= 0.209 s

19. t =s

v=

30.0 m

345 m / s= 0.0870 s

16.6

1. l =gT 2

4π 2 =(9.80 m / s2 )(1.00s)2

4π 2 ×100cm

1m= 24.8cm

2. l =T 2g

4π 2=

(0.85 s)2 (9.80 m / s2 )

4π 2= 0.18 m

3. 1.25

2 2 2.249.80 /

l mT s

g m sπ π= = =

4. 2

2.002 2 1.57

32.2 /

l ftT s

g ft sπ π= = =

5. l =gT 2

4π 2 =(32.2 ft / s 2)(2.25s)2

4π 2 = 4.13 in.

6. l =gT 2

4π 2 =(9.80 m / s2 )(0.700s)2

4π 2 = 0.122 m

7. T = 2πlg

= 2π1.50 ft

32.2 ft / s2 = 1.36s

8. T = 2πlg

= 2π0.350 m

9.80m / s 2 = 1.19 s

9. Let l = 2 l in T = 2πlg

; thenT = 2π2 lg

= 2 2πlg

,or 2 timesthe original period .

10. Let T = 2T in l =gT 2

4π 2 ; then l =g(2T )2

4π 2 =g(4T 2)

4π 2 =gT 2

π 2 , or 4 times the original length.

11. T = 2πl

g= 2π

0.25 m

9.80 m / s2= 1.0 s



135

12. (a) Longer.

(b) l =T 2g4π 2 =

(2.40s)2 (9.80 m / s2 )4π 2 = 1.43m

13. T = 2πl

g= 2π

0.25 m

9.80 m / s2= 1.0 s

14. l =T 2g

4π 2=

(0.85 s)2 (9.80 m / s2 )

4π 2= 0.18 m

Chapter 16 Review Questions 1. a, b, c 2. c 3. a 4. b

5. b 6. a 7. c, d

8. Interference is the result of two or more waves traveling through the same region at the same time. Diffraction is the bending of a single wave passing near an obstacle with an opening nearly the same size as the wavelength.

9. Two or more waves add together to form a larger wave is constructive interference. The formation of a smaller wave is destructive interference.

10. Sound would not be heard if some obstacle came between the stereo speakers and you and there were no nearby reflecting surfaces.

11. Waves passing through a break in a sea wall. 12. It increases. 13. A wave is a periodic disturbance. A pulse is a one-time-disturbance. 14. A sharp explosion creates a sound pulse. 15. The speed of sound increases. The higher kinetic energy of air molecules at high temperatures leads

to a higher velocity of sound through the air. 16. A seismograph detects slight vibrations of the earth’s surface by detecting the relative motion of the

earth’s surface and a massive object. 17. The speed of sound is higher in water than in air. The higher density of water provides a higher

speed of sound. 18. Motion toward an oncoming sound wave increases the frequency at which the maximum pressure

regions in the sound wave strike an observer, thereby producing a higher-frequency sound. Motion away from an oncoming sound wave decreases the frequency at which the maximum pressure regions strike an observer, thereby producing a lower-frequency sound.

19. Sympathetic vibrations occur when an object vibrates at its natural resonance frequency in response to the vibration of a nearby object at some other frequency. Forced vibration occurs when an object vibrates at the same frequency as another nearby vibrating object.

20. Resonance occurs when an object vibrates at its natural frequency in response to the vibration at the same frequency of a nearby object.

21. The light from these stars is shifted toward the red as a result of their motion away from the earth. 22. Amplitude is maximum displacement. 23. Period is the time required for one full vibration. Frequency is the number of complete vibrations

per unit of time. 24. No; the period of a pendulum is independent of its mass.



136

Chapter 16 Review Problems 1. T = 1/ f = 1/ (355 × 103 / s) = 2.82 × 10−6 s or 2.82 μs

2. f = 1/T = 1/ (0.320 s) = 3.13Hz 3. (a) 8

14

7

3.00 10 // 6.67 10

4.50 10

m sf c Hz

mλ

−= = = ×

×

3. (b) T = 1/ f = 1/ (6.67 × 1014 / s) = 1.50 × 10−15 s

4. v = λf = (0.654 m)(8.97 / s) = 5.87m / s

5. λ = v / f =5.78 m / s4.65/ s

= 1.24 m

6. f =85

1.3s= 65Hz

7. λ = c / f =3.00 × 108 m / s

65.5 × 106 / s= 4.58m

8. v = 331m / s + (0.61(m / s)°C)T = 331m / s + (0.61(m / s)°C )(85°C) = 383m / s

9. v = 331m / s + (0.61(m / s)°C)T = 331m / s + (0.61(m / s)°C )(−35°C) = 310 m / s

10. t = s /v =1450 m

150 0 m / s= 0.967s

11. t = s /v =2(22km)

1.50 km / s= 29 s

12. ′ f = fv

v − vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 525Hz

1090 ft / s1090 ft / s − 139 ft / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 602 Hz

13. (a) 345 /

4950 5360345 / 26.4 /s

v m sf f Hz Hz

v v m s m s′ = = =

− −

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

(b) 345 /

4950 4600345 / 26.4 /s

v m sf f Hz Hz

v v m s m s′ = = =

+ −

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

14. f = ′ f v − vs

v⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 425Hz

338 m / s− 45 m / s338 m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 368Hz

15. f =cλ

=3.00 × 108 m / s5.00 × 10−7 m

= 6.00 × 10 14 Hz

′ f = fv

v + vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = (6.00 × 1014 Hz)

331m / s331m / s+ 24 m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 5.59 × 1014 Hz

16. T = 2πlg

= 2π0.450 m

9.80 m / s2 = 1.35s

17. l =gT 2

4π 2 =(32.2 ft / s 2)(0.700s)2 12 in.

1 ft⎛ ⎝ ⎜

⎞ ⎠ ⎟

4π 2 = 4.80 in.



137


1. (a) l =gT 2

4π 2 =(9.80 m / s2 )(1s)2

4π 2 = 0.248 m

(b) No, since “g” is different, the clock will keep a different time.

(c) l =gT 2

4π 2 =(1.62 m / s2 )(1s)2

4π 2 = 0.0411m

2. (a) 1 wavelength

(b) V = f ⋅ λ = (0.20 Hz)(2800 ft) = 560 ft / s

(c) λ = v / f = 560 ft / s2(0.20Hz)( )= 1400 ft

3. (a) λ = c / f = 3.00 × 108 m / s550 × 103 Hz = 545m

λ = c / f = 3.00 × 108 m / s1650 × 103 Hz = 182 m

AM: 182 m to 545 m

λ = c / f = 3.00 × 108 m / s88 × 106 Hz = 3.41m

8

6

3.00 10 //108 10 2.78

m sc fHz m

λ ×= =× =

FM: 2.78 m to 3.41 m

(b) FM wavelengths are closer to the sizes of openings of tunnels and underpasses. As stated in the text, “wave diffraction is commonly observed only when the opening is nearly the same size as the wavelength.”

4. (a) ( )2 2 5

2 2

10 104 4 (30.4 ) 1.16 10

W WP I A r m W

m mπ π= ⋅ = = = ×⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(b) 5

2 2

1.16 102.50

4 (60.8 )

P W WI

A m mπ

×= = =

(c) I =PA

=P

4π (2r)2 =14

P4πr 2

25 % of the initial intensity.



138

5. v = 331m / s +0.61m / s

°C(23.3°C) = 345m / s

(a) ′ f = fv

v ± vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟

836 Hz = 765Hz345m / s

345m / s − vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟

vs = 29m / s

(b) ′ f = fv

v ± vs

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 765Hz

345Hz345m / s + 29 m / s

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 706 Hz



139

Chapter 17 Chapter 17 17.3

1.

F =

kq1q

2

r 2=

9.00 × 109 N m2

C 2( )−8.00 × 10−5 C( )2

0.250m( )2= 922N

2. q =

Fr 2

k=

8.00N( ) 0.100m( )2

9.00 × 109 Nm2 / C 2= 2.98 × 10−6 C

3. r =

kq1q

2

F=

9.00 × 109 Nm2 / C 2( ) 3.0 × 10−6 C( ) 6.0 × 10−6 C( )940N

= 0.013m = 1.3cm

4. r =

kq1q2

F=

9.00 × 109 Nm2 / C 2( ) 3.0 × 10−8 C( ) 5.0 × 10−7 C( )0.045N

= 0.055m = 5.5cm

5.

q

1=

Fr 2

kg2

=850N( ) 1.2 × 10−3 m( )2

9.00 × 109 Nm2 / C 2( ) 9.0 × 10−6 C( )= 1.5 × 10−8 C

6. r =

kq1q2

F=

9.00 × 109 Nm2 / C 2( ) 6.00 × 10−6 C( )2

25.0N= 0.114m = 11.4cm

7. (a)

F

AB=

kqAq

B

rAB( )2 =

9.00 × 109 Nm2 / C 2( ) 3.00 × 10−6 C( ) 5.50 × 10−6 C( )0.400m( )2

= +0.928N

F

CB=

kqBq

C

rCB( )2

=9.00 × 109 Nm2 / C 2( )5.50 × 10−6 C( ) −4.60 × 10−6 C( )

0.350m( )2= −1.86N

FT= F

AB+ F

CB= +0.928N + 1.86N( )= 2.78N ;1.86N is pos because it is in the pos direction.

(b) FT= F

BA+ F

CA

F

BA=

kqAq

B

rBA( )2

=9.00 × 109 Nm2 / C 2( ) 3.00 × 10−6 C( ) 5.50 × 10−6 C( )

0.400m( )2= +0.928

F

CA=

kqAq

C

rCA( )2

=9.00 × 109 Nm2 / C 2( ) 3.00 × 10−6 C( ) −4.60 × 10−6 C( )

0.750m( )2= −0.221N

FT= F

BA+ F

CA= +0.928N − 0.221N = +0.707N



140

(c) FT= F

AC+ F

BC

F

AC=

kqAq

C

rAC( )2

=9.00 × 109 Nm2 / C 2( ) 3.00 × 10−6 C( ) −4.60 × 10−6 C( )

0.750m( )2= −0.221N

F

BC=

kqBq

C

rBC( )2

=9.00 × 109 Nm2 / C 2( ) 5.50 × 10−6 C( ) −4.60 × 10−6 C( )

0.350m( )2= −1.86N

FT= F

AC+ F

BC= −0.221N + −1.86N( )= −2.08N

8. FT= F

AB+ F

CB

F

AB=

kqAq

B

rAB( )2

=9.00 × 109 Nm2 / C 2( ) 5.00 × 10−6 C( ) 4.50 × 10−6 C( )

0.250m( )2= 3.24N

F

CB=

kqBq

C

rCB( )2

=9.00 × 109 Nm2 / C 2( ) 4.50 × 10−6 C( ) −4.20 × 10−6 C( )

0.400m( )2= −1.06N

FT= F

AB+ F

CB= 3.24N + 1.06N( )= 4.30N ;1.06 is pos because it is directed in the pos direction.

17.4

1. E =

Fq1

=0.600N

4.00 × 10−5C= 1.50 × 104 N / C 2. E =

Fq1

=0.600N

2.00 × 10−8C= 3.00 × 106 N / C

3. E =

Fq1

=2.50 × 10−4 N5.00 × 10−4 C

= 0.500N / C 4. E =Fq1

=3.00 × 10−4 N7.50 × 10−4 C

= 0.400N / C

5. q1 =

FE

=8.00 × 10−4 N0.450N / C

= 1.78 × 10−3C 6. q1 =FE

=6.20 × 10−4 N0.370N / C

= 1.68 × 10−3C

7. F = Eq1 = 1.75 × 104 N / C( ) 3.68 × 10−5C( )= 0.676N

8. F = Eq1 = 3.00 × 106 N / C( ) 4.00 × 10−5C( )= 120N

17.5

1. R =

plA

=2.83 × 10−6 Ωcm( ) 7800cm( )

2.07 × 10−2 cm2 = 1.07Ω 2.

R = pl = 0.0262Ω / ft( ) 315 ft( )= 8.25Ω

3. ρ =

Rl

=9.57Ω

580 ft= 0.0165Ω / ft 4. ′R =

′l Rl

=500 ft( )0.651Ω( )

100 ft= 3.26Ω

5. R =

plA

=1.72 × 10−6 Ωcm( ) 47,500cm( )

2.07 × 10−2 cm2 = 3.95Ω 6. R =plA

=1.72 × 10−6 Ωcm( )10,000cm( )

2.07 × 10−2 cm2 = 0.831Ω



141

7. R =

plA

=2.83 × 10−6 Ωcm( ) 5000cm( )

2.07 × 10−2 cm2 = 0.684Ω 8. t =Rρ

=3.00Ω

0.0262Ω / ft= 115 ft

9. A =

ρlR

=1.72 × 10−6 Ωcm( ) 6000cm( )

0.788Ω= 0.0131cm2 10. t =

Rρ

=5.62Ω

0.0262Ω / ft= 215 ft

17.6

1. I =

VR

=115V24.0Ω

= 4.79A 2. R =

VI

=12.0V2.50A

= 4.80Ω 3. V = IR = 14.0A( )15.7Ω = 220V

4. R =

VI

=220V15.0A

= 14.7Ω 5. R =

VI

=115V

0.750A= 153Ω 6. I =

VR

=115V75.0Ω

= 1.53A

7. R =

VI

=220V12.5A

= 17.6Ω 8. I =

VR

=115V50.0Ω

= 2.30A 9. I =VR

=220V175Ω

= 1.26A

10. R =

VI

=115V3.50A

= 32.9Ω 11. (a) I =VR

=10V

150Ω= 0.067 A

(b) V = ′I R = 3I( )R = 3 V / R( )R = 3V = 3 10V( )= 30V (c) I =

VR

=10V75Ω

= 0.13A

12. (a) R =

VI

=32.0V4.25A

= 7.53Ω (b) R =

VI

=32.0V8.50A

= 3.76Ω 13. R =VI

=120V6.40A

= 18.8Ω

14. R =

VI

=115V2.50A

= 46.0Ω 15. I =VR

=120V65.0Ω

= 1.85A 16. I =VR

=120V20.0Ω

= 6.00A

17.7

1. R = R1 + R2 + R3 = 20.0Ω + 5.00Ω + 65.0Ω = 13.50Ω 2. I =VR

=24.0V

13.50Ω= 1.78A

3. R = R1 + R2 + R3 = 15.0Ω + 21.0Ω + 24.0Ω = 60.0Ω 4. I =VR

=12.0V60.0Ω

= 0.200A

5. I =

VR

=18.0V

15.0Ω + 9.00Ω= 0.750A 6. V1 = IR1 = 0.750A( ) 15.0Ω( )= 11.3V

7. E = I R1 + R2 + R3 + R4( )= 7.00A( ) 10.0 + 17.0 + 4.00 + 23.0( )Ω = 378V

8. V = IR3 = 7.00A( ) 4.00Ω( )= 28.0V 9. R1 =

V1

I=

115V5.00A

= 23.0Ω

10. R3 = R − R1 − R2 = 23.0Ω − 7.00Ω − 3.00Ω = 13.0Ω

11. R1 =

V1

I=

27.0V2.50A

= 10.8Ω R2 =V2

I=

35.5V2.50A

= 14.2Ω R3 =V3

I=

47.5V2.50A

= 19.0Ω

12. V1 = IR1 = 3.75A( ) 8.75Ω( )= 32.8V R2 =

V2

I=

75.0V3.75A

= 20.0Ω

V3 = IR3 = 3.75A( ) 3.25Ω( )= 12.2V



142

13. R1 =

V1

1=

13.5V4.85A

= 2.78Ω V2 = IR2 = 4.85A( ) 8.25Ω( )= 40.0V

R3 =

V3

I=

16.5V4.85A

= 3.40Ω

17.8

1. (a)

1R

=1R1

+1R2

=1

11.0Ω+

17.00Ω

R = 4.28Ω

(b) I =

ER

=50.0V4.28Ω

= 11.7 A

(c) I1 =

V1

R1

=50.0V11.0Ω

= 4.55A

(d) I2 =

V2

R2

=50.0V7.00Ω

= 7.14A

2. (a) I2 =

V2

R2

=10.0V5.00Ω

= 2.00A (b) I3 =V3

R3

=10.0V8.00Ω

= 1.25A

(c) I1 =

V1

R1

=10.0V2.00Ω

= 5.00A (d) I = I1 + I2 + I3 = 2.00A + 1.25A + 5.00A = 8.25A

(e)

1R

=1R1

+1R2

+1R3

=1

2.00Ω+

15.00Ω

+1

8.00Ω

R = 1.21Ω

3. (a) R =

EI

=25.0V4.00A

= 6.25Ω

1R3

=1R

−1R1

−1R2

=1

6.25Ω−

115.0Ω

−1

12.0Ω

R3 = 100Ω

(b) I1 =

V1

R1

=25.0V15.0Ω

= 1.67 A

(c) I3 =

V3

R3

=25.0V100Ω

= 0.250A



143

4. (a)

1R

=1R1

+1R2

+1R3

+1R4

=1

13.0Ω+

17.00Ω

+1

15.0Ω+

121.0Ω

R = 2.99Ω (b)

E = IR = 10.0A( ) 2.99Ω( )= 29.9V

(c) 29.9 V

(d) I1 =

V1

R1

=29.9V13.0Ω

= 2.30A I2 =V2

R2

=29.9V7.00Ω

= 4.27 A I3 =V3

R3

=29.9V15.0Ω

= 1.99A

I4 =

V4

R4

=29.9V21.0Ω

= 1.42A

17.9

1. (a) R2 and R3 (b)

1R

=1R2

+1R3

=1

4.00Ω+

112.0Ω

R = 3.00Ω

2. R = R1 + Req = 6.00Ω + 3.00Ω = 9.00Ω

3. I =

ER1

=80.0V9.00Ω

= 8.89A

4. V1 = I1R1 = 8.89A( ) 6.00Ω( )= 53.3V

5. (a) I3 =

V3

R3

=80.0V − 53.3V

12.0Ω= 2.23A (b) I2 =

V2

R2

=80.0V − 53.3V

4.00Ω= 6.68A

6.

1Req

=1

R2 + R3

+1R4

=1

10.0Ω + 4.00Ω+

120.0Ω

Req = 8.24Ω

7. R = R1 + Req + R5 = 5.00Ω + 8.24Ω + 8.00Ω = 21.24Ω

8. I1 =

ER

=115V

21.24Ω= 5.41A

9. V = E − I1R1 − I5R5 = 115V − 5.41A( ) 5.00Ω( )− 5.41A( ) 8.00Ω( )= 45V

10. I1 =

V2,3

R2,3

=45V

14.0Ω= 3.2A

11. 11. I5 =

ER

=115V

21.24Ω= 5.41A

12. V3 = I3R3 = 3.2A( ) 4.00Ω( )= 13V

13.

1Req

=1

R2 + R3

+1R4

=1

24.0Ω + 6.00Ω+

115.0Ω

Req = 10.0Ω



144

14.

1Req

=1R5

+1R6

=1

3.00Ω+

110.0Ω

Req = 2.31Ω

15. R = 5.00Ω + 10.0Ω + 2.31Ω = 17.3Ω

16. E = IR = 3.00A( ) 17.3Ω( )= 51.9V

17. V = IR = 3.00A( ) 10.0Ω( )= 30.0V

18. V4 = IR = 3.00A( ) 10.0Ω( )= 30.0V

19. V6 = IR = 3.00A( ) 2.31Ω( )= 6.93V

20. I6 =

VR

=6.93V10.0Ω

= 0.693A

21.

1Req

=1R2

+1R3

=1

20.0Ω+

15.00Ω

Req = 4.00Ω

1′Req

=1R1

+1

Req + R4

=1

5.00Ω+

14.00Ω + 6.00Ω

′Req = 3.33Ω

R = ′Req = R5 = 3.33Ω + 10.0Ω = 13.3Ω

22. I5 =

ER

=80.0V13.3Ω

= 6.02A

23. V5 = I5R5 = 6.02A( ) 10.0Ω( )= 60.2V

24. I1 =

E − V5

R1

=80.0V − 60.2V

5.00Ω= 3.96A

I4 = I5 − I1 = 6.02A − 3.96A = 2.06A

V4 = I4 R4 2.06A( ) 6.00Ω( )= 12.4V

25. I2 =

V2

R2

=80.0V − 60.2V − 12.4V

20.0Ω= 0.370A

17.10 1. V I R Batt 12.0 V 9.00A 1.33 Ω

R1 12.0 V 6.00A 2.00 Ω

R2 12.0 V 3.00A 4.00 Ω



145

2. V I R Batt 36.0 V 2.00 A 18.0 Ω

R1 8.00 V 2.00 A 4.00 Ω

R2 12.0 V 2.00 A 6.00 Ω

R3 16.0 V 2.00 A 8.00 Ω 3. V I R Batt 36.0 V 6.00 A 6.00 Ω

R1 36.0 V 2.00 A 18.0 Ω

R2 36.0 V 3.00 A 12.0 Ω

R3 36.0 V 1.00 A 36.0 Ω 4. V I R Batt 12.0 V 2.00 A 6.00 Ω

R1 7.20 V 1.20 A 6.00 Ω

R2 4.80 V 1.20 A 4.00 Ω

R3 12.0 V 0.800 A 15.0 Ω 5. V I R Batt 50.0 V 5.00 A 10.0 Ω

R1 25.0 V 2.00 A 12.5 Ω

R2 25.0 V 2.00 A 12.5 Ω

R3 10.0 V 3.00 A 3.33 Ω

R4 40.0 V 3.00 A 13.3 Ω 6. V I R Batt 24.0 V 6.00 A 4.00 Ω

R1 8.00 V 6.00 A 1.33 Ω

R2 16.0 V 4.00 A 4.00 Ω

R3 16.0 V 2.00 A 8.00 Ω 7. V I R Batt 36.0 V 9.00 A 4.00 Ω

R1 12.0 V 6.00 A 2.00 Ω 2R 12.0 V 3.00 A 4.00 Ω 3R 24.0 V 6.00 A 4.00 Ω 4R 24.0 V 3.00 A 8.00 Ω



146

8. V I R Batt 30.0 V 5.00 A 6.00 Ω 1R 6.00 V 3.00 A 2.00 Ω 2R 6.00 V 2.00 A 3.00 Ω 3R 15.0 V 5.00 A 3.00 Ω 4R 8.00 V 1.00 A 8.00 Ω 5R 8.00 V 4.00 A 2.00 Ω 6R 1.00 V 5.00 A 0.02 Ω 9. V I R Batt 80.0 V 12.0 A 6.67 Ω 1R 18.0 V 4.00 A 4.50 Ω 2R 18.0 V 2.00 A 9.00 Ω 3R 18.0 V 6.00 A 3.00 Ω 4R 48.0 V 12.0 A 4.00 Ω 5R 8.00 V 4.00 A 2.00 Ω 6R 8.00 V 8.00 A 1.00 Ω 7R 6.00 V 12.0 A 0.500 Ω 10. V I R Batt 46.0 V 10.0 A 4.60 Ω 1R 18.0 V 3.00 A 6.00 Ω 2R 18.0 V 4.00 A 4.50 Ω 3R 18.0 V 3.00 A 6.00 Ω 4R 28.0 V 3.00 A 9.30 Ω 5R 28.0 V 7.00 A 4.00 Ω 11. V I R Batt 65.0 V 5.00 A 13.0 Ω 1R 10.0 V 0.500 A 20.0 Ω 2R 10.0 V 1.00 A 10.0 Ω 3R 10.0 V 2.50 A 4.00 Ω 4R 10.0 V 1.00 A 10.0 Ω 5R 25.0 V 5.00 A 5.00 Ω 6R 30.0 V 5.00 A 6.00 Ω 17.12 1. ( ) ( )1 2 1.50 0.250 0.0450 1.49E E E V A V= − = − Ω =

2. ( ) ( )1 2 11.8 0.500 0.150 11.9E E E V A V= + = + Ω =

3. 12.0 11.6 1.330.300

E V VI AR− −

= = =Ω



147

4. (a) 0.850I A= (b) 1 2 3 1.50 1.50 1.50 4.50E E E E V V V V= + + = + + = (c) 1 2 3 0.0500 0.1500 0.0500 0.0500 0.1500r r r r= + + = Ω + Ω + Ω + Ω = Ω

5. (a) 1 2 3 4 5 0.750 0.750 0.750 0.750 0.750 3.75I I I I I I A A A A A A= + + + + = + + + + =

(b) E = VIt = 12.0v( ) 205A( ) 2.50s( )= 6150

Jc

∗Cs

∗ s = 6150J

(c) 0.100 0.02005

onerR

numberΩ

= = = Ω

6. 1 2

1

1.50 1.50 0.4057.50

E E V VI AR+ +

= = =Ω

7. 1 6.00 0.16037.5

E VI AR

= = =Ω

8. 11

9.00 6.00 1.501.20

E VR RI A

= − = − Ω = Ω

9. ( )

1 2

2 1.500.120

10.0 15.0VE EI A

R R+

= = =+ Ω + Ω

12. 1 2 10.0 15.0 25.0R R R= + = Ω + Ω = Ω 17.13 1. ( ) ( )110 8.70 957P VI V A W= = =

2. ( ) ( )22 3.50 6.70 82.1P I R A W= = Ω =

3. 75.0 0.682110

P WI AV V

= = =

4. 110 1610.682

V VRI A

= = = Ω

5. 750 6.82110

P WI AV V

= = =

6. ( ) ( ) ( )1cos 750 40.0 $0.07 / $2.101000

cents kWt Pt W h kWhkWh W

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

7. ( ) ( ) ( ) ( )1cos 6 50.0 25.0 $0.075 / $0.561000


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

8. ( ) ( )

cos $3.84 $0.0533.00 24.0

cents tkWh Pt kWh h

= = =

9. 1000 1200 1100 30.0 ;110

P W W WI A YesV V

+ += = =

10. ( ) ( ) ( )

cos $0.50 6.67/ 1.00 $0.075 /

tt hcents kWh kWh kWh

= = =

11. ( ) ( ) ( ) ( ) 1cos 1.50 110 2.00 $0.08 / $0.0261000

cents kWt Pt A V h kWhkWh W

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



148

12. ( ) ( ) ( ) ( ) 1cos 2.50 110 3.00 $0.07 / $0.05781000


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

13. ( ) ( ) ( ) ( ) 1cos 3.00 110 2.00 $0.07 / $0.0461000


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

14. 60 0.55110

P WI AV V

= = =

15. (a) microwave, 2 light bulbs, 2 florescent bulbs (b) Projection TV, personal computer, and 2 light bulbs.

16. (a)microwave, hand drill, personal computer, 12” TV, 3 light bulbs, 3 florescent bulbs. (b) Two projection TV’s, 5 12” TV’s.

17. ( ) ( )5.90 0.400 2.36P VI V A W= = =

18. ( ) ( )4.40 1.10 4.84P VI V A W= = =

19. ( ) ( )120 0.500 60.0P VI V A W= = =

20. ( ) ( )12.0 2.00 24.0P VI V A W= = =

21. ( ) ( ) ( )6.00 0.500 300 900W Pt VIt V A s J= = = =

22. ( ) ( ) ( )12.0 210 10.0 25,200 25.2W Pt VIt V A s J or kJ= = = =

23. (a) ( ) ( )120 2.00 240P VI V A W= = = (b) ( ) ( ) ( )0.240 7.00 / 30 50.4W Pt kW h day days kWh= = =

(c) ( ) ( )cos 50.4 &0.11/t kWh kWh=

24. (a) ( ) ( )120 15.0 1800 1.80P VI V A W kW= = = =

(b) ( ) ( ) ( )1.80 5.00 / 30 270W Pt kW h day days kWh= = =

(c) ( ) ( )cos 270 $0.11/ $29.70t kWh kWh= =

25. (a) 3115 9.58 10 9.5812,000

V VI Aor mAR

−= = = ×Ω

(b) ( ) ( )3115 9.58 10 1.10P VI V A W−= = × =

(c) ( ) ( ) ( ) ( )3cos 1.10 10 $0.09 / 24 / 30 $0.07t kW kWh h day days−= × =

26. E = VIt = 12.0v( ) 230 A( ) 2.00s( )= 5520

Jc

∗Cs

∗ s = 5520J

27. E = VIt = 12.0v( ) 205A( ) 2.50s( )= 6150

Jc

∗Cs

∗ s = 6150J



149

Chapter 17 Review Questions 1. b 2. c 3. c 4. b 5. e 6. a and e

7. a 8. c 9. b 10. c 11. c 12. b and c

13. Materials can become charges when a charged object is brought nearby, inducing a polarization (separation) of a charge on the material. If one side of the material is touched by another object, the charge at the one side of the material can then be “drained” off, leaving the charge at the other side of the material.

14. Protons, electrons, and neutrons. 15. Protons and neutrons. 16. Electrons are located in the charge clouds surrounding the nucleus. 17. Positive and negative. Protons carry a positive charge; electrons carry a negative charge. 18. A charged object is brought into contact with the electroscope, thereby providing some of that charge

to the electroscope. 19. A charged object is brought near the conducting ball on an electroscope causing a polarization

(separation) of charge on the electroscope. The conducting ball is touched by a “ground”, allowing one type of charge to leave the electroscope and go to the ground, leaving the other charge behind.

20. Coulomb’s Law states that the force between two charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the separation between the two charges

21. An electric field at a point represents the magnitude and direction of the force that would be exerted on a single unit of charge if placed at that point.

22. Lightning is the discharge of built-up static charge on a portion of cloud. 23. Current 24. (a) ampere (b) volt (c) ohm 25. It decreases the resistance by a factor of 4. 26. The voltage drop across a segment of a circuit equals the product of the current through that segment

and the resistance of that segment of the circuit. 27. The current flows through each portion of a series circuit. The current is divided among different

segments of a parallel circuit. 28. The equivalent resistance in a series circuit is the sum of the resistance in the circuit. In a parallel

circuit, the equivalent resistance is given by the reciprocal of the sum of all resistances. 29. The highest range. 30. Water flow is split and flows in parallel through different segments of a water distribution system. In a

similar manner, current is divided and flows parallel through resistors connected in parallel. 31. The current decreases by a factor of 2. 32. The current increases by a factor of 2. 33. The resistance increases by a factor of 2. 34. Electrical charges move from regions of higher potential to regions of lower potential. Chemical

reactions in batteries raise charges to higher potential energy. These charges can flow through a circuit and do work on circuit elements (create heat, light, or motion).

35. Chemical energy is transferred into electrical potential energy in the dry cell. Charges flow through the two lamps, giving up their energy in the form of light and heat, whereas secondary cells are.

36. Primary cells are not rechargeable. 37. An electric current flows in the reverse direction through a secondary cell, causing the normal

chemical reaction to proceed in reverse, thus storing charge in the battery. 38. The electrolyte causes a chemical reaction at the plates and conducts current between the plates. 39. An electrolyte causing a chemical reaction at the plates that releases energy to force electrical charge to

move through the battery and outside the circuit. 40. It decreases the voltage outside the circuit. 41. watt 42. Power is the product of the voltage and the current.



150

43. We pay for our energy use. Energy used is the total work done. Power is the instantaneous use of electricity; energy is the power times the time that the power is used.

44. Power is the square of the voltage divided by the resistance. 45. Power increases by a factor of 2. 46. The power increases by a factor of 2. 47. The power decreases by a factor of 4. 48. The cost increases by a factor of 2. Chapter 17 Review Problems

1. ( ) ( )

( )

6 629 41 2

2 2 23

4.50 10 4.50 109.00 10 8.10 10

1.50 10

c cq q NmF k Nr c m

μ μ− −

−

− × − ×= = × = ×

×

2. q =

Fr 2

k=

0.750N( ) 0.100m( )2

9.00 × 109 Nm2 / c2 = 9.13× 10−7 c

3. ( ) ( ) ( )9 2 2 8 7

1 29.00 10 / 2.50 10 5.00 10

0.06710.0250

m c c ckq qr m

F N

− −× × ×= = =

4. 56

0.500 2.00 10 /2.50 10

F NE N Cq C−= = = ×

′ ×

5. 5. 68

0.0500 3.33 101.50 10

F NE Cq c−= = = ×

′ ×

6. 4 52.50 10 / 4.25 10 1.06F Eq N C C N−′= = × × =

7. ( ) ( )6

2 2

2.83 10 85501.17

2.07 10

cm cmlRA cmρ

−

−

× Ω= = = Ω

×

8. ( ) ( )0.743 560

3.47120

ftR lRl ft

Ω′= = = Ω

′

9. ( ) ( )6

2 2

1.72 10 13, 4001.11

2.07 10

cm cmlRA cmρ

−

−

× Ω= = = Ω

×

10. 3.97 1450.0273 /

Rl ftftρ

Ω= = =

Ω

11. ( ) ( )6

21.79 10 5540

0.01050.943

m cmlA cmRρ

−× Ω= = =

Ω

12. 115 7.4715.4

V VI AR

= = =Ω

13. 220 25.18.75

V VRI A

= = = Ω

14. 115 0.491243

V VI AR

= = =Ω

15. 1 2 3 4 3.40 6.54 8.32 1.34 19.60R R R R R= + + + = Ω + Ω + Ω + Ω = Ω

16. 12.0 0.61219.60

V VI AR

= = =Ω

17. ( ) ( ) ( )1 2 3 4 1.34 11.5 16.5 2.43 5.76 48.5E I R R R R A V= + + + = Ω + Ω + Ω + Ω =

18. 24.5 16.21.51

E VRI A

= = = Ω



151

19. 3 1 2 16.2 11.3 1.45 3.5R R R R= − − = Ω − Ω − Ω = Ω

20. 1 2

1 1 1 1 15.65 4.32eqR R R

= + = +Ω Ω

2.45eqR = Ω

21. 110 44.92.45

E VI AR

= = =Ω

22. 11

1

110 19.55.65

V VI AR

= = =Ω

23. 22

2

110 25.54.32

V VI AR

= = =Ω

24. 1 2

1 1 1 1 14.32 6.35R R R

= + = +Ω Ω

2.57R = Ω 3 2.57 2.43 5.00eqR R R= + = Ω + Ω = Ω

25. 350.0 10.05.00

E VI AR

= = =Ω

26. 11

1

50.0 24.3 5.954.32

V V VI AR

−= = =

Ω

2 1 10.0 5.95 4.05I I I A A A= − = − =

27. 1 2

1 1 1 1 14.35 3.75R R R

= + = +Ω Ω

2.01R = Ω

1 4 3

1 1 1 1 12.01 0.765 1.56R R R R

= + = +′ + Ω + ΑΩ Ω

1.00R′ = Ω 5 1.00 3.42 4.42eqR R R′= + = Ω + Ω = Ω

28. 575.5 17.14.42

E VI AR

= = =Ω

29. ( ) ( )5 5 5 17.1 3.42 58.5V I R A= = Ω = Ω 30. 3 5 75.5 58.5 17.0V E V V V V= − = − =

4 317.017.1 6.201.56

VI I I A A= − = − =Ω

( ) ( )4 4 4 6.20 0.765 4.74V I R A V= = Ω = 1 3 4 17.0 4.74 12.3V V V V V V= − = − =

11

1

12.3 2.834.35

V VI AR

= = =Ω

31. ( ) ( )1 1 1 2.83 4.35 12.3V I R A V= = Ω =



152

32. V I R Batt 35.0V 4.70A 7.45 Ω 1R 5.00V 2.75A 1.82 Ω 2R 5.00V 1.95A 2.56 Ω 3R 13.2V 4.70A 2.80 Ω 4R 7.50V 0.97A 7.73 Ω 5R 7.50V 3.73A 2.01 Ω 6R 9.30V 4.70A 1.98 Ω 33. ( ) ( )1.44 0.135 0.0550 1.43V E Ir V A V= − = − Ω =

34. ( ) ( )12.0 0.858 0.245 12.2E V Ir V A V= + = + Ω =

35. (a) ( )1 2 3 5 6 6 0.658 3.95I I I I I I A A= + + + + = = (b) 6.00E V=

(c) 0.0987 0.01656

onerr

nΩ

= = = Ω

36. 1 2

1 2

4.53 5.62 0.52810.5 8.73

E E V VI AR R

+ += = =

+ Ω + Ω

37. 9.00 6.671.35

V VRI A

= = = Ω

38. ( ) ( )22 2.45 6.55 39.3P I R A W= = Ω =

39. 150 1.36110

P WI AV V

= = =

40. ( ) ( ) ( ) 1cos 150 135 $0.05 $1.011000


⎛ ⎞= = × =⎜ ⎟⎝ ⎠

41. ( ) ( ) ( )

cos $0.45 1000 2400.40 110 $0.043/ 1

t Wt hcents A V kWh kWPkWh

= = × =⎛ ⎞⎜ ⎟⎝ ⎠

42. 100 0.91110

P WI AV V

= = =


1. (a) ( ) ( )

( )

19 1929 81 2

2 2 211

1.60 10 1.60 109.00 10 8.23 10

5.29 10

C Cq q NmF k Nr C m

− −−

−

− × − ×= = × = − ×

×

(b) ( ) ( )

( )

31 27211 471 2

2 2 211

9.11 10 1.67 106.67 10 3.63 10

5.29 10

kg kgm m NmF G Nr kg m

− −− −

−

× ×= = × = ×

×

(c) We would need an extremely large amount of charge on our bodies to feel that attraction. The earth has so much mass that we are noticeably attracted to the earth via gravity.



153

2. (a) 5 23.12 10 9.80 /wF mg kg m s−= = × = 43.06 10 N−×

(b) 43.06 10F Fw N−= = ×

(c) ( )242

92 2

9 812

3.06 10 0.1038.38 10

9.00 10 4.31 10

N mFrq ckq Nm C

C

−−

−

×= = = ×

× − ×

(d) As the distance between the particles decreases, the force increases, which results in a stronger electric field in the sawdust.

3. (a) 45

110 2.20 105.00 10

V VI AR

−= = = ×× Ω

(b) 110 1.10100

V VI AR

= = =Ω

4. (a) 1000 8.33120microwave

P WI AV V

= = =

40.0 0.333120light

P WI AV V

= = =

550 4.58120computer

P WI AV V

= = =

(b) 120 14.48.33microwave

V VRI A

= = = Ω

120 3600.333light

V VRI A

= = = Ω

120 26.24.58computer

V VRI A

= = = Ω

5. ( )2 110

17.3700

VVRP W

= = = Ω

2l l lR A rA R Rρ ρ ρπ= = =

6

41.00 10 6.00 3.32 1017.3 3.14

l m mr mRρπ

−−× Ω

= = = ×Ω



154

Chapter 18 18.2

1. B =μο I2πR

=(4π × 10−7 T m / A)(15.0 A)

2π (0.250m)= 1.20 × 10−5 T

2. B =μο I2πR

=(4π × 10−7 T m / A)(7.50A)

2π (0.500m)= 3.00 × 10−6 T

3. I =2πRBμο

=2π(2.00 m)(5.75 × 10−6 T )

4π × 10−7 T m / A= 57.5 A

4. (a) B =μο I2πR

=(4π × 10−7 T m / A)(675 A)

2π(5.00 m)= 2.70 × 105 T

(b) tan Θ =B

BE=

2.70 × 10−5 T5.20 × 10−5 T

Θ = 27.4°

5. B =μο I2πR

=(4π × 10−7 T m / A)(3.00 A)

2π(0.350 m)= 1.71 × 10−6 T

6. I =2πRB

μο=

2π(2.50 m)(3.50 × 10−6 T )4π × 10−7 T m / A

= 43.8 A

7. B = μο In =(4π × 10−7 T m / A)(5.00 A)(1000 )

0.320m= 0.0196T

8. I =B

μο n=

(0.100T )(0.350 m)(4π × 10−7 T m / A)(3000 )

= 9.28 A

9. I =B

μο n=

(1.25 × 10−3 T )(0.150m)(4π × 10−7 T m / A)(600 )

= 0.249 A

10. I =B

μο n=

(0.100 T )(0.200 m)(4π × 10−7 T m / A)(250 0)

= 6.37 A

11. B = μο In = (4π × 10−7 T m / A)(2.50 A)(100 0 turns / 0.250m) = 0.0126T

12. I =B

μο n=

1.30 × 10−3 T(4π × 10−7 T m / A)(100 0 turns / 0.100m)

= 0.103 A

13. I =B

μοn=

1.50x10−3 T

(4π × 10−7 T m / A)(750 turns / 0.150 m)= 0.239A

14. I =B

μοn=

1.00T

(4π × 10−7 T m / A)(2500 turns / 0.150 m)= 4.77A



155


1. d

2. b

3. b

4. tesla

5. A tightly wound solenoid with turns per unit length carrying a large current produces a high magnetic field. A magnetic core such as iron significantly increases the magnetic field.

6. Use Ampere’s rule to find the direction of the flux line from any single turn in the solenoid. The magnetic field direction of the solenoid is in this direction.

7. The current-carrying coil produces a magnetic field that causes the magnetic domains in the magnetic material to align in the direction of the coil’s field. This produces a stronger induced field in the core.

8. A moving magnet induces a current in the generator’s coil.

9. The commutator is a split ring that allowed the current produced by a generator always to flow in the same direction.

10. An induced magnetic field in the motor’s electromagnet is repelled by the permanent magnet, causing the rotor to spin. The commutator in a dc motor allows the current through the electromagnet to change polarity, causing the rotor to continue spinning.

11. A synchronous motor rotates at a frequency that depends on the number of coils and the frequency of the ac power source. The motor has a number of poles along the stator that cause the rotor to spin at a fixed frequency.

12. A universal motor can operate on either dc or ac. The induction motor can operate only on ac.

13. The stator is a static magnet. The armature is an electromagnet that is free to rotate.

14. An electromagnet produces a strong magnetic field when a current is run through a solenoid. This magnetic field in turn induces a stronger magnetic field in a magnetic core.

15. The magnetic field increases by a factor of 2.

16. The field does not change as long as the length of the solenoid is much greater than the original diameter.

17. The magnetic field increases by a factor of 4.

18. The flux lines can be found by placing a small compass or magnetic filings near the magnet.

19. A spinning armature in the field of a stator crosses the magnetic flux lines of the stator, thereby inducing a current to flow in the armature coil. As the armature rotates and is reversed in the field, the direction of the current in the coil is reversed.


1. B =μο I2πR

=(4π × 10−7 T m / A)(1.38 A)

2π (0.255m)= 1.08 × 10−6 T

2. B =μο I2πR

=(4π × 10−7 T m / A)(8.95 A)

2π (0.365m)= 4.90 × 10−6 T

3. I =2πRBμο

=2π(1.75m)(4.75 × 10−6 T )

4π × 10−7 T m / A= 41.6 A



156

4. B = μο In =(4π × 10−7 T m / A)(4.55 A)(200 0)

0.452m= 0.0253T

5. I =B

μο n=

(0.235T )(0.182m)(4π × 10−7 T m / A)(2750)

12.4 A

6. B =μο I2πR

=(4π × 10−7 T m / A)(500 A)

2π (7.00m)= 1.43 × 10−5 T tan Θ =

BBE

=1.43 × 10−5 T5.20 × 10−5 T

Θ = 15.4°


1. (a) B =μο I2πR

=4π × 10−7 Tm / A 8.25 A

2π 0.0835m= 1.98 × 10−5 T

(b) 7

7

4 10 / 8.253.17

2 2 5.20 10

I Tm A AR m

B Tομ π

π π

−

−

×= = =

×

2. (a) 6.00

0.40015.0

v VI A

R= = =

Ω

B =μο I2π R

=4π × 10−7 T m / A 0.400 A

2π 0.0103m= 7.77 × 10−6 T

(b) 5.20 × 10−5 T7.77 × 10−6 T

= 6.69 TimesStronger

The wire’s magnetic field is 14.9% of the strength of the earth’s magnetic field. 3. The magnetic field of the inner cable cancels out the opposite magnetic field orientation of the outer

braid. 4. (a) For both solenoids, N is on top and S is on the bottom. (b) “b”

5. (a) 8

2 4 2

1.72 10 20.00.525

(4.57 10 )

m mR

A r m

ρ ρ

π π

−

−

× Ω= = = = Ω

×

l l

(b) I =VR

=4.50 v

0.525Π= 8.57 A

(c) C = 2πr = 2π 0.0325m = 0.204 m

number of loops =

lC

=20.0 m

0.204 m= 98.0 loops

(d) number of loops x “diameter of wire” = (98 loops) ( )4 24.57 10 2 8.96 10m m− −× = ×

(e) ( ) ( )( )7 34 10 / 8.57 98.0 1.05 10B I n T m A A loops Tομ π − −= = × = ×



157

Chapter 19 19.1

1. Emax =e

sinθ=

27.0Vsin 35.0°

= 47.1V

2. Emax =e

sinθ=

82.0Vsin 65.0°

= 90.5°

3. max max sin (165 )(sin 45.0 ) 117e E V Vθ= = ° =

4. i = Imax sinθ = (8.00A)(sin 60.0°) = 6.93A

5. Imax =i

sinθ=

6.50 Asin 45.0°

= 9.19 A

6. Emax =e

sinθ=

14.5Vsin 51.0°

= 18.7V

7. e = Emax sinθ = (145V)(sin 35.0°) = 83.2V

8. i = Imax sinθ = (5.75A)(sin 80.0°) = 5.66 A

9. Imax =i

sinθ=

4.00 Asin 45.0°

= 5.66 A

10. Emax =e

sinθ=

450 Vsin 55.0°

= 549V

11. max

4.32sin ; 70.3

4.59

i A

I Aθ θ= = = °

12. max

1.23sin ; 21.1

3.41

i A

I Aθ θ= = = °

13. E = 0.707 Emax = (0.707)(2250V) = 1590V

14. Imax = 1.41 I = (1.41)(6.00 A) = 8.46 A

15. E = 0.707 Emax = (0.707)(165V ) = 117V

16. Imax = 1.41I = (1.41)(4.00 A) = 5.64 A

17. I = 0.707 Imax = (0.707)(17.0 A) = 12.0A

18. E = 0.707 Emax = (0.707)(1150V) = 813V

19. Imax = 1.41 I = (1.41)(8.50 A) = 12.0 A

20. Imax = 1.41 I = (1.41)(7.00A) = 9.87 A

21. E = 0.707 Emax = (0.707)(135V ) = 95.4 V

22. I = 0.707 Imax = (0.707)(125A) = 88.4 A

23. (a) V = 0.707Vmax = (0.707)(34.0V) = 24.0V (b) I = 0.707 Imax = (0.707)(0.170 A) = 0.120

(c) R =VI

=24.0V

0.120 A= 200 Ω

24. V = 0.707 Vmax = (0.707)(170 V) = 120 V

25. (a) I = 0.707 Imax = (0.707)(0.700 A) = 0.495 A (b) R =VI

=120 V

0.495A= 242Ω

26. Pavg = VI = (0.707Vmax)(0.707 Imax) = 0.500Pmax so, Pmax = 2P = 2(150 W ) = 300 W

27. Pavg = VI = (0.707Vmax)(0.707 Imax) = 0.500Pmax so, Pmax = 2P = 2(100 0W ) = 200 0W

19.2

1. R =PI 2 =

350 W(4.00 A)2 = 21.9Ω

2. P = I 2R = (6.00 A)2(12.0Ω) = 432W

3. P = VI = (500 V)(7.00 A) = 350 0W

4. R =V 2

P=

(110 V)2

450 W= 26.9Ω

5. P = VI = (220 V)(6.00 A) = 1320W

6. I =PR

=375W32.0Ω

= 3.42 A

7. P = I 2R = (7.00 A)2 (12.0Ω) = 588W

8. R =V 2

P=

(110 V)2

750 W= 16.1Ω



158

9. P = I 2 R = (5.00 A)2(110 Ω) = 2750W

10. P = I 2 R = (5.55 A)2 (19.5Ω) = 601 W

11.

NP =N SVP

VS=

(15.0 turns)(30.0V)45.0V

= 10.0 turns

12.

VS =VP NS

NP=

(250 V)(275 turns)730 turns

= 94.2 turns

13. VS =VP IP

IS=

(39.0V)( 6.00 A)4.00 A

= 58.5V

14.

NS =N PVS

VP=

(65.0 turns)( 230 0V)115V

= 130 0 turns

15.

NP =N SVP

VS=

(30.0 turns)(115V)11.5V

= 300 turns

16. VS =VP NS

NP=

(12.0V)( 220 0 turns)20.0 turns

= 1320V

17. IS =I PVP

VS=

(9.00 A)(39.0V)58.5V

= 6.00 A

18. P = VI = (110 V)(15.0 A) = 1650W

19. (a) IS =I PVP

VS=

(8.00 A)(110 V)15, 0 00V

= 0.0587 A (b) PP = VPIP = (110 V)(8.00 A) = 880 W

20. VP =PO

IP=

990W0.45A

= 2200V

21. V =PI

=775W5.00 A

= 155V

22. (a) IP =I SVS

VP=

(14.0 A)(120 V)660 0V

= 0.255A (b) P = VS IS = (14.0A)(120 V) = 1680W

23. (a) VS =VP NS

NP=

(720 0V)(125 turns)750 0 turns

= 120 V (b) IP =VSI S

VP=

(120 V)(36.0 A)720 0V

= 0.600 A

24. (a) IS =IP NP

NS

=(20.0 A)(175turns)

750turns= 4.67A (b) IP =

VSI S

VP=

(1350V )(2.00 A)90.0V

= 30.0 A

(c) PP = VPIP = (90.0V)(30.0 A) = 270 0W (d) PS = VSIS = (1350V)(2.00 A) = 270 0W

25. NS =VS NP

VP

=(3000V )(150turns)

110v= 4090V

26. IS =IP NP

NS

=(20.0 A)(175turns)

750turns= 4.67A

19.3 1. XL = 2πfL = 2π (60.0Hz)(3.00 × 10−3H ) = 1.13Ω

2. XL = 2πfL = 2π (75.0Hz)(20.0 × 10−3H ) = 9.42Ω

3. XL = 2πfL = 2π (10.0 × 103 Hz)(70.0 × 10−3 H) = 440 0Ω

4. XL = 2πfL = 2π (8.00 × 103 Hz)(8.00 × 10−3 H) = 402Ω



159

5. XL = 2πfL = 2π (15.0 × 106 Hz)(425 × 10−6 H) = 4.01 × 104 Ω = 40.1kΩ

6. XL = 2πfL = 2π (125 × 106 Hz)(655 × 10−6 H) = 5.14 × 105 Ω

7. I =E

XL=

E2πfL

=14.0V

2π (125Hz)(30.0 × 10−3 H)= 0.594 A

8. I =E

XL=

E2πfL

=145V

2π (125 × 103Hz)(1.00 × 10−3 H)= 0.185A

9. I =E

XL=

E2πfL

=50.0V

2π (2.00 × 103Hz)(5.00 × 10−3 H)= 0.796 A

10. I =E

XL=

E2πfL

=75.0V

2π (7.00 × 106Hz)(30.0 × 10−3 H)= 5.68 × 10−5 A

11. I =E

XL=

E2πfL

=105V

2π (2.00 × 106Hz)(72.0 × 10−6 H)= 0.116 A

12. I =E

XL=

E2πfL

=65.0V

2π (25.0 × 106Hz)(525 × 10−6 H)= 7.88 × 10−4 A

19.4

1. (a) Z = R2 + (2πfL)2 = (200 Ω)2 + 2π (1.25 × 103 Hz)(10.0 × 10−3 H )[ ]2= 215Ω

(b) tanφ =XL

R=

2πfLR

=2π (1.25 × 103 Hz)(10.0 × 10−3 H)

200 Ω= 0.393

φ = 21.4°

(c) I =VZ

=45.0V215Ω

= 0.209 A

2. (a) Z = R2 + (2πfL)2 = (12.0Ω)2 + 2π (900 Hz)(1.00 × 10−3 H)[ ]2= 13.3Ω

(b) tanφ =XL

R=

2πfLR

=2π (900 Hz)(1.00 × 10−3 H)

12.0Ω= 0.471

φ = 25.2°

(c) I =VZ

=10.0V13.3Ω

= 0.752 A



160

3. (a) Z = R2 + (2πfL)2 = (1.00 × 103 Ω)2 + 2π (10.0 × 103 Hz)(50.0 × 10−3 H)[ ]2= 330 0Ω

(b) tanφ =XL

R=

2πfLR

=2π (10.0 × 103 H )(50.0 × 10−3 H )

1.00 × 103 Ω= 3.14

φ = 72.3°

(c) I =VZ

=15.0V330 0Ω

= 4.55 × 10−3 A = 4.55mA

4. (a) Z = R2 + (2πfL)2 = (2.00 × 103 Ω)2 + 2π (5.00 × 103 Hz)(70.0 × 10−3 H)[ ]2= 2970Ω

(b) tanφ =XL

R=

2πfLR

=2π (5.00 × 103 Hz)(70.0 × 10−3 H)

2.00 × 103 Ω= 1.10

φ = 47.7°

(c) I =VZ

=12.0V2970Ω

= 4.04 × 10−3 A = 4.04 mA

5. (a) Z = R2 + (2πfL)2 = (300 Ω)2 + 2π (3.00 × 103 Hz)(2.00 × 10−3 H )[ ]2 = 302Ω

(b) tanφ =XL

R=

2πfLR

=2π (3.00 × 103 Hz)(2.00 × 10−3 H)

300 Ω= 0.126

φ = 7.2°

(c) I =VZ

=6.00V302Ω

= 1.99 × 10−2 A = 19.9 mA

19.5

1. XC =1

2πfC=

12π (1.00 × 103 Hz)(20.0 × 10−6 F)

= 7.96 Ω

2. XC =1

2πfC=

12π (100 Hz)(7.00 × 10−3 F)

= 0.227Ω

3. XC =1

2πfC=

12π (100 Hz)(0.600 × 10−6 F)

= 2650Ω

4. XC =1

2πfC=

12π (2.50 × 106 Hz)(30.0 × 10−3 F)

= 2.12 × 10−6Ω

5. XC =1

2πfC=

12π (0.250 × 106 Hz)(0.800 × 10−6 F)

= 0.796 Ω

6. XC =1

2πfC=

12π (60.0Hz)(15.0 × 10−6 F)

= 177 Ω

7. XC =1

2πfC=

12π (60.0Hz)(45.0 × 10−6 F)

= 58.9 Ω



161

8. XC =1

2πfC=

12π (100 Hz)(6.00 × 10−3 F)

= 0.265 Ω

9. XC =1

2πfC=

12π (300 Hz)(330 × 10−6 F)

= 1.61 Ω

10. XC =1

2πfC=

12π (120 Hz)(222 × 10−6 F)

= 5.97 Ω

19.6

1. (a) Z = R2 +1

2πfC⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= (1.00 × 103 Ω)2 +1

2π (100 Hz)(1.00 × 10−6 F)

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

= 1880Ω

(b) tanφ =XC

R= (1 /2πfC ) / R =

12πfCR

=1

2π (100 Hz)(1.00 × 10−6 F)(1.00 × 103 Ω)= 1.59

φ = 57.9°

(c) I =EZ

=100 V1880Ω

= 0.0532 A = 53.2mA

2. (a) Z = R2 +1

2πfC⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= (375 Ω)2 +1

2π(1.00 × 103 Hz)(5.00 × 10−6 F)

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

= 376Ω

(b) tanφ =1

2πfCR=

12π(1.00 × 103 Hz)(5.00 × 10−6 F)(375Ω)

= 0.0849

(c) I =EZ

=20.0V376Ω

= 0.0532 A = 53.2 mA

φ = 4.85°

3. (a) Z = R2 +1

2πfC⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= (4.80 × 103Ω)2 +1

2π(1.75 × 103 Hz)(45.0 × 10−6 F)

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

= 480 0Ω

(b) tanφ =1

2πfCR=

12π(1.75× 103 Hz)(45.0 × 10−6 F)(4.80 × 103 Ω)

= 4.21 × 10−4

φ = 0.0241°

(c) I =EZ

=15.0V480 0Ω

= 3.13 × 10−3 A = 3.13 mA



162

4. (a) Z = R2 +1

2πfC⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= (145 × 10−3 Ω)2 +1

2π (72.5 × 103 Hz)(10.0 × 10−6 F)

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

= 0.263Ω

(b) tanφ =1

2πfCR=

12π(72.5× 103 Hz)(10.0 × 10−6 F)(145 × 10−3 Ω)

= 1.52

φ = 56.6°

(c) I =EZ

=7.00 × 10−3 V

0.263Ω= 0.0266 A = 26.6 mA

5. (a) Z = R2 +1

2πfC⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= (10.0 × 10−3 Ω)2 +1

2π (10.0 × 103 Hz)(5.00 × 10−6 F)

⎛

⎝ ⎜ ⎞

⎠ ⎟

2

= 3.18Ω

(b) tanφ =1

2πfCR=

12π(10.0× 103 Hz)(5.00 × 10−6 F)(10.0 × 10−3 Ω)

= 318

φ = 89.8°

(c) I =EZ

=15.0× 10−3 V

3.18Ω= 4.72 × 10−3 A = 4.72 mA

19.7

1. XL = 2πfL = 2π (60.0Hz)(50.0mH ) = 18.8Ω

XC =1

2πfC=

12π (60.0Hz)(50.0μF)

= 53.1Ω

Z = R2 + (XL − XC )2 = (25.0Ω)2 + (18.8Ω) − (53.1Ω)[ ]2= 42.4 Ω

I =EZ

=5.00V42.4 Ω

= 0.118 A

2. XL = 2πfL = 2π (1.00kHz)(10.0 mH) = 62.8Ω

XC =1

2πfC=

12π (1.00 kHz)(0.200 μF)

= 796Ω

Z = R2 + (XL − XC )2 = (225Ω)2 + (62.8Ω − 796Ω) 2 = 767Ω

I =EZ

=15.0V767Ω

= 0.0196 A =19.6 mA

3. XL = 2πfL = 2π (10.0kHz)(10.0 mH) = 628Ω

XC =1

2πfC=

12π (10.0 kHz)(30.0 mH)

= 5.31 × 10−4 Ω

Z = R2 + (XL − XC )2 = (1.00kΩ)2 + (628Ω − 5.31 × 10−4 Ω) 2 = 1180Ω

I =EZ

=15.0V1180Ω

= 0.0127 A =12.7mA



163

4. XL = 2πfL = 2π (60.0Hz)(0.700 H) = 264 Ω

XC =1

2πfC=

12π (60.0 Hz)(30.0 μF)

= 88.4 Ω

Z = R2 + (XL − XC )2 = (1.00kΩ)2 + (264Ω − 88.4 Ω) 2 = 1020Ω

I =EZ

=8.00V1020Ω

= 7.84 × 10−3 A = 7.84 mA

5. XL = 2πfL = 2π (120 Hz)(0.600 H) = 452Ω

XC =1

2πfC=

12π (120 Hz)(35.0 μF)

= 37.9 Ω

Z = R2 + (XL − XC )2 = (150 Ω)2 + (452Ω − 37.9Ω) 2 = 440 Ω

I =EZ

=6.00V440 Ω

= 0.0136 A = 13.6 mA

6. XL = 2πfL = 2π (60.0Hz)(0.550 H) = 207Ω

XC =1

2πfC=

12π (60.0 Hz)(5.00 × 10−6F)

= 531 Ω

Z = R2 + (XL − XC )2 = (225Ω)2 + (207Ω − 531Ω) 2 = 394 Ω

I =EZ

=7.50V394Ω

= 0.0190 A = 19.0 mA

7. XL = 2πfL = 2π (100 Hz)(0.735 H) = 462Ω

XC =1

2πfC=

12π (100 Hz)(4.50μF)

= 354 Ω

Z = R2 + (XL − XC )2 = (175Ω)2 + (462Ω − 354Ω) 2 = 206Ω

I =EZ

=5.00V206Ω

= 0.0243A = 24.3mA

8. XL = 2πfL = 2π (600 Hz)(0.400 H) = 1510Ω

XC =1

2πfC=

12π (600 Hz)(100 μF)

= 2.65 Ω

Z = R2 + (XL − XC )2 = (575Ω)2 + (1510Ω − 2.65Ω) 2 = 1610Ω

I =EZ

=100 V1610Ω

= 0.0621 A = 62.1mA



164

9. XL = 2πfL = 2π (1.00kHz)(45.0 mH) = 283Ω

XC =1

2π (1.00kHz)(35.0μF)= 4.55 Ω

Z = R2 + (XL − XC )2 = (450 Ω)2 + (283Ω − 4.55Ω) 2 = 529Ω

I =EZ

=25.0V529Ω

= 0.0473A = 47.3 mA

10. XL = 2πfL = 2π (500 Hz)(0.500H) = 1570Ω

XC =1

2πfC=

12π (500 Hz)(500 μF)

= 0.637 Ω

Z = R2 + (XL − XC )2 = (375Ω)2 + (1570Ω − 0.637Ω) 2 = 1610Ω

I =EZ

=55.0V1610Ω

= 0.0342 A = 34.2mA

19.8

1. f =1

2π LC=

12π (1.00μH )(4.00μF)

= 79,600 Hz = 79.6kHz

2. f =1

2π LC=

12π (2.00μH )(35.0μF)

= 19, 0 00Hz = 19.0 kHz

3. f =1

2π LC=

12π (2.50μH )(7.00μF)

= 38, 0 00Hz = 38.0 kHz

4. f =1

2π LC=

12π (2.65μH)(35.0μF)

= 16, 500Hz = 16.5kHz

5. f =1

2π LC=

12π (42.5μH)(40.0μF)

= 3860Hz = 3.86kHz

6. f =1

2π LC=

12π (75.0μH )(25.0μF)

= 3680 Hz = 3.68 kHz

7. f =1

2π LC=

12π (43.5μH)(33.0μF)

= 420 0 Hz = 4.20 kHz

8. f =1

2π LC=

12π (37.5μH)(10.0μF)

= 8220 Hz = 8.22kHz

9. f =1

2π LC=

12π (100 μH )(8.00μF)

= 5630Hz = 5.63kHz

10. f =1

2π LC=

12π (300 μH )(3.75μF)

= 4750Hz = 4.75kHz



165

19.10 1. actual power = (apparent power)(power factor) = (12,600 kVA)(0.850) = 10,700kW

2. actual power = (apparent power)(power factor) = (12,800 kVA)(0.910) = 11,600 kW

3. apparent power = actual powerpower factor

=120 , 000kW

0.900= 133, 000kVA


=1, 90 0, 000 kW

0.800= 2, 380,000 kVA

5. actual power = (apparent power)(power factor) = (11,500 kVA)(0.880) = 10,100 Kw


=2,350, 000 kW

0.850= 2, 760,000 kVA




=350 , 000 kW

0.860= 407,000 kVA


=1, 250,000 kW

0.820= 1, 520,000 kVA

11. power factor = actual power

appartent power=

55,800 kW63,400

= 0.880

12. power factor = actual power

appartent power=

587, 000 kW645, 000kVA

= 0.910

Chapter 19 Review Questions 1. d

2. c

3. a and e

4. a, b, and c

5. The maximum current is the maximum instantaneous current. The effective value of an altering current is the number of amperes that produce the same amount of heat in a resistance as an equal number of amperes of a steady direct current.

6. The maximum voltage is the maximum instantaneous voltage. The instantaneous voltage is the voltage at any instant of time.

7. Power is the product of the effective values of the voltage and current.

8. Power is the square of the effective value of the voltage divided by the resistance.

9. The output voltage doubles.

10. Henry

11. Inductive reactance allows the analysis of circuits containing inductors.



166

12. The inductive reactance is directly proportional to the frequency.

13. The current lags the voltage in an inductive circuit.

14. Energy is stored in the form of potential energy associated with a sheet of positive charge on the one side of the capacitor and a sheet of negative charge on the other side.

15. The current leads the voltage in a capacitive circuit.

16. The frequency and reactance of a capacitor are inversely proportional.

17. Resonance occurs when the inductive reactance equals the capacitive reactance. The current is then given by its maximum value.

18. A diode allows current to flow in one direction but not in the reverse direction.

19. Amplification produces an increase in the value of a voltage or current in a circuit. Rectification produces a current or voltage in only one direction.

20. No, the phase angle depends on the frequency.


1. Emax =e

sinθ=

95.4Vsin 62°

= 110V

2. e = Emax = sinθ = (185V)sin 41° = 120V

3. e = Emax = sinθ = (175V)sin 23° = 68V

4. E = 0.707 Emax = 0.707(135V ) = 95.4 V

5. Imax = 1.41 I = 1.41(6.35 A) = 8.95 A

6. P = I 2R = (6.87 A)2(15.4 Ω) = 727W

7. P = VI = (110 V)(4.50 A) = 495W

8. P = I 2R = (7.65A)2 (22.3Ω) = 1310W

9. (a) NS =VS N P

VP=

(2050V )(75.0 turns)115V

= 1340 turns (b) IS =I PN P

NS=

(4.55 A)(75.0 turns)1340 turns

= 0.255 A

(c) P = IPVP = (4.55 A)(115V) = 523W

P = ISVS = (0.255 A)(2050V ) = 523W

10. XL = 2πfL = 2π (60.0Hz)(48.0mH ) = 18.1Ω

Z = R2 + XL2 = (23.0Ω)2 + (18.1Ω)2 = 29.3Ω

(a) I =EZ

=115V29.3Ω

= 3.92 A

(b) tanφ =XL

R=

18.1Ω23.0Ω

= 0.787; φ = 38.2°

(c) E = IXL = (3.92 A)(18.1Ω) = 71.0V



167

11. XL = 2πfL = 2π (60.0Hz)(43.2mH ) = 16.3Ω

Z = R2 + XL2 = (47.5Ω)2 + (16.3Ω)2 = 50.2Ω

(a) I =EZ

=115V50.2Ω

= 2.29 A

(b) tanφ =XL

R=

16.3Ω47.5Ω

= 0.343; φ = 18.9°

(c) E = IXL = (2.29 A)(47.5Ω) = 109V

12. XC =1

2πfC=

12π (60.0Hz)(57.4μF)

= 46.2Ω

Z = R2 + XL2 = (19.5Ω)2 + (46.2Ω)2 = 50.1Ω

(a) I =EZ

=110 V50.1Ω

= 2.20 A

(b) tanφ =XC

R=

46.2Ω19.5Ω

= 2.369; φ = 67.1°

(c) V = IR = (2.20 A)(19.5Ω) = 42.9V

(d) V = IXC = (2.20 A)(46.2Ω) = 102V

13. XC =1

2πfC=

12π (60.0Hz)(38.5μF)

= 68.9Ω

Z = R2 + XC2 = (21.6Ω)2 + (68.9Ω)2 = 72.2Ω

(a) I =EZ

=110 V72.2Ω

= 1.52 A

(b) V = IR = (1.52 A)(21.6Ω) = 32.8V

(c) tanφ =XC

R=

68.9Ω21.6Ω

= 3.190; φ = 72.6°

14. XL = 2πfL = 2π (60.0Hz)(62.0mH ) = 23.4 Ω

XC =1

2πfC=

12π (60.0Hz)(25.0μF)

= 106Ω

Z = R2 + (XL − XC )2 = (175Ω)2 + (23.4 Ω − 106Ω)2 = 194 Ω

I =EZ

=110 V194Ω

= 0.567 A



168

15. XL = 2πfL = 2π (60.0Hz)(65.0mH ) = 24.5Ω

XC =1

2πfC=

12π (60.0Hz)(35.0μF)

= 75.8Ω

Z = R2 + (XL − XC )2 = (115Ω)2 + (24.5Ω − 75.8 Ω)2 = 126Ω

I =EZ

=110 V126Ω

= 0.873A

16. f =1

2π LC=

12π (3.70μH )(7.50μF)

= 30,200 Hz = 30.2kHz

XL = 2πfL = 2π (60.0Hz)(3.70μH ) = 1.39 × 10−3 Ω

XC =1

2πfC=

12π (60.0Hz)(7.50μF)

= 354 Ω

Z = R2 + (XL − XC )2 = (525Ω)2 + (1.39 × 10−3 Ω − 354Ω)2 = 633Ω

I =EZ

=110 V633Ω

= 0.174 A

17. f =1

2π LC=

12π (4.50μH )(4.70μF)

= 34, 600Hz = 34.6 kHz

XL = 2πfL = 2π (60.0Hz)(4.50μH ) = 1.70 × 10−3 Ω

XC =1

2πfC=

12π (60.0Hz)(4.70μF)

= 564 Ω

Z = R2 + (XL − XC )2 = (25.0Ω)2 + (1.70 × 10−3 Ω − 564Ω)2 = 565Ω

I =EZ

=110 V565Ω

= 0.195A


=2.90 × 106 kW

0.850= 3.41 × 106 kVA

Chapter 19 Applied Concepts 1. (a) P = I V = 11.8 A 110 V = 130 0W

(b) Imax = 1.41I = 1.41 11.8 A =16.6 A

(c) max sin180 0i I= ° = ; Because the current is reversing at this point.



169

2. (a) 307 9.00

; 23.0120

turnsP SP PS

S S P

VN VV NN turns

V N V V= = = =

(b) PP = IPVP = 0.953 A 120 V = 114W

IS

IP=

N P

N S= IS =

N P IP

NS=

307 turns 0.953 A23 turns

= 12.7 A

PS = IS VS = 12.7 A 9.00V =114 N

(c) The power is conserved in the transformer.

3. (a) P = I V I = PV =

8.50 × 104 W220 V

= 386 A

(b) P = I V I =PV

=8.50 × 104 W

600 0V= 14.2 A

(c) P = I 2R = (386 A)2 (0.250 A) = 37,200W

P = I 2R = (14.2 A)2 (0.250 A) = 50.4W

(d) The 600 0V power line is better becauseit only looses 50.4W power

4. (a) 125

2 0.3322 2 (60.0 )

LL

XX fL L H

f Hzπ

π π

Ω= = = =

(b) 65.5

125

P VP I R I

R= = =

Ω

(c) P = I ⋅V = (0.524 A)(65.5A) = 34.3W

5. (a) f =1

2π LCC =

1f 2 4π 2L

=1

(8.80 × 105 Hz)2 4π 2(0.275H)= 1.19 × 10−13 F

(b) 14

2 2 6 2

1 1 19.04 10

4 (1.01 10 ) 4 0.2752f C F

f L Hz HLC π ππ−= = = = ×

×

(c) Tuning to a high frequency lowers the capacitance because the plates are similar.



170

Chapter 20 20.2 1.

s = ct = 3.00 × 108 m / s( ) 5.00s( )= 1.50 × 109 m

2.

s = ct = 3.00 × 108 m / s( ) 6.40s( )= 1.92 × 109 m

3. t =

sc

=20,000mi

186,000mi / s= 0.108s

4. t =sc

=2.40 × 105 mi

186,000mi / s= 1.29s

5. t =sc

=9.30 × 107 mi186,000mi / s

= 500s

6. s = ct = 3.00 × 108 m / s( ) 2.50 × 10−5 s

2⎛

⎝⎜⎞

⎠⎟= 3750m = 3.75km

7. t =

sc

=3000mi

186,000mi / s= 0.0161s = 16.1ms

8. t =

sc

=100mi

3.00 × 108 m / s= 3.33× 10−7 s

9. t =sc

=18,250mi

186,000mi / s= 0,0981s

10. s = ct = 3.00 × 108 m / s( ) 1.24 × 10−3s

2⎛

⎝⎜⎞

⎠⎟= 186,000m = 186km

11. t =

sc

=0.95m

3.00 × 108 m / s= 3.2 × 10−9 s 12. t =

sc

=18,250mi

186,000mi / s= 0.0981s

13 smin = rM − rE = 218 × 106 km − 143× 106 km = 75 × 106 km = 7.5 × 1010 m

t =

sc

=7.50 × 1010 m

3.00 × 108 m / s= 250s (4.17 min)

smax = rM + rE = 218 × 106 km + 143 × 106 km = 3.61× 1011 m

t =

sc

=3.61× 1011 m

3.00 × 108 m / s= 1200s (20.0 min)

14. s = ct = 186,000mi / s( ) 4.31yr( ) 1.25( )×

365.25d1yr

×24h1d

×3600s

1h= 3.16 × 1013 mi

15. (a) smin = rJ − rE = 725 × 106 km − 143× 106 km = 582 × 106 km = 5.82 × 1011 m

t =

sc

=5.82 × 1011 m

3.00 × 108 m / s= 1940s (32.3min)

(b) smax = rJ + rE = 725 × 106 km + 143× 106 km = 868 × 106 km = 8.68 × 1011 m

t =

sc

=8.68 × 1011 m

3.00 × 108 m / s= 2890s (48.2 min)

16. s = ct = 3.00 × 108 m / s( ) 0.330s( )= 9.90 × 107 m



171

17. 17. s = ct = 3.00 × 108 m / s( )1.28s( )= 3.84 × 108 m

18. t =

sc

=1.50 × 108 m

3.00 × 108 m / s=

1.50 × 1011 m3.00 × 108 m / s

= 500s ×1min60s

= 8.33min

20.3

1. f =

cλ

=3.00 × 108 m / s4.55 × 10−5 m

= 6.59 × 1012 / s = 6.59 × 1012 Hz

2. f =

cλ

=3.00 × 108 m / s9.70 × 10−10 m

= 3.09 × 1017 / s = 3.09 × 1017 Hz

3. λ =

cf

=3.00 × 108 m / s9.70 × 1011 Hz

= 3.09 × 10−4 m

4. 4. λ =

cf

=3.00 × 108 m / s24.2 × 106 Hz

= 12.4m

5. λ =

cf

=3.00 × 108 m / s45.6 × 106 Hz

= 6.58m

6. λ =

cf

=3.00 × 108 m / s

415Hz= 7.23 × 105 m

7. f =

cλ

=3.00 × 108 m / s6.59 × 1012 m

= 4.55 × 10−5 Hz

8. f =

cλ

=3.00 × 108 m / s

9.23 × 103 m= 3.25 × 104 Hz = 32.5kHz

9. λ =

cf

=3.00 × 108 m / s1400 × 103 Hz

= 214m

10. λ =

cf

=3.00 × 108 m / s100 × 106 Hz

= 3.00m

11. f =

cλ

=3.00 × 108 m / s

85.5m= 3.51× 106 Hz = 3.51MHz

12. f =

cλ

=3.00 × 108 m / s3.25 × 10−8 m

= 9.23 × 1015 Hz

13. f =

cλ

=3.00 × 108 m / s

4.0 × 10−7 m= 7.5 × 1014 Hz

14. f =

cλ

=3.00 × 108 m / s

7.5 × 10−7 m= 4.0 × 1014 Hz



172

15. f =

cλ

=3.00 × 108 m / s

4.0 × 10−7 m + 7.5 × 10−7 m( )/ 2= 5.2 × 1014 Hz

16. f =

cλ

=3.00 × 108 m / s

237m= 1270kHz

20.4 1.

E = hf = 6.626 × 10−34 Js( ) 8.95 × 1010 Hz( )= 5.93× 10−23 J

2. f =

Eh

=3.96 × 10−22 J

6.626 × 10−34 Js= 5.98 × 1011 Hz

3. E = hf = 6.626 × 10−34 Js( ) 4.55 × 108 Hz( )= 3.01× 10−25 J

4. f =

Eh

=2.00 × 10−24 J

6.626 × 10−34 Js= 3.02 × 109 Hz

5. f =

Eh

=5.50 × 10−26 J

6.626 × 10−34 Js= 8.30 × 107 Hz = 83.1MHz

6. E = hf = 6.626 × 10−34 Js( ) 2.50 × 1012 Hz( )= 1.66 × 10−21 J

7. f =

Eh

=3.65 × 10−23 J

6.626 × 10−34 Js= 5.51× 1010 Hz

8. E = hf = 6.626 × 10−34 Js( ) 9.20 × 1016 Hz( )= 6.10 × 10−17 J

9. E = hf = 6.626 × 10−34 Js( ) 4.0 × 1014 Hz( )= 2.7 × 10−19 J (f from 17.3, # 14)

10. E = hf = 6.626 × 10−34 Js( ) 7.5 × 1014 Hz( )= 5.0 × 10−19 J (f from 17.3, # 13)

11. E = hf = 6.626 × 10−34 Js( ) 5.2 × 1014 Hz( )= 3.4 × 10−19 J

12. E = hf = 6.626 × 10−34 Js( ) 6.50 × 105 Hz( )= 4.31× 10−28 J

13. f =

Eh

=9.37 × 10−28 J

6.626 × 10−34 Js= 1410kHz

20.5

1. 48.0cd ×

4π l m1cd

= 603l m

2. 342cd ×

4π l m1cd

= 4300l m

3. 765l m ×1cd

4π l m= 60.9cd

4. 432l m ×1cd

4π l m= 34.4cd



173

5. 75.0cd ×

4π l m1cd

= 942l m

6. 650l m ×

1cd4π l m

= 51.7cd

7.

E =I

4πr 2 =900l m

4π 7.00 ft( )2 = 1.46 ft − candles

8.

E =I

4πr 2 =741l m

4π 6.50m( )2 = 1.40lux

9. E =I

4πr 2 =893l m

4π 3.25 ft( )2 = 6.73 ft − candles

10.

I = 4πr 2 E = 4π 9.00m( )24.32lux( )= 4400l m

11. I = 4πr 2 E = 4π 6.00 ft( )2

10.5 ft − candles( )= 4750l m

12. I = 4πr 2 E = 4π 9.85 ft( )2

5.50 ft − candles( )= 6710l m

13. I = 4πr 2 E = 4π 4.50m( )2

2.39lux( )= 608l m

14. 14. I = 4πr 2 E = 4π 6.50m( )2

5.28lux( )= 2800l m

15. (a) Decrease (b)

19

16. 1− cos 35.0° = 0.181or18.1%

17. ET = E1 + E2 + E3

E1 =I1

4πr12 =

150l m

4π 2.00m( )2 = 2.98lux

E2 =150l m

4π 2.70m( )2 = 1.64lux

E3 =150l m

4π 2.98m( )2 = 1.34lux

Therefore ET = 5.96lux

18. ET = E1 + E2

I

4π 1.40m( )2 +I

4π 1.96m( )2 = 3.54l m / m2

I = 57.7 l m

19. ET = E1 + E2

I

4π 0.880m( )2 +I

4π 1.12m( )2 = 5.86l m / m2

I = 35.3 l m

20. E =I

4πr 2 =1850l m

4π 3.35m( )2 = 13.1lux

Chapter 20 Review Questions 1. a, c, and e

2. c

3. d

4. c

5. Yes; the wavelength varies inversely with the frequency.

6. The energy is directly proportional to the frequency.

7. The intensity falls off as 1 divided by the square of the distance.



174

8. The speed of light is measured by determining the time light takes to travel a measured distance and using the relationship v = s/t.

9. It always travels at the same velocity in a vacuum. It has a different velocity (slower) in any media.

10. Electromagnetic radiation.

11. Max Planck

12. Albert Einstein and Max Planck

13. Olaus Roemer

14. By measuring the time difference for the start of the eclipse of the moons of Jupiter as viewed from different parts or earth’s orbit.

15. Candela and lumen 16. Luminous intensity is the brightness of a light source.


1. s = ct = 3.00 × 108 m / s( ) 21.5h( )×

3600s1h

= 2.32 × 1013 m

2. s = ct = 186,000mi / s( ) 3.78 × 10−5 s

2⎛

⎝⎜⎞

⎠⎟= 3.52mi

3. t =

sc

=2 0.245mi( )

186,000mi / s= 2.63× 10−6 s or2.63μs 4. t =

sc

=0.274mu

186,000mi / s= 1.47 × 10−6 s or1.47μs

5. t =

sc

=22,500mi

186,000mi / s= 0.121s 6. λ =

cf

=3.00 × 108 m / s

1230kHz= 244m

7. f =

cλ

=3.00 × 108 m / s

46.5m= 6.45 × 106 Hzor6.45MHz

8. f =

cλ

=3.00 × 108 m / s5.415 × 10−8 m

= 5.54 × 1015 Hz 9.

E = hf = 6.626 × 10−34 Js( )1.45 × 1011 Hz( )= 9.61× 10−23 J

10. f =

Eh

=4.75 × 10−23 J

6.626 × 10−34 Js= 7.17 × 1010 Hz

11. E = hf = 6.626 × 10−34 Js( ) 8.25 × 1015 Hz( )= 5.47 × 10−18 J

12. I = 4πr 2 E = 4π 6.75 ft( )2

3.75 ft − candles( )= 2150l m

13. I = 4πr 2 E = 4π 9.25m( )2

4.86lux( )= 5230l m

14. I = 4πr 2 E = 4π 2 9.25m( )⎡⎣ ⎤⎦

24.86lux( )= 20,900l m

15. smin = sJ − sM = 725 × 106 km − 215 × 106 km = 940 × 106 km = 9.40 × 1011 m

t =

sc

=9.40 × 1011 m3.00 × 08 m / s

= 3130s (52.2 min)

smin = sJ − sM = 725 × 106 km − 215 × 106 km = 510 × 106 km = 5.10 × 1011 m

t =

sc

=5.10 × 1011 m3.00 × 08 m / s

= 1700s (28.3min)



175

16. ET = E1 + E2

I

4π 0.454m( )2 +I

4π 0.538m( )2 = 8.46l m / m2

I = 12.8 l m

17.

ET = E1 + E2 + E3 =125l m

4π 1.85m( )2 +125l m

4π 1.92m( )2 +125l m

4π 2.43m( )2 = 7.29l m / m2 = 7.29lux


1. (a) s = ct t =

sc

=3470mi

186,000mi / s= 0.0187s

(b)

a2 + b2 = c2 c = a2 + b2 = 2.20 × 104 mi( )2

+ 1735mi( )2= 2.21× 104 mi × 2(up and down) = 4.42 × 104 mi

s = ct t =

sc

=4.42 × 104 mi

186,000mi / s= 0.238s

2. (a) s = ct = 3.00 × 108 m / s 1.28s = 3.84 × 108 m = 3.84 × 105 m = 2.39 × 105 mi

(b) s = ct t =

sc

=1.50 × 1011 mi

3.00 × 108 m / s= 500s = 8.33min

(c) s = ct = 3.00 × 108 m / s 4.31yr = 3.00 × 108 m / s 1.36 × 108 s = 4.08 × 1016 m = 2.53 × 1013 mi

3. (a) c = λ f λ =

cf

=3.00 × 108 m / s88.0 × 106 Hz

= 3.41m

λ =

cf

=3.00 × 108 m / s108 × 106 Hz

= 2.78m

(b) E = hf = 6.626 × 10−34 Js 88.0 × 106 Hz = 5.83 × 10−26 J

E = hf = 6.626 × 10−34 Js 108 × 106 Hz = 7.16 × 10−26 J (c) The higher the frequency, the greater the energy.

4. c = λ f λ =

cf

=3.00 × 108 m / s54.0 × 106 Hz

= 5.56m

14

λ =14

5.56m = 1.39m

λ =

cf

=3.00 × 108 m / s88.0 × 106 Hz

= 3.41m

14

λ =14

3.41m = 0.852m

5. (a) E =

I

4πr 2; I = 4πr 2 E = 4π 2.50m( )2

180lx = 14,100lm

(b) I = 4π 2r( )2E = 4π 5.00m( )2

180lx = 56,500lm (4 x the original)



176

Chapter 21 21.5

1. 1f

=1so

+1si

=1

1.65cm+

16.00cm

; f = 1.29cm

2. 1so

=1f

−1si

=1

15.0cm−

13.00cm

; so = −3.75cm

3. 1so

=1f

−1si

=1

10.0cm−

114.5cm

; so = 32.2cm

4. 1so

=1f

−1si

=1

−5.00cm−

110.0cm

; so = −10.0cm

5. 1f

=1so

+1si

=1

7.35cm+

117.0cm

; f = 5.13cm

6. so =−siho

h i=

− 6.00cm( ) 4.50cm( )2.75cm

= −9.82cm

7. hi =−siho

so=

− 13.0cm( ) 12.0cm( )25.0cm

= −6.24cm

8. so =−siho

h i=

− 15.5cm( ) 2.50cm( )3.50cm

= −11.1cm

9. hi =−siho

so

=

− − 0.76m( ) 1.40m( )6.20m

= 0.17m

10. 1si

=1f

−1so

=1

16.0cm−

110.5cm

; si = −30.5cm

ni =−siho

so=

− −3.05cm( ) 30.0cm( )10.5cm

= 87.1cm

11. 1f

=1so

+1si

=1

20.0cm+

120.0cm

; f = 10.0cm

12. 1f

=1so

+1si

=1

12.6cm+

1−6.00cm

; f = −11.5cm

13. hi =−siho

so

=

− −1.50m( ) 1.20m( )7.60m

= 0.237m

14. hi =−siho

so

=

− − 0.76m( ) 1.40m( )6.20m

= 0.17m

15. 1f

=1so

+1si

=1

24.5cm+

1−3.55cm

; f = −4.15cm

16. 1f

=1so

+1si

=1

34.4cm+

1−5.66cm

; f = −6.77cm

17. 1f

=1so

+1si

=1

6.50cm+

1−2.30cm

; f = −3.56cm

18. 1so

=1f

−1si

=1

13.5cm−

111.5cm

; so = 77.6cm

19. (a) ho =−hiso

si=

− −2.75cm( ) 7.33cm( )5.03cm

= 4.01cm

(b) 1f

=1so

+1si

=1

7.33cm+

15.03cm

; f = 2.98cm

(c) 1si

=1f

−1so

=1

2 2.98cm( ) −1

7.33cm; si = 31.9cm

21.10

1. n =sinisin r

=sin 31.5°sin 25.6°

= 1.21 2. sin r =sini

n=

sin14.6°2.40

; r = 6.0°

3. Speed of light in liquid = cn

=3.00 × 108 m / s

1.50= 2.00× 108 m / s

4. n =sinisin r

=sin38.0°sin24.5°

= 1.48

5. n =1

sin ic=

1sin 42.4°

= 1.48

6. sin ic =1n

=1

2.45; ic = 24.1°

7. 1si

=1f

−1so

=1

15.0cm−

148.0cm

; si = 21.8cm



177

8. (a) 1f

=1so

+1si

=1

20.0cm+

19.00cm

; f = 6.21cm

(b) hi =−siho

so=

− 9.00cm( ) 2.50cm( )20.0cm

= −1.13cm

9. 1so

=1f

−1si

=1

5.00cm+

11.50cm

; so = −2.14cm 10. 1f

=1so

+1si

=1

40.0cm+

11.90cm

; f = 1.81cm

11. (a) 1f

=1so

+1si

=1

15.0cm+

17.50cm

; f = 5.00cm

(b)

hi =−siho

so=

− 7.50cm( ) 5.00cm( )15.0cm

= −2.50cm

12. (a) 1si

=1f

−1so

=1

26.0cm−

118.0cm

; si = −58.5cm

(b)

hi =−siho

so=

− −58.5cm( ) 4.50cm( )18.0cm

= 14.6cm

13. 1si

=1f

−1so

=1

19.5cm−

15.76cm

; si = −8.17cm

hi =−siho

so=

− −8.17cm( ) 1.45cm( )5.76cm

= 2.06cm

14. 1si

=1f

−1so

=1

14.5cm−

110.5cm

; si = −38.1cm

hi =−siho

so=

− −38.1cm( ) 2.35cm( )10.5cm

= 8.53cm

15. hi

ho= 2 =

si

so; si 2so

1f

=1so

+1si

=1

13.3cm / 2+

113.3cm

; f = 4.43cm

Chapter 21 Review Questions 1. a

2. d

3. a and b

4. Parallel light rays reflected off a rough surface may scatter in many different directions (diffusion) whereas parallel light rays reflect off a smooth surface as parallel rays.

5. At a smooth surface, the incident angles of a light ray is the same as the angle of the reflected ray.

6. At a smooth surface, the normal to the surface and both the incident and reflected rays lie in the same plane.

7. They are virtual, erect, and lie as far behind the mirror as the object is in front of the mirror.



178

8. A real image can be shown on a screen. A virtual image cannot.

9. Concave mirrors curve away from the observer, convex mirrors curve toward the observer.

10. For large apertures, not all parallel rays are reflected through the focal point. This produces a fuzzy or aberrant image.

11. The image distance is increased.

12. The image distance is increased.

13. b

14. b

15. Converging lenses convert parallel rays into converging rays. Diverging lenses convert parallel rays into diverging rays.

16. Light propagating in an optical fiber, light reflected into a swimming pool from an underwater light.

17. Light passing at an angle through an interface from a medium of low optical density to a medium of higher optical density is bent toward the normal to the interface.

18. The speed of light is lower in the high index material.

19. A wave passing at an angle from one medium to another with a different wave velocity will be bent toward or away from the normal, depending on whether the speed is higher or lower in the new material.

20. Virtual

21. Real or virtual

22. They roughen the surface and scatter the light.

23. It appears shallower because of the refraction of light at the surface.

24. Light travels in a straight line unless it is reflected of refracted.

25. Light traveling at an angle greater than the critical angle is reflected from one side of the fiber to another as it travels down the length of the fiber.

26. When the object is closer to the lens than the focal length.

27. When the object is located a distance from the lens equal to the focal length.

28. When the object is located outside the focal point.


1. 1f

=1so

+1si

=1

3.50cm+

17.25cm

; f = 2.36cm

2. 1si

=1f

−1so

=1

25.0cm−

18.50cm

; si = −12.9cm

3. hi =−siho

so=

− 7.50cm( ) 6.50cm( )14.0cm

= −3.48cm

4. hi =−siho

so=

4.35m( ) 3.75m( )7.35m

= −2.22m

5. (a) 1si

=1f

−1so

=1

21.4cm−

123.4cm

; si = 250 cm (b) hi =−siho

so=

− 250 cm( ) 43.0cm( )23.4cm

= −459cm

6. 1f

=1so

+1si

=1

45.3cm=

145.3cm

; f = 22.7cm

7. n =sinisin r

=sin 41.0°sin 29.0°

= 1.35



179

8. speed of light in liquid = cn

=3.00 × 108 m / s

1.44= 2.08× 108 m / s

9. n =1

sin ic=

1sin 45.6°

= 1.40

10. sin ic =1n

=1

1.50;i c = 41.8°

11. 1si

=1f

−1so

=1

12.0cm−

136.0cm

; si = 18.0cm

12. (a) 1f

=1so

+1si

=1

20.0cm+

112.0cm

; f = 7.50cm

(b) hi =−siho

so=

− 12.0cm( ) 4.50cm( )20.0cm

= −2.70cm

13. 1so

=1f

−1si

=1

4.00cm−

17.20cm

; so = 9.00cm

14. hi

ho= 3 =

si

so; si = 3so

1f

=1so

+1si

=1

25.0cm / 3+

125.0cm

; f = 6.25cm

15. 1f

=1so

+1si

=1

5.33m+

1−3.44m

; f = −9.70cm

16. 1si

=1f

−1so

=1

16.0cm−

110.5cm

; si = −30.5cm

ni =−siho

so=

− −3.05cm( ) 30.0cm( )10.5cm

= 87.1cm

17. speed of light in diamond = speed of light in vacuum

n=

3.00× 108 m / s2.417

= 1.24 × 108 m / s

18. sin ic =1

1.50;i c = 41.8°

19. 1f

=1so

+1si

=1

39.3cm+

117.8cm

; f = 12.3cm

20. (a) 1si

=1f

−1so

=1

8.70cm−

119.3cm

; si = 15.8cm

hi =−siho

so=

− 15.8cm( ) 13.2cm( )19.3cm

= −10.8cm

(b) 1si

=1f

−1so

=1

8.70cm−

12 19.3cm( ) ; si = 11.2cm

hi =−siho

so=

− 11.2cm( ) 13.2cm( )38.6cm

= −3.83cm




180

1. (a) f =radius

2=

38.0cm2

= 19.0cm (b) 1si

=1f

−1so

=1

19.0cm−

112.5cm

; si = −36.5cm

(c) upright (d) hi =−siho

so=

− −36.5cm( ) 25.0cm( )12.5cm

= 73.0cm

2. (a) 1si

=1f

−1so

=1

−0.750m−

111.5m

; si = −0.704m (b) virtual

(c) hi =−siho

so=

− −0.704m( ) 0.255m( )11.5m

= 0.0156m (d) Better for increasing the field of view.

3. (a) sin r =sin inglass

=sin 30°1.575

; i = 18.5° (b) sin r =nglass sini

nwater=

1.575sin 18.5°1.33

; r = 22.1°

(c) Angle will be smaller because light refracts toward the normal line when entering a denser material.

4. (a) sin ic =1n

=1

2.417;i c = 24.44° (b) larger than 24.44° (c) sin ic =

1n

=1

1.92;i c = 31.4°

(d) Diamond; zircon will allow light to escape when angles are less than 31.4°

5. (a) 1si

=1f

−1so

=1

0.0600m−

19.20m

= 0.0604m

(b) hi =−siho

so=

−0.0604m 2.0m9.20m

= −0.0131m = −13.1mm



181


1. Red, orange, yellow, green, blue, and violet.

2. The color of light is determined by its wavelength or its frequency.

3. (a) Infrared (b) Ultraviolet

4. Black. The cloth in the dress absorbs all colors except green and reflects only green. The red light is absorbed and no light is reflected.

5. (a) Red stars on a black field, rest of the flag would be red. (b) Blue field with no stars, rest of flag would have blue and black stripes.

6. Red, green, and blue.

7. Two colors that combine to form white are called complimentary colors. The complement of red is cyan; the compliment of green is magenta; and the compliment of blue is yellow.

8. The primary pigments are the complements of the three primary colors.

9. Cyan, magenta, and yellow for the colors and black for the shadow areas and definition.

10. The sky is blue because sunlight is scattered as it passes through the atmosphere. The amount of scattering depends on the wavelength of the light. The longer wavelengths of red, orange, and yellow are scattered much less than the shorter wavelengths of blue and violet. Since our eyes are not very sensitive to violet light, the scattered blue light in the atmosphere dominates our vision so that we see a blue sky.

11. At sunset, sunlight travels through significantly more of the atmosphere than at other times of the day. This results in even more scattering of the shorter wavelengths of violet, blue, and green which are taken out before reaching us with some orange but primarily red, the lightest wavelength and least scattered, most readily passing through the atmosphere to reach our eyes for a red sunset.

12. Clouds consist of many different sized water droplets. The small droplets tend to scatter the short- wavelength (blue) light, while the medium droplets tend to scatter the medium-wavelengths (green) light, and the large droplets tend to scatter the long-wavelength (red) light. The overall result is that all of the colors are scattered and combined so that we see a general white reflection.

13. The ocean is blue because the water surface acts like a mirror and reflects the blue color of the sky and the water tends to reflect and scatter the short-wavelength yellow, green, blue, and violet light and absorb the long-wavelength orange and red light.

14. False.

15. Diffraction.

16. Diffraction pattern.

17. Thomas Young

18. Constructive interference.

19. Destructive interference.

20. Polarized sunglasses restrict the light waves passing through to a single plane rather than normal sunlight this is emitted in all directions with many orientations.



182

Chapter 23 23.3

1. E = −kZ2

n2 = −13.595eV( ) 1( )2

12 = −13.595eV 2. E = −kZ2

n2 = −13.595eV( ) 1( )2

32 = −1.511eV

3. −1.511eV( )− −13.595eV( ) = 12.084eV ( from#1 and 2)

4. λ =cf

=c

E /h=

chE

=3.00× 108 m / s( )6.262× 10−34Js( )

12.084eV × 1.602 × 10−19J1eV

= 1.03× 10−7 m

5. E = −kZ2

n2 = −13.595eV( ) 1( )2

32 = −1.511eV

E = −kZ2

n2 = −13.595eV( ) 1( )2

22 =−3.399eV+1.888eV

λ =cf

=c

E /h=

chE

=3.00× 108 m / s( )6.626× 10−34Js( )

1.888eV × 1.602 × 10−19J1eV

= 6.57 × 10−7 m

23.4 1. Neon 2. Helium 3. Fluorine 4. Sodium 23.5 1. (a) 12 (b) 6 (c) 6 (d) 6 (e) 12 (f) 6 2. (a) 13 (b) 7 (c) 6 (d) 7 (e) 13 (f) 6 3. (a) 48 (b) 22 (c) 26 (d) 22 (e) 48 (f) 26 4. (a) 141 (b) 58 (c) 83 (d) 58 (e) 141 (f) 83 5. (a) 11 (b) 22.9898 μ 6. (a) 26 (b) 55.847μ 7. (a) 82 (b) 207.2μ 8. (a) 17 (b) 35.453 μ 9. 14C 10. 234U 11. Bromine 12. Cadmium 13. Argon 14. Iron 23.6

1. 232.037131μ ×1.6605 × 10-27 kg

1μ= 3.8530× 10−25kg

2. 228.028716μ×1.6605× 10-27kg

1μ= 3.7864 × 10−25kg



183

3. 7 protons, 14 neutrons, 7 electrons mi = 7 1.6726× 10−27 kg( )+ 14 1.6749 × 10−27 kg( )+ 7 9.1094 × 10−31kg( )= 3.5163× 10−26 kg

Δm = 3.5163× 10−26 kg −14.00307μ ×1.6605× 10−27 kg

1μ= 1.1911× 10−26kg

E = Δmc2 = 1.1911× 10−26 kg( )2.998× 108 m / s( )2= 1.071× 10−9 J ×

1MeV1.602× 10−13 J

= 6680MeV

4. 92 protons, 140 neutrons, 92 electrons mi = 92 1.6726× 10−27 kg( )+ 140 1.6749× 10−27kg( )+ 92 9.1094 × 10−31kg( )= 3.8845× 10−25kg

Δm = 3.8845× 10−25 kg − 232.037131μ ×1.6605× 10−27 kg

1μ= 3.8845× 10−25kg − 3.8530× 10−25kg = 3.15× 10−27 kg

E = Δmc2 = 13.15× 10−27 kg( )2.998× 108 m / s( )2= 2.83× 10−10 J ×

1MeV1.602 × 10−13J

= 1770MeV

5. 90 protons, 138 neutrons, 90 electrons mi = 90 1.6726× 10−27 kg( )+ 138 1.6749× 10−27kg( )+ 90 9.1094 × 10−31kg( )= 3.8175× 10−25kg

Δm = 3.8175× 10−25 kg − 228.028716μ ×1.6605× 10−27kg

1μ= 3.11× 10−27 kg

E = Δmc2 = 3.11× 10−27 kg( )2.998× 108 m / s( )2= 2.80 × 10−10J ×

1MeV1.602× 10−13 J

= 1750MeV

6. 8.7 MeV 7. 8.5 MeV 8. 7.7 MeV 23.7

1. T1/ 2 =0.693

λ=

0.6931.72× 104 / s

= 4.03× 10−5 s 2. T1/ 2 =0.693

λ=

0.6938.25× 10−6 / s

= 84,0 00s

3. λ =0.693T1/ 2

=0.69330.8s

= 2.25× 10−2 / s

ratio remaining after t seconds:

NNo

= e−λt = e− 2.25×10−2 /s( )240 s( )

= 4.52× 10−3

50.0g 4.52× 10−3( )= 0.226g

4. λ =0.693T1/ 2

=0.6933.82d

= 0.181 /d

ratio remaining after t days:

NNo

= e−λt = e− 0.181/ d( ) 10.0d( ) = 0.164

75.0g 0.164( ) = 12.3g

5. λ =0.693T1/ 2

=0.69330.8s

= 0.0225/ s


NNo

= e−λt = e− 0.0225/ s( ) 10.0s( ) = 0.799 = 79.9 %



184

6. λ =0.693T1/ 2

=0.693

4.47 × 109 yr= 1.55× 10−10 / yr


NNo

= e−λt = e− 1.55×10−10 / yr( )975yr( )

= 99.99 %

% remaining = 99.99% % delayed = less than 0.01%

7. λ =0.693T1/ 2

=0.693

4.50 × 109 yr= 1.54 × 10−10 / yr

N = N oe−λt = 5.50× 1020( )e− 1.54 ×10 −10 /yr( )2.45×109 yr( )= 3.77 × 1020atoms

8. λ =0.693T1/ 2

=0.693

5730yr= 1.21× 10−4 / yr

N = N oe−λt = 3.75× 1021( )e− 1.21×10 −4 /yr( )10 0 0yr( )

= 3.32 × 1021 atoms

9. λ =0.693T1/ 2

=0.693

5730yr= 1.21× 10−4 / yr

Ratio remaining after t years :

NNo

= e−λt = e− 1.21×10−4 / yr( )300 0yr( )

= 0.696

% remaining = 69.6% % decayed = 30.4%

10. λ =0.693T1/ 2

=0.6932.35s

= 0.295/ s


NNo

= e−λt = e− 0.295/s( ) 1.00s( ) = 0.745

% remaining = 74.5% % decayed = 25.5% 23.9

1. λ =0.693T1/ 2

=0.693

10.0 min= 0.0693 /min

A = Aoe−λt = 1.24Ci( )e− 0.0693/ min( ) 20.0min( ) = 0.310Ci

2. λ =0.693T1/ 2

=0.693

5370yr= 1.29 × 10−4 / yr

A = Aoe−λt = 2.64Ci( )e

− 1.29×10−4 /yr( )400 0yr( )= 1.58Ci

3. λ =0.693T1/ 2

=0.693

14.95h= 0.0464 /h

A = Aoe−λt = 0.476Ci( )e− 0.0464 / h( ) 36.5h( ) = 0.0875Ci

4. λ =0.693T1/ 2

=0.693

20.4 min= 0.0340 /min

A = Aoe−λt = 3.98Ci( )e− 0.0340 /min( ) 10.3h( ) 60 min/h( ) = 2.98× 10−9Ci

5. λ =0.693T1/ 2

=0.693

1590yr= 4.36 × 10−4 / yr

A = Aoe−λt = 10.0Ci( )e

− 4.36 ×10 −4 / yr( )125yr( )= 9.47Ci



185

6. λ =0.693T1/ 2

=0.6938.06d

= 0.0860 /d

A = Aoe−λt = 4.00Ci( )e− 0.0860 /d( ) 0.25d( ) = 3.91Ci

7. λ =0.693T1/ 2

=0.693

1.50 × 10−4 s= 4620 / s

A = Aoe−λt = 75.0× 10−36 Ci( )e− 4620 /s( ) 5.00×10−6 s( )= 73.3μCi

Chapter 23 Review Questions 1. a, b, and c 2. c 3. c 4. The ground state is the lowest allowed energy level for the electron in the atom. The excited states are

the higher energy levels, which are unstable. An electron in an excited state will in time decay through lower-energy-level excited states to the ground state. The ground state has the smallest “orbital distance” from the nucleus.

5. In the Bohr theory of the atom, the energy levels of the electrons are restricted to certain values; that is, the energy is quantized. The energy levels in a hydrogen atom are given by the equation E = −kZ 2 / n2 .

6. Protons and neutrons are both neucleons that exert the attractive strong force on other atoms. They both have similar masses, although the neutron is slightly more massive. The proton is a stable particle as an individual nucleon. The neutron, however, is unstable by itself. The proton has a positive charge, equal tin magnitude to that of the electron. The neutron is uncharged.

7. The electric force can either be attractive or repulsive and is exerted between charged particles. The strong force is always attractive and is exerted between nucleons, whether charged or not. The strong force is a very short range force. The electric force is exerted over large distances.

8. The nucleus would expand in size due to repulsive electrical force between protons. The nucleus might even break apart.

9. Mass and energy are equivalent forms. Mass can be changed into energy under the proper conditions, and vice versa.

10. All three atoms have 6 protons in the nucleus and 6 electrons in the orbital shells. They all exhibit the chemical properties of carbon. They each have a different number of neutrons in the nucleus and therefore have a different mass.

11. An electron volt is the energy gained or lost by an electron passing through a potential difference of 1 volt.

12. The neutron is the “glue” that binds the positively charged neutrons together in the nucleus. Without neutrons, the positively charged neutrons would be repelled by their similar electrical charge.

13. An alpha ray is composed of particles that have a double positive charge and are composed of two protons and two neutrons. The α particles are identical to the nucleus of a helium atom.

14. A γ ray is composed of very energetic photons of uncharged electromagnetic radiation. 15. A β ray is composed of negatively charged particles (electrons). 16. Enrico Fermi discovered that the neutron bombardment of uranium can result in the formation of

lighter nuclei that are approximately one-half the mass of uranium. 17. A self-sustaining chain reaction is a nuclear reaction in which a sufficient number of neutrons are

produced that will cause subsequent nuclear reactions to continue at a fixed rate. 18. Nuclear fusion reactions are reactions in which nuclei bombard each other, producing heavier nuclei.

The original nuclei “fuse” together. 19. In nuclear fission reactions, nuclei are bombarded by nuclear particles such as neutrons or α particles,

causing the original nuclei to split apart. 20. 6.25% 21. Molecules in plant or animal cells can be damaged as a result of changes in the chemicals caused by

nuclear reactions produced by the radiation. Some of these damaged molecules may be in the genetic material, which might cause cells produced by the division of this cell to be defective.



186

22. It is used for diagnostic purposes by the injection or ingestion of radioactive tracer materials that may allow the identification of cancerous or defective cells. It can also be used to treat cancer by bombardment of the cancer cells, therefore destroying the cancerous material.


1. ( ) ( )22

2 2

13.595 10.5438

5eVkZE eV

n= − = − = −

2. ( ) ( )22

2 2

13.595 11.511

3eVkZE eV

n= − = − = −

( ) ( )0.5438 1.511 0.967eV eV eV− − − =

3. He, Ne, Ar, Kr, Xe, Rn

4. Cu, Ag, Au, Fe, Ti, Cr,, Mn, etc.

5. 27

261.6605 1015.003065 2.4913 101x kgu x kgu

−−× =

6. 27

261.6605 1018.998404 3.1547 101x kgu x kgu

−−× =

7. 27

251.6605 10165.930292 2.7553 101x kgu x kgu

−−× =

8. 38 protons, 48 neutrons, 38 electrons

( ) ( ) ( )27 27 31 2538 1.6726 10 48 1.6749 10 38 9.1094 0 1.4399 10im kg kg kg kg− − − −= × + × + × = ×

27

25 25 271.6605 101.4399 10 85.909266 1.4265 10 1.34 101

kgm kg u kg kgu

−− − −×

= × − × = × = ×V

( ) ( )22 27 8 1013

11.34 10 2.998 10 / 1.20 10 7491.602 10

MeVE mc kg m s J MeVJ

− −−= = × × = × × =

×V

9. 8.7 MeV

10. 8.3 MeV

11. 8.1 MeV

12. 1/ 2

0.693 0.693 0.0450 /15.4

dayT days

λ = = =

( ) ( )( )0.0450 / 54.013 128.25 10 7.26 10day daystoN N e e atomsλ −−= = × = ×

13. 1/ 2

0.693 0.693 0.0862 /8.04

dayT days

λ = = =

( ) ( )( )0.0862 / 34.418 178.33 10 4.29 10day daystoN N e e atomsλ −−= = × = ×



187

14. 1/ 2

0.693 0.693 0.0238 /29.1

yrT yr

λ = = =

Ratio remaining after t years: ( )( )0.0238 / 2.40 0.944yr yrt

o

N e eN

λ −−= = =

% remaining: 94.4%

% decayed: 5.6%

15. 1/ 2

0.693 0.693 0.0450 /15.4

dayT days

λ = = =

Ratio remaining after t days: ( )( )( )0.0450 / 2.00 1 / 24 0.9963day h day ht

o

N e eN

λ −−= = =

% remaining: 99.63%

% decayed: 0.37%

16. 1/ 2

0.693 0.693 0.0450 /15.4

dayT days

λ = = =

( ) ( )( )0.0450 / 43.32.43 0.346day daystoA A e Ci e Ciλ −−= = =

17. 1/ 2

0.693 0.693 0.0693/ min10.0minT

λ = = =

( ) ( )( )0.0693/ min 43.0 min3.79 0.193toA A e Ci e Ciλ −−= = =

18. 1/ 2

0.693 0.693 0.0340 / min20.4minT

λ = = =

( ) ( )( )0.0340 / min 95.4 min9.41 0.367toA A e Ci e Ciλ −−= = =

19. 109

1/ 2

0.693 0.693 1.5 10 /4.5 10

yrT yr

λ −= = = ××

( ) ( )( )10 91.5 0 / 1 106.75 5.8yr yrtoA A e Ci e Ciλ

−− × ×−= = =

20. 1/ 2

0.693 0.693 0.0259 / min26.8minT

λ = = =

( ) ( )( )0.0259 / min 60.0 min650.0 10 10.6toA A e Ci e Ciλ μ−− −= = × =



188


1. (a) ( ) ( )22

42 2

13.595 10.84969

4

eVkZE E eVn

−−= = = = −

( ) ( )22

22 2

13.595 13.3988

2

eVkZE E eVn

−−= = = = −

0.84969 3.3988 2.5491 ( #4 #2)eV eV eV from to− + =

(b) 14

3419

2.5491 6.163 1016.626 10

1.602 10

E eVf HzeVh J s

J−

−

= = = ×⎛ ⎞× ⋅ ⎜ ⎟×⎝ ⎠

(c) 83.00 0 /c m s= ×

(d) 8

714

3.00 10 / 4.87 106.163 10

c m s mf Hz

λ −×= = = ×

×

2. 8

157

3.00 10 / 2.46 101.22 10

c m sf Hzmλ −

×= = = ×

×

( ) ( )34 1519

16.626 10 * 2.46 10 10.21.602 10trans

eVE hf J s Hz eVJ

−−

⎛ ⎞= = × × =⎜ ⎟×⎝ ⎠

1 10.2 13.595 3.40n transE E E eV eV eV−= + = + = −

2

2

kZEn

−=

( ) ( )22 13.595 12.00

3.40

eVkZnE eV

−−= = =

−

3. (a) 13

121.602 028.40 5.00 101

JE MeV JMeV

−−⎛ ⎞×

= = ×⎜ ⎟⎝ ⎠

(b) 2 protons, 2 neutrons, 2 electrons (c) 2 2 2e p nm m m m= + +

( ) ( ) ( )31 27 272 9.1094 0 2 1.6726 10 2 1.6749 10kg kg kg− − −= × + × + ×

276.6968 10 kg−= ×

(d) ( )

1229

2 28

5.00 10 5.56 103.00 10 /

E Jm kgc m s

−−×

= = = ××

V

27 29 276.6968 10 5.56 10 6.64 10f om m m kg kg kg− − −= − = × − × = ×V

4. (a) 1/ 2

0.693 0.693 0.181/3.82

dT d

λ = = =

( )( )

2324

0.181/ 10

2.45 10 1.50 10o t d d

N atomsN atomse eλ− −

×= = = ×

5. ( ) ( )

1/ 2

0.6930.693 32.414.95

0.0875 0.0875 0.393o tht hT

A CiA Cie

eeλ− ⎛ ⎞ ⎛ ⎞−− ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= = = =



189

Chapter 24 24.2 1.

E = mc2 = 2.30 × 0−3 kg 3.00 × 108 m / s( )2

= 2.07 × 1014 J

E = mc2 = 2.30 × 10−3 kg 3.00 × 108 m / s( )2

= 2.07 × 1014 J

2. E = mc2 = 1.30kg 3.00 × 108 m / s( )2

= 1.17 × 1017 J

3. m = E / c2 =

600J3.00 × 108( )m / s

= 6.67 × 10−15 kg

4. m = E / c2 =

67.0J3.00 × 108( )m / s

= 7.44 × 10−16 kg

Chapter 24 Review Questions 1. c 2. d 3. Physics is the same for moving and non-moving objects. 4. Perhaps you are in a car and flip a coin. The coin flips into the air and falls back into your hand the

same as it would if you were standing on the side of the road. 5. It would travel at 14 m/s because the ball was already traveling at 10 m/s while attached to a bike. 6. The speed of light is not relative, but is constant. 7. The light would only travel at 3.00 × 108 m / s . 8. Time passes slower than someone not traveling close to the speed of light. 9. Relative to the ground (considering I am not walking and reading), just the time dimension. 10. That energy and mass are equivalent, just in different forms. Mass has energy and energy has mass. 11. Spatially we exist in three dimensions (x, y, z), yet we also move through the dimension of time.

Space and time are needed to locate a particular occurrence in the universe. 12. Gravity and acceleration are the same. 13. Light can be warped around massive objects. 14. The light curved around the warping of space-time that was created by the mass of the sun and the

moon. Chapter 24 Review Problems 1. v = 2.00mi / h

2. v = 6.70mi / h

3. v = 65.0mi / h − 2.00mi / h = 63.0mi / h

4. E = mc2 = 9.10 × 10−31 kg 3.00 × 108 m / s( )2

= 8.19 × 10−14 J

5. E = mc2 = 1.67 × 10−27 kg 3.00 × 108 m / s( )2

= 1.50 × 10−10 J

6.

m =Ec2 =

9.80J

3.00 × 108 m / s( )2 = 1.09 × 10−16 kg



190

Chapter Tests



191

CHAPTER 1: THE PHYSICS TOOLKIT 1. The metric prefix for 0.010 is .

2. The metric symbol, or abbreviation, for milli is .

Write the following numbers in scientific notation:

3. 45,000

4. 0.0000742

Write the following number in decimal form:

5. 3.28 x 10-4

6. 2.65 x 108

Fill in each blank: (Round to three significant digits when necessary.)

7. 1.2 km = m

8. 6.9 L = ml

9. 750 μm = mm

10. 670 mm2 = cm2

11. 6500 cm3 = m3

12. 15 kg = lb

13. 96 in. = yd

14. 15 ft3 = m3

Determine the accuracy (the number of significant digits) in each measurement:

15. 20,900 m

16. 0.04060 s

17. 3.6 x 10-12 km

Determine the precision of each measurement:

18. 23.9 m

19. 14,050 ft



192

20. 5.0 x 104s

21. Use the rules of measurement to add 18.5 m; 2070 cm; 95.25 cm; 0.045 m.

Use the rules for multiplication and division of measurements to find the value of:

22. ( )( )( )2

36.0 451.35kg m

s

23. Find the area of a rectangle 22.6 m long and 4.60 m wide.

24. Find the volume of a rectangular box 18.0 cm long, 9.0 cm wide, and 6.00 cm high.

Express each of the following values using Scientific Notation:

25. 0.0001042 mi.

26. 6,800,000 in.

27. 0.00481 L

28. 0.000000015 gal

29. 127,500 cm

30. 0.00008207 km

31. 89,600,000 g

32. 18,000 in.

Perform the following conversions:

33. 47.5 ft3 = gal = L

34. 200 m2 = ft2 = cm2

35. 923 m3 = yd3 = gal

36. 55.0 gal = L = ft3



193

CHAPTER 2: PROBLEM SOLVING

1. What is the difference between a subscript and an exponent?

2. What is the difference between a formula and a working equation?

3. What is the purpose of estimation when problem solving?

4. Solve for m in the formula F = ma

5. Solve for t in the formula s = ½ (vf + vi)t

6. Solve for vf in the formula s = ½ (vf + vi)t

7. Solve for h in PE = mgh

8. Given A = ½ bh, if b = 10.0 cm and h = 12.2 cm, what is A?

9. A cone has a volume of 315 cm3 and a radius of 7.50 cm. What is its height?

10. A right triangle has a side of 82.4 mm and a side of 19.6 mm. Find the length of the hypotenuse.

11. Given a cylinder with a radius of 14.4 cm and a height of 16.8 cm, find the lateral surface area.

12. A rectangle has a perimeter of 80.0 cm. One side has a length of 28.0 cm. What is the length of the adjacent side?

13. The formula for the volume of a cylinder is V = πr2h . If V = 4520 m3 and h = 36.0 m, find r.

14. The formula for the area of a triangle is A = ½ bh. If b = 3.12 m and A = 82.6 m2, find h.

15. A rectangular parking lot measures 80.0 m by 75.0 m. If the parking lot needs three sections that each measure 8.00 m by 8.00 m for tree plantings, how much area is left for parking spaces?



194

CHAPTER 3: VECTORS

1. Explain the difference between a scalar and a vector quantity.

2. Explain the difference between distance and displacement.

3. Using a scale of 1.0 cm = 25 km, find the length of the vector that represents a displacement of 450 km.

Use graph paper to find the resultant of each set of displacements in the following problems.

4. 45 km due east, then 45 km due north, and then 45 km due east.

5. 85 km due south, then 25 km 45° south of east, then 45 km due east, and then 45 km at 25° north of east.

Use a calculator to find each trigonometric ratio rounded to four significant figures in the following problems.

6. sin 18.5°

7. tan 67.25°

8. cos 25.075°

Use a calculator to find each angle rounded to the nearest tenth of a degree for the following problems.

9. sin A = 0.4639

10. tan B = 0.1679

11. cos A = 0.8462

12. A right triangle has legs of 24.0 cm and 36.0 cm. Find the length of the hypotenuse.

13. A right triangle has a hypotenuse with length of 85.0 cm and one leg with 55.5 m. Find the length of the other leg.

14. A right triangle has one acute angle that measures 34.5°. What is the measure of the other acute angle?



195

Solve each triangle (find the missing angles and sides):

15.

16.

17. An antenna is attached to a vertical pole. If a guy wire is attached 40.0 ft from the bottom of the pole and to an anchor on level ground 25.0 ft from the base, what angle does the guy wire make with level ground?

18. A conveyor is used to lift grain to a bin. The angle of elevation of the conveyor is 48.5°. If the grain is elevated 15.0 m, how long is the conveyor?

19. Find the x- and y-components of the resultant vector R and graph the resultant vector R.

Vector x-component y-component

A +4 -6

B -6 -3

C -5 +2

D -1 +4

20. Find the x- and y-components of the vector v = 45.0 m/s at 205.0° in standard position.



196

CHAPTER 4: MOTION

1. While driving at 75 km/h, how far can you travel in 1.25 h?

2. If Grandma lives 225 km away, how long will it take to drive to her house if you average 95 km/h?

3. A trucker drove from San Francisco to Washington, DC (2865 mi). The entire trip required 77 h and 30 min. If 16 h and 30 min of that time was spent sleeping and 3 h 40 min was used for meals and refueling, what was the average speed while on the road?

Change 95.0 km/h to the following speed units in the following problems:

4. m/s

5. mi/h

6. ft/s

7. A sports car accelerates from a standing start to 65 mi/h in 4.61 s. a. Find its acceleration in ft/s. b. How far can it travel in that time?

8. A motorcycle traveling at 45.0 mi/h accelerates to 60.0 mi/h in 5.00 s. a. Find its acceleration in ft/s. b. How far can it travel in that time?

9. An airplane can average 450 mi/h with enough fuel for a flight of 3.5 h. How far can it travel?

10. An archer shoots an arrow into a piece of wood. The arrow is traveling at 120 km/h when it strikes the wood. The arrow penetrates 3.8 cm into the wood before stopping. What is its average acceleration (in m/s) into the wood?

11. A rock is dropped from a bridge to the water below. It takes 3.2 s for the rock to hit the water.

a. Find the speed (in m/s) of the rock as it hits the water. b. How high (in m) is the bridge above the water?

12. A bullet is fired vertically upward from a gun with an initial velocity of 225 m/s. a. How high does the bullet go? b. How long does it take to reach its maximum height? c. How long is it in flight?

13. A ball is thrown straight up with an initial velocity of 20.0 ft/ s from a deck that is 30.0 ft above the ground.

a. How long does it take to reach its maximum height? b. What maximum height above the deck does it reach? c. At what speed does it hit the ground?



197

d. What total length of time is the ball in the air?

14. A penny is dropped from a 70Ō ft high bridge. a. How long does it take for the penny to reach the ground below? b. What is the velocity of the penny upon impact?

15. It takes 4.20 s for a ball that was thrown down from the top of a building to reach the sidewalk below. The ball’s velocity upon striking the sidewalk is 148 mi/h.

a. How tall is the building? b. What is the initial velocity of the ball?

16. A dragster runs the quarter mile in 8.96 s. a. What is the car’s velocity at the finish line? (Provide an answer in both ft/s

and mi/h) b. What is the car’s average acceleration? (Provide an answer in both ft/s2)

17. A stone is dropped from a cliff to the stream below. It takes 6.28 seconds for the stone to hit the water.

a. How fast, in m/s, is the stone traveling when it hits the water? b. How high, in m, is the cliff above the stream?

18. A baseball is hit out to centerfield. Explain how the vertical and horizontal components of its velocity change on the way up and on the way back down to the ground.

19. Using the same baseball as in the previous problem, explain how the vertical and horizontal components of the acceleration change during the ball’s flight.

20. A cannon launches a cannonball horizontally off a cliff that is 25.5 m high. If the cannonball is fired with a velocity of 28.3 m/s, how far from the edge of the cliff will the ball land?

21. The cannon in the previous problem is brought down off the cliff. If the cannon fires again, with a velocity of 28.3 m/s and at an angle of 30°, how far will the cannon ball travel?

22. Explain what happens to the range of projectiles as the angle of launching changes from 0° to 45°. What happens to the range of projectiles as the angle of launch changes from 45° to 90°?



198

CHAPTER 5: FORCE 1. Explain inertia. 2. Find the acceleration produced by a total force of 165 N on a mass of 4.50 kg. 3. Find the force necessary to give a 1750- kg car an acceleration of 11.0 m/s 2. 4. Explain the difference between starting friction and sliding friction. Which has a

greater coefficient of friction? 5. A wooden crate is dragged across a concrete floor. What are two things that can

be done to decrease the coefficient of friction between these two surfaces? 6. The coefficient of sliding friction on concrete is 0.30. What weight of steel can

be pulled across a concrete floor by a winch with a 450 lb capacity? 7. A sled weighing 3350 N is pulled over snow at uniform speed by exerting a force

of 1200 N. Find the coefficient of friction. 8. Find the net force including its direction of the following: 175 N to the left; 234

N to the right; 75 N to the right; 225 N to the left; 45 N to the right. 9. A cyclist and her bicycle have a combined mass of 75.0 kg. If the frictional force

acting against the motion of the bicycle is -23.5 N, what force must the cyclist apply to maintain a constant velocity?

10. The cyclist and her bicycle (combined mass of 75.0 kg) apply a force of 34.2 N.

If the frictional force is still –23.5 N, what will be the acceleration of the bicycle? 11. A small railroad engine is able to exert a 17,500 lb pulling force. The coefficient

of friction for steel wheels on steel rail is 0.40. How many rail cars weighing 45,000 lb each can this engine pull on level track?

12. Find the acceleration of a car of mass 1650 kg acted upon by a driving force of

2350 N that is opposed by a frictional force of 450 N. 13. What is the difference between mass and weight? 14. A scale tells a person that he weighs 167 lbs. Since weight is a force, why isn’t

the person accelerating? 15. Find the weight of a person with mass 68.0 kg. 16. Find the mass of a motorcycle that weighs 775 lb.



199

17. Find the weight of a person with mass 63.0 kg on the moon where g = 1.63 m/s 2. 18. Explain the law of action and reaction. Provide an example of this law. 19. A 64.5 kg person steps off a 129 kg rowboat with a force of 34.0 N. What is the

force that is applied to the person by the rowboat? 20. Referring to the previous problem, which object will have the larger acceleration?



200

CHAPTER 6: MOMENTUM

1. Provide a non-mathematical definition for momentum. 2. Which object has a greater momentum, a stationary tractor-trailer or a tossed

tennis ball? Explain your answer. 3. Find the momentum of a tennis ball of mass 60.0 g served with velocity 65.0 m/s. 4. Explain how a tennis racket can apply an impulse to the tennis ball. 5. If a tennis racket strikes a ball (m = 60.0 g; v = 65.0 m/s) for a period of 7.25 x

10-2 s, what force does the tennis racket apply to the ball? What is the impulse of this interaction?

6. Find the mass of a projectile when fired at 275 m/s if it is to have the same

momentum as a 145- g projectile fired at 325 m/s. 7. Using the concept of impulse, explain the benefits of “crumple zones” in

automobile collisions. 8. Why does a bouncing ball experience a greater impulse than a ball that strikes the

surface and stops? 9. Explain the difference between elastic and inelastic collisions. 10. Provide an example of an elastic collision and an inelastic collision.

Use the following information to answer the questions below: A 75.0-g bullet is fired at a muzzle velocity of 476 m/s from a gun with a mass 4.75 kg and a barrel length with 60.0 cm.

11. How long is the bullet in the barrel? 12. What is the force of the bullet while it is in the barrel? 13. Find the impulse given to the bullet while it is in the barrel.

14. Find the bullet’s momentum as it leaves the barrel.

Use the following information to answer the questions below: A person is driving an automobile at 95.0 km/h and throws a bottle of mass 0.600 kg straight out the window.

15. With what momentum does the bottle hit a roadway sign?



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16. With what momentum does the bottle hit an oncoming automobile traveling in the opposite direction at 85.0 km/h?

17. With what momentum does the bottle hit an automobile passing and traveling in

the same direction at 115.0 km/h?



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CHAPTER 7: CONCURRENT AND PARALLEL FORCES

1. Find the sum of the following set of forces acting on the same point in a straight line: 745 N (right); 372 N (left); 427 N (left); 555 N (left); 965 N (right).

2. Forces of 348 N and 295 N act on the same point. a. What is the magnitude of the maximum net force the two forces can exert together? b. What is the magnitude of the minimum net force the two forces can exert together?

Find the sum of each set of vectors. Give angles in standard position.

3. 4. 5. Forces of F1 = 975 N, and F2 = 745 N, and F3 = 1175 N are applied to the same

point. The angle between F1 and F2 is 60.0°. The angle between F2 and F3 is 30.0° . F2 is between F1 and F3. Find the resultant force.

6. Three forces, each of magnitude 1450 lb act on the ame point. The angle between

adjacent forces is 45.0°. Find the resultant force. 7. A given wire can support 4550 lb before it breaks. How many 575 lb weights can

it support without breaking?



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Find the forces F1 and F2 that produce equilibrium in each free-body diagram. 8. 9. 10. Find the tension in the cable and the compression in the support of the sign shown

below.



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11. Find the tension in each cable.

12. Determine the Resultant and Equilibrium for each of the following vector systems.

13. Determine the tension of ‘T’ of the cable and the compression ‘C’ on the boom assembly of the crane.



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14. The first condition of equilibrium states that a. all parallel forces must be zero. b. all perpendicular forces must be zero. c. all frictional forces must be zero.

15. In the second condition of equilibrium

a. clockwise and counterclockwise torques are unequal. b. clockwise and counterclockwise torques are equal. c. there are no torques.

16. If the sum of the opposing forces is zero, can there still be rotation? 17. Is it possible to select different points of rotation in a nonconcurrent force

problem? 18. A hanging sign weighs 3950 N. If the tension of one vertical support cable is

980 N, what is the tension in the other vertical support cable? 19. A platform supports a bricklayer and bricks together weighing 450 lb. If the

compression force in one end support is 290 lb, what is the supporting force in the other?

20. If ∑ F = F1 + F2 when∑ F =110.0N and F2 = 76.0N , find F1 . 21. If ∑Tclockwise = ∑Tcounterclockwise and∑Tclockwise = 40.0N( ) 5.00m( ), find ∑Tcounterclockwise . 22. Two ladders which support the ends of a scaffold support 100.0 kg each. A 75.0-

kg worker is on the scaffold with a pile of bricks. Find the mass of the bricks. 23. Solve for Fw :

12.0 ft( ) 6.00lb( )+ 20.0 ft( ) Fw( )= 10.0 ft( ) 9.00lb( )+ 20.0 ft( )10.0lb( ) 24. How far from the heavy end of a 68.0-cm long bat would its center of gravity be if

it is at one-fourth of the length of the bat? 25. A bridge with mass 9000 kg supports a truck with mass 3000 kg that is stopped

in the middle of the bridge. What mass must each pier of the bridge support? 26. If the truck in the preceding problem stops 9.00 m from the end of the 32.0-cm

bridge, what mass must each pier of the bridge support?



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CHAPTER 8: WORK AND ENERGY 1. Find the work done by a person pulling a wagon containing 60.0 lb of bricks across

36.0 ft of level floor by exerting a constant force of 21.0 lb. 2. How much work (in ft lb) is done by a large bucket that can lift 4.5 tons of dirt 14 ft

into the air and dump it into a 5.0 yd 3 dump truck? 3. A mechanical lift raises a 325 kg automobile 2.00 m for servicing. How much work

was done by the lift? 4. A pallet is pulled 125 m across a floor by a cable that makes an angle of 45° with the

floor. If 1150 N is exerted on the cable, how much work is done? 5. How much power is required to lift a 575 lb weight 30.0 ft in 4.00 s? 6. How much power is required to lift a 225 kg body 15.0 m in 5.00 s? 7. How much time is required to lift a 1275 lb object 45.0 ft with a motor that produces

15.0 hp? 8. How much time is required to lift a 525 kg object 25.0 m with a motor that produces

2.50 kW? 9. What maximum mass can be lifted 45.0 m in 1.00 min by a 1.50 kW motor? 10. A pump is capable of developing 5.0 kW of power. How many liters of water per

minute can be lifted a distance of 25 m? (1 L of water has a mass of 1 kg.) 11. Find the potential energy of a 1250 lb counter balance in an elevator that is raised

45.0 ft from its lowest point. 12. Find the kinetic energy of a 310 kg wrecking ball moving at 4.0 m/s. 13. Find the kinetic energy of a 5.0 ton truck traveling at 65 mi/h. 14. Water is pumped at the rate of 275 m 3/min from a lake into a tank on a hill 75.0 m

above the lake. (1 m 3 = 1000 L) a. What power (in kW) must be delivered by the pump? b. What horsepower rating must this pump motor have? c. What is the increase in potential energy of the water each minute? d. Assuming 25% of the pump power is lost, what power (in kW) must be delivered by the pump?

15. A pile driver weighs 975 lb and falls from a height of 45.0 ft. Find its velocity as it hits the pile.



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16. A 55.0 g bullet is fired vertically with an initial velocity of 123 m/s. a. What is its velocity at its highest point of travel? b. What maximum height does it reach? c. At what velocity does it hit the ground?

17. A 2.00 lb rock is dropped from a bridge to the water 50.0 ft below. a. What is its velocity as it hits the water? b. How long does it take to hit the water? c. What is its kinetic energy as it hits the water?

18. A 2.50 N rock is thrown downward from a cliff at 10.0 m/s. a. What is its velocity after falling 25.0 m? b. What is its velocity as it hits the water 125 m below? c. What is its kinetic energy as it hits the water?

19. A farmer lifts a dozen 100 lb sacks of grain from the ground to the bed of his truck 3.50 ft above the ground. What is the total work done?

20. A 1.0 ton beam is raised 35 ft in 2.0 minutes.

a. How much work was done? b. How much power (in horsepower) did it take?

21. An electric motor running on a 120 volt receptacle is drawing 8.00 Amps of current. What is the motor’s power? Express your answer in both watts and horsepower.



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CHAPTER 9: ROTATIONAL MOTION 1. Convert 15.0 revolutions to (a) radians and (b) degrees. 2. A motorcycle wheel turns 100 π rad during 45.0 s. What is the angular velocity of

the wheel? 3. A trailer tire turns at 72.0 rpm and has a radius of 15.0 cm. What is the linear speed

of a point on the outside of the tire? 4. A woman changes a flat tire with a tire iron 50.0 cm long. She exerts a force of 53.0

N. How much torque does she produce? 5. A torque of 81.0 ft lb is produced by a torque arm of 3.00 ft. What force is applied? 6. A model helicopter pulls into a tight circular curve of radius 30.0 m. The 300 g

helicopter is traveling at 90.0 km/h. What is the helicopter’s centripetal force? 7. A toy is spun in a circle with a centripetal force of 2.40 lb. The toy has mass of 0.650

slug and is attached to a string 2.00 ft long. At what linear velocity does it travel? 8. Explain the difference between inertia and the moment of inertia. 9. A spinning figure skater tucks her arms and legs closer to the axis of rotation while

spinning. Her body rotates much faster in the tucked position than the non-tucked or extended position. This is an example of what physical law?

10. A boy riding his scooter creates a torque of 1.20 ft lb with an angular speed of 45.0

rpm. How much power does he produce? 11. A motorcycle generates 500 W of power. The torque is 50.0 N m. What is the

angular velocity? 12. A large wheel of diameter 5.00 cm turns a smaller one of diameter 2.00 cm. The

smaller turns at 15.0 rpm. Find the rpm of the larger. 13. A shaft is driven by two gears. The first gear with 40 teeth revolves at 60.0 rpm. The

second gear revolves at 80.0 rpm. How many teeth does the second gear have? 14. One gear with thirteen teeth rotates at 115 rpm and turns a second gear with 26 teeth.

Find the rpm of the second gear. 15. A gear train has 9 shafts. Do the first and last gears rotate in the same direction? 16. A pulley of diameter 15.0 cm is driven by a motor and revolves at 100 rpm. The

pulley drives a second one with diameter 10.0 cm. Find the rpm of the second pulley.



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17. A pulley with diameter 5.00 cm rotates at 100 rpm. Find the diameter of the second

pulley if it is driven at 250 rpm by the first. 18. How does the presence of an idler gear affect the relationship between a driver gear

and a driven gear in a gear train? 19. When the number of shafts in a gear train is four, do the first and last gears rotate in

the same or opposite direction? 20. Why may a gear that is both a driver gear and a driven gear be omitted from a

computation?



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CHAPTER 10: SIMPLE MACHINES

1. What is the purpose of using simple machines?

2. If more than one simple machine is used in a mechanism, it is called a complex machine. Provide three examples of complex machines.

3. Define efficiency for a simple machine.

4. Is it possible to have a 100% efficient machine? Explain.

5. A sport fisherman catches a swordfish on his fishing pole. What class lever is the fishing pole if the end of the pole is positioned in a fixed fulcrum? Will the fishing pole have a Mechanical Advantage greater or less than 1.00?

6. With respect to simple machines, explain why trucks and busses have larger steering wheels than sports cars.

7. A mechanic lifts the front of a car with a jack. Find the mechanical advantage if a force of 30.0 lb on the jack handle lifts the car that weighs 960 lb.

8. What could be done to the jack to increase the mechanical advantage of the jack in the previous problem?

9. A worker exerts 85.0 lb on a lever to lift a weight. The distance from the worker to the fulcrum is 8.00 ft; the distance from the weight to the fulcrum is 2.00 ft. Find the weight lifted. Find the mechanical advantage of the lever.

10. A worker uses a wheelbarrow to haul 225 lb of brick. The handles are located 4.00 ft from the axle; the center of the load is 1.50 ft from the axle. How much upward force does the worker exert to lift the handles of the wheelbarrow?

11. A winch has a handle with a radius of 18 in. and a drum of radius 3.0 in. How much force is required to lift a 120 lb weight?

12. A system consisting of two fixed and two movable pulleys has a mechanical advantage of 4.00. If a force of 124 N is exerted, what weight can be raised?

13. Find the greatest mechanical advantage of a pulley system with two fixed and two movable pulleys.

14. Find the mechanical advantage of a ramp 18.0 m long and 3.00 m high. Find the force required to pull a box of mass 975 kg up the ramp. Disregard friction.

15. A winch capable of pulling 250 lb must pull a 2800 lb load up a ramp of height 5.5 ft. Find the length of the shortest ramp for this job.



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16. A house jack must lift 112,000 lb. The screw pitch is 0.125 in. and the screw diameter is 3.00 in. Find the length of the handle to lift the load when a person applies a force of 75.0 lb to the handle.

17. If the pitch in the previous problem is increased, would that increase or decrease the mechanical advantage of the house jack?

18. Find the mechanical advantage of a pipe vise if it has a handle diameter of 16 in., a screw diameter of 0.75 in., and a screw pitch of 0.125 in.

19. A load is to be pulled up an inclined plane using a pulley system. The inclined plane is 25.0 ft long and 5.00 ft high. The pulley system has 2 fixed pulleys and 1 movable pulley. Find the mechanical advantage of this compound machine.

20. A direct drive winch with a handle radius of 12 in. and a drum diameter of 3.0 in. is used to pull the end of a lever that is 8.0 ft long. The fulcrum is 6.0 ft from the end on which the winch is pulling. If a 240 lb load is attached to the short end of the lever, how much effort is required at the winch handle to lift the load?



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CHAPTER 11: UNIVERSAL GRAVITATION & SATELLITE MOTION

1. What are the two main factors that determine gravitational attraction?

2. The distance between two objects doubles, how does that effect the gravitational force between the objects?

3. The mass of one objects doubles, how does that effect the gravitational force between the objects?

4. Explain what the “G” in the Universal Gravitation formula represents.

5. Two 65.0 kg ice skaters are standing 2.00 m away from each other in the ice. What is the gravitational attraction between the two skaters?

6. Using the gravitational force from the previous problem, what will be the acceleration of the skaters towards each other as a result of this attraction?

7. What is the gravitational force felt between the earth and the sun? (The sun’s mass = 1.99 x 1030 kg, the earth’s mass = 5.97 x 1024 kg, the distance between the sun and the earth = 1.50 x 1011m)

8. What is the gravitational force felt between the earth and the moon? (The earth’s mass = 5.97 x 1024 kg, the moon’s mass = 7.35 x 1022 kg, the distance between the earth and the moon = 3.84 x 108 m)

9. Explain why the gravitational force between the sun and the earth is much stronger than the gravitational force between the earth and the moon even though the earth and the moon are much closer to one another than the sun and the earth.

10. What is an orbit and what force is responsible for maintaining a satellite’s orbit?

11. What two variables determine the orbital velocity of a satellite?

12. Find the orbital velocity of the earth as it orbits the sun.

13. Find the orbital velocity of the moon as it orbits the earth.

14. What would happen to the orbital velocity if the distance between the objects increased?

15. What role does the satellite’s mass play in calculating orbital velocity?

16. Find the orbital period of the earth as it orbits the sun.

17. Find the orbital period of the moon as it orbits the earth.



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18. What would happen to the orbital period if the distance between the objects increased?

19. What role does the satellite’s mass play in calculating orbital period?

20. Comets that orbit the sun tend to have major elliptical orbits. What happens to the gravitational force between the comet and the sun when the comment is positioned closer to the sun?



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CHAPTER 12: MATTER

1. Define tension and provide an example of where tension can be found.

2. Define compression and provide an example of where compression can be found.

3. Define shearing and provide an example of where shearing can be found.

4. Define torsion and provide an example of where torsion can be found.

5. Define bending and provide an example of where bending can be found.

6. A force of 32.5 N stretches a spring of 0.500 cm. What force will stretch a similar spring 1.60 cm?

7. A force of 7.50 N is applied to a spring whose spring constant is 0.298 N/cm. Find its change in length.

8. The vertical steel columns of an office building each support a weight of 50, 000 lb, with each being compressed 0.0260 in. What is the spring constant of the steel? What would the compression be if the weight were 60, 000 lb?

9. Find the weight density of a block 7.00 in. x 6.50 in. x 8.00 in. that weighs 500 lb.

10. A cylindrical piece of aluminum is 6.00 cm tall and 2.00 cm in radius. How much does it weigh?

11. A piece of metal has a mass of 9.00 kg. If it displaces water that fills a container 10.0 cm x 10.0 cm x 10.0 cm, what is this mass density of the metal?

12. A block of wood is 27.7 in. x 36.3 in. x 12.4 in. and weighs 197 lb. Find its weight density.

13. Find the volume (in cm3) of 800 g of mercury.

14. Find the volume (in m3) of 2000 g of hydrogen at 0° C at 1 atm of pressure.

15. Find the mass of 1000 m3 of oxygen at 0° C and 1 atm of pressure.

16. Find the weight of1000 ft3 of water.

17. What is the weight density of a block having dimensions of 4.27 in. x 3.87 in. x 5.44 in. weighing 0.872 lb?

18. Find the weight density 1.00 qt of liquid weighing 4.00 lb.



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19. A quantity of liquid weighs 4.65 lb and has a weight density of 50.0 lb/ft3. What is its volume?

20. The density of a metal is 700 kg/m3. Find its specific gravity.

21. A solid displaces 5.00 gal of water and has a weight density of 135 lb/ft3. What is its weight?



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CHAPTER 13: FLUIDS

1. What is the hydraulic principle (or Pascal’s Principle)?

2. How does an airplane wing provide lift?

3. Describe the difference between absolute and gauge pressure.

4. Is the pressure on a small piston different from the pressure on a large piston in the same hydraulic system? Are the forces on the two pistons the same?

5. Why must the thickness of a dam be larger at the bottom than at the top?

6. Find the pressure (in kPa) at the bottom of a water-filled drum 4.00 m high.

7. Find the depth in a lake at which the pressure is 300 lb/in.2.

8. Find the height of a water column where the pressure at the bottom of the column is 297 kPa.

9. What is the total force exerted on the bottom of a rectangular tank 9.00 ft2 by 4.00 ft deep?

10. Find the water pressure (in kPa) at the 35.0 m level of a 55.0 m water tower.

11. Find the total force on the bottom of a cylindrical water tower 55.0 m high and a 8.00 in radius.

12. Find the total force on the side of a cylindrical water tower 35.0 m high and 8.00 m in radius.

13. Find the total force on the side of a rectangular water trough 1.25 m high by 1.55 m by 3.00 m.

14. Find the water pressure on the ground to supply the third floor (9.00 m up) with a pressure of 300 N/cm2.

15. A submarine is submerged to a depth of 1650 ft in the Atlantic Ocean. What air pressure (in kPa) is needed to blow water out of the ballast tanks?

16. The area of the large piston in a hydraulic jack is 4.75 in.2. The area of the small piston is 0.564 in.2. What force must be applied to the small piston if a weight of 1000 lb is to be lifted?

17. What is the mechanical advantage of the hydraulic jack in the preceding problem?

18. The MA of a hydraulic jack is 400 . What force must be applied to lift an automobile weighing 12,500 N?



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19. Find the absolute pressure in a motorcycle tire with a gauge pressure of 200 kPa.

20. Find the gauge pressure of a tank with absolute pressure of 655 kPa.

21. A rock weighs 50.0 N in air and 42.6 N in water. What is the buoyant force on the rock?



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CHAPTER 14: TEMPERATURE AND HEAT TRANSFER

1. Distinguish between heat and temperature.

2. Which would cool a hot object better: 10 kg of water at 0° C or 10 kg of ice at 0° C? Explain.

3. Describe why automotive cooling systems are designed to operate at elevated levels.

4. Change 320 °R to degrees Fahrenheit.

5. Change 24° C to Kelvin.

6. Change 500 Co to degrees Fahrenheit.

7. Change 1000 Fo to degree Celsius.

8. Why do aluminum bleachers feel colder on a cold day than wood bleachers at the same temperature?

9. Find the amount of heat in cal generated by 100J of work.

10. Find the amount of heat in kcal generated by 7000J of work.

11. Find the amount of work that is equivalent to 10, 000Btu .

12. Find the heat flow during 5.00 h through a glass window of thickness 0.15 in. with an area of 35 ft2 if the average outer surface temperature is 22°F and the average inner temperature is 48°F.

13. How many Btu of heat must be added to 900 lb of steel to raise its temperature from 20.0°F to 500 Fo ?

14. A 100 kg steam boiler is made of steel and contains 200 kg of water at 5.00° C. How much heat is required to raise the temperature of both the boiler and water to 100 °C?

15. A brass rod 50.0 cm long expands 0.0734 cm when heated. Find the temperature changes.

16. The diameter of a hole drilled through aluminum at 20.0°C is 0.750 cm. Find the diameter of the hole at 90.0°C.

17. Find the increase in volume 50.0 L of acetone when heated from 40.0°C to 75.5°C.

18. Explain why burns caused by steam are more dangerous than burns caused by hot water.



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19. How many kcal of heat are required to vaporize 25.0 kg of water at 100°C?

20. How many Btu of heat are required to melt 9.00 lb of ice at 32.0°F?



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CHAPTER 15: PROPERTIES OF GASES

1. If the temperature of a gas is constant and the volume is increased, the pressure will .

2. If the temperature of gas is constant and the pressure is increased, the volume will .

3. If the pressure on a gas is constant and the temperature is increased, the volume will .

4. If the pressure on a gas is constant and the volume is increased, the temperature will .

5. A gas occupies 15.0 ft3 at 40.0°F. Find the volume of this gas at 90.0°F if the pressure is held constant.

6. Some hydrogen occupies 120 ft at 60.0°F. Find the temperature when its volume is 132 ft3 if the pressure remains constant.

7. Some propane occupies 150 cm3 at 12.4°C. Find the temperature when it volume is 132 ft3 if the pressure remains constant.

8. Some air at 275 kPa absolute pressure occupies 50.0 m3. Find its absolute pressure if its volume is doubled at constant temperature.

9. Some helium at 20.0 psi gauge pressure occupies 40.0 ft3. Find its volume at 32.4 psi if the temperature remains constant.

10. Given 435 in.3 of nitrogen at 1340 psi (absolute) at 72.0° F. Find the volume at 1150 psi at 45.0° F.

11. A welding tank has a gauge pressure of 2000 psi at 40.0° F. What is the new gauge pressure if the temperature rises to 100 ° F?

12. Find the temperature if the gauge pressure in the preceding problem falls to 1750 psi.

13. An ideal gas occupies a volume of 5.00 L at STP. What is its gauge pressure (in kPa) if the volume is halved and the absolute pressure is doubled?

14. An ideal gas occupies a volume of 6.00 L at STP. What is its gauge pressure (in kPa) if the volume is doubled and the temperature is increased to 50.0° C?

15. A volume of 1200 L of helium at 4000 Pa is heated from 45.0°C to 75.0°C increasing the pressure to 6000 Pa. What is the resulting volume of the gas?



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CHAPTER 16: WAVE MOTION AND SOUND

1. The minimum distance between particles in a wave that have the same displacement and are moving in the same direction is called .

2. Which of the following refers to the time required for a single wave to pass a given point?

a. The period b. The frequency c. The wavelength d. None of the above

3. Which of the following refers to the number of complete waves passing a given point per unit of time?

a. The period b. The frequency c. The wavelength d. None of the above

4. An example of a transverse wave is a . a. sound wave b. water wave c. interference

5. Explain the difference between interference and diffraction.

6. Explain the difference between constructive and destructive interference.

7. How does the speed of sound differ in water and air? Explain the reason for this difference.

8. Explain the Doppler Effect.

9. Explain Resonance.

10. Find the period of a wave with frequency 400 kHz.

11. Find the frequency of a wave whose period is 0.320 s.

12. What is the frequency of a light wave whose wavelength is 5.00 x 10-7 m and velocity is 3.00 x 108 m/s?

13. Find the period of the wave in the previous problem.

14. What is the speed of a wave whose frequency is 9.00 Hz and wavelength is 0.654 m?

15. Find the frequency of a wave produced by a generator that emits 100 pulses in 2.00 s.



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16. What is the wavelength of an electromagnetic wave whose frequency is 70.0 MHz?

17. Find the speed of sound (in m/s) at 75.0°C at 1 atm pressure in dry air.

18. Find the speed of sound (in m/s) at –35.0°C at 1 atm pressure in dry air.

19. How long will it take a sound wave to propagate (travel) through 1500 m at 25.0°C?

20. A train traveling at a speed of 80.0 mi/h approaches an observer at a station and sounds a 525 Hz whistle. What frequency will be heard by the observer? Assume that the sound velocity in air is 1090 ft/s.

21. A car is traveling toward you at 70.0 mi/h. The car horn produces a sound at a frequency of 5000 Hz. What frequency do you hear?

22. What frequency do you hear if the car in the preceding problem is traveling away from you?

23. Find the period of a pendulum that has a length of 40.0 cm.

24. Find the length (in cm) of a pendulum with a period of 1.20 s.



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CHAPTER 17: BASIC ELECTRICITY

1. Two charges of equal magnitude are 25.5 cm apart. If the force between the charges is 155 N, what is the magnitude of the charges?

2. Two charges, one with magnitude +4.85 x 10-6 C and the other with magnitude –3.93 x 10-6 C, are separated by a distance of 0.485 cm.

a. What is the magnitude of the electrostatic force on the charge? b. What is the direction of the electrostatic force?

3. A charge of –8.25 μC is placed 0.0800 cm from another charge, q. The force between them is 4.25 N. What is the magnitude of charge q?

4. Three charges are located along the y-axis. Charge A (-2.50 μC) is located at the origin. Charge B (+3.95 μC) is located at y = -32.5 cm. Charge C (+8.25 μC) is located at y = 15.5 cm.

a. What is the total force (magnitude and direction) on charge A? b. What is the total force (magnitude and direction) on charge B?

5. A charge of +3.55 μC and another charge of –8.35 μC have an attractive force between them of 355 N. How far are these charges?

6. The force of repulsion between two identical negative charges is 3.95 N when the charges are 35.0 cm apart. What is the magnitude of these charges?

7. The force of attraction between two charges is 2.75 N. The charges are 0.212 m apart. The magnitude of one charge is twice the magnitude of the other. Find the magnitude of each charge.

8. Three charges are located on the x-axis. Charge A (+1.75 μC) is located at x = +11.5 cm. Charge B (-9.38 μC) is located at the origin. Charge C (+3.70μC) is located at x = -9.30 cm. Find the magnitude and direction of the total force

a. on charge A. b. on charge B. c. on charge C.

9. Find the resistance of 845 ft of No. 24 copper wire if it has resistance of 0.0262 Ω/ft.

10. Find the resistance of 38.5 m of No. 20 aluminum wire at 20° C. (ρ = 2.83 x 10-6 Ω cm, A = 2.07 x 10-2 cm2)

11. Find the resistance of 215 m of No. 20 copper wire at 20° C. (ρ = 1.72 x 10-6 Ω cm, A = 2.07 x 10-2 cm2)

12. Find the length of a copper wire with a resistance of 0.0262 Ω/ft and total resistance of 1.75 Ω.



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13. What current does a 205 Ω resistance draw of 115 V?

14. What voltage is required to produce a current of 2.85 A in the circuit below?

15. What is the value of R1 in the circuit below?

16. Find the value of V in the circuit below?

17. Three resistors (1.50 Ω, 3.25 Ω, and 2.05 Ω) are connected in a series with a 36.0 V source.

a. Find the current through the circuit. b. Find the equivalent resistance. c. Find the voltage drop across the 3.25 Ω resistor.



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18. What emf is needed to produce a current of 9.35 A in the circuit below?

19. Use the diagram to answer the following questions: a. Find the equivalent resistance. b. Find the current

in this circuit. c. Find the voltage

across R1. d. Find the voltage

across R2. e. Find the voltage

across R3.

20. Use the diagram to answer the following questions: a. Find the voltage across

R1. b. Find the current through

R2. c. Find the voltage across

R3. d. Find the voltage V.

21. Determine the listed quantities for the circuit below.

I = mA

V1 = V

V2 = V



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22. Calculate the five currents labeled in the circuit below.

IT = mA

I1 = mA

I2 = mA

I3 = mA

I4 = mA

23. Complete the table below as it applies to the given circuit.

24. The voltage applied to a circuit is 8.85 V when the current through the battery is 0.350 A. What is the internal resistance of the battery if the emf of the battery is 9.00V?

25. The emf of a battery is 1.50 V. If the internal resistance is 0.250 Ω and the voltage applied to the circuit is 1.41 V, what is the current through the battery?

26. Two 1.50 V cells, each with internal resistance of 0.0850 Ω, are connected in series to form a battery that provides a current of 0.430 A to the external circuit?

a. Find the current through each cell. b. What is the emf of the battery? c. What is the voltage applied to the external circuit?

27. Three 1.50 V cells, each with internal resistance of 0.950 Ω, are connected in parallel.

A current of 0.155 A flows through the external circuit. a. Find the current through each cell. b. Fin the voltage applied to the external circuit.



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28. Find the wattage rating of a soldering iron that draws 2.30 A on a 110 V line.

29. How many amperes will a 55 W lamp draw on a 110 line?

30. Find the cost to operate a 75 W lamp for 125 h if the cost of energy is $0.04/kWh.

31. How long could you operate a 150 W lamp for $2.50 if the cost of energy is $0.035/kWh?

32. Find the cost of operating a 2.25 A motor on a 110 V line for 30 days if the cost of energy is $0.045/kWh.



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CHAPTER 18: MAGNETISM

1. Sketch the magnetic field lines surrounding two non-similar magnetic poles facing each other.

2. An electrical engineer needs to determine the direction of a magnetic field that surrounds a current-carrying wire. What rule would the engineer use?

3. Is the magnetic field that surrounds a current-carrying wire is located perpendicular or parallel to the wire?

4. Describe what happens to the magnetic field surrounding a current-carrying wire if the current is reversed?

5. Describe how an electromagnet functions?

6. The unit of magnetic field strength is known as the _________.

7. Find the magnetic field at a position 2.25 m away from a long wire carrying a current of 15.0 A.

8. What is the current in a wire if the magnetic field is 3.75 x 10-6 T at a distance of 1.75 m from the wire?

9. Two parallel wires are 0.250 m apart; each carries a current of 18.5 A. a. Find the magnetic field halfway between the two wires if the currents flow in

opposite directions. b. Find the magnetic field halfway between the two wires if the currents flow in

the same direction. 10. A solenoid of diameter 12.4 cm has 775 turns of wire over its 14.5 cm length and

carries a current of 11.5 A. Find the magnetic field at the center of the solenoid.

11. A solenoid has 4525 turns, is 22.5 cm long, and carries a current of 6.75 A. What is the magnetic field at the center of the solenoid?

12. A solenoid has 5230 turns, is 15.5 cm long, and has a magnetic field at its center of 4.75 x 10-5 T. What is the current through the solenoid?

13. How would a person go about determining the direction of a magnetic field in a solenoid?

14. If the current in a solenoid is reduced to half the original amount, how is the strength of its magnetic field altered?

15. The length of a tightly turned solenoid is doubled in length (using the same amount of wire). How is the strength of its magnetic field altered?



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16. Describe the difference between a motor and a generator.



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CHAPTER 19: ALTERNATING CURRENT ELECTRICITY

1. The maximum ac current on a line is 12.5 A. Find the instantaneous current at Θ = 33°.

2. What is the maximum voltage in an ac circuit where the instantaneous voltage at 51° is 14.5 V?

3. A power plant generator develops a maximum voltage of 195 V. What is the effective voltage?

4. Find the effective value of an ac voltage whose maximum voltage is 185 V?

5. What is effective current?

6. What is instantaneous current?

7. Does a step down transformer have more coils on the primary or secondary coil?

8. A step-down transformer on a 115 V line provides a voltage of 15.0 V. If the primary coil has 2510 turns, how many turns are in the secondary?

9. If the voltage in the secondary coil of a transformer is 25.0 V and the current in it is 5.35 A, what power does it supply?

10. Is it possible to have a 100% efficient transformer? What happens to some of the energy in the transformer?

11. High voltage power lines allow electricity to flow at low current. Why would this be beneficial to the power company?

12. What is an inductor?

13. In a series circuit with a resistance of 55.0 Ω, and an inductance of 75.0 mH, and a 115 V, 60.0 Hz power supply, find:

a. the impedance. b. the phase angle. c. the current.

14. In a series circuit with a capacitor of 75.5 μF, a 19.5 Ω resistor, and a 110 V source

of 60.0 Hz, find: a. the impedance. b. the phase angle. c. the current.

15. In a series circuit with a capacitor of 12.0 μF, a 95.5 Ω resistor, and a 12.0 mH

inductance, and a 110 V source of 60.0 Hz, find:



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a. the impedance. b. the current.

16. In a series circuit with a capacitor of 15.0 μF, a 73.0 Ω resistor, and a 117 mH

inductance, and a110 V source of 60.0 Hz, find: a. the impedance. b. the current.

17. What needs to occur in a circuit for resonance to occur?

18. What is the function of a diode in an alternating current circuit?

19. Find the resonant frequency of a circuit with an inductance of 1.50 μH and a capacitance of 4.55 μF.

20. Find the apparent power produced by a generating station whose actual power is 2,500 kW if the power factor is 0.830.



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CHAPTER 20: LIGHT

1. What is a photon?

2. Is light considered to be a wave or a particle?

3. “Laser Tape Measures” are often used to measure distances. If a laser light takes 6.67 x 10-8 s to leave the unit, hit a wall, and return to the unit, how far away is the wall?

4. How far away (in km) is an airplane if the radar wave takes 9.36 x 10-3 s to go and return to the radar unit?

5. How long does it take a radar signal to go and return to the originating radar unit from an airplane 275 mi away?

6. What is a light-year?

7. Find the frequency of an electromagnetic wave with wavelength 2.75 x 10-8 m.

8. Find the frequency of a microwave with a wavelength of 1.00 x 10-2 m.

9. What main variable determines the energy of an electromagnetic wave?

10. What is the energy of a photon with frequency 1.55 x 109 Hz.

11. Find the frequency of a photon with energy 5.24 x 10-22 J.

12. A light source’s intensity is doubled. What will happen to the illumination E on the screen?

13. If a light source is moved twice as far from its projection screen, what will happen to the illumination E on the screen?

14. Find the illumination E and a surface located 1.50 m from a source with an intensity of 375 lm.

15. Find the intensity of a light source that produces an illumination of 7.50 ft-candles at a distance of 8.45 ft from the source.



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CHAPTER 21: REFLECTION AND REFRACTION

1. State the difference between opaque and translucent.

2. If an object is placed 15.5 cm from a plane mirror, how far behind the mirror will the image be located?

3. State the difference between a real and a virtual image.

4. Explain what concave mirrors are used for and provide an example where a concave mirror might be used.

5. Explain what convex mirrors are used for and provide an example where a convex mirror might be used.

6. An object 16.0 cm tall is located 9.45 cm from a concave mirror with focal length 18.5 cm.

a. Where is the image located? b. How tall is the image?

7. An object 3.50 m tall is located 6.80 m from a large mirror. If an image is formed

3.05 m from the mirror, how tall is the image?

8. Is a magnifying glass a converging lens or a diverging lens?

9. An object 1.75 cm tall is placed 13.5 cm from a converging lens. A real image is formed 7.86 cm from the lens.

a. What is the focal length of the lens? b. How tall is the image?

10. The focal length of a lens is 4.00 cm. How far from the lens must an object be

placed to produce an image 6.15 cm from the lens?

11. An object 2.45 cm tall is placed 113 cm from a converging lens with a focal length of 34.5 cm.

a. What is the location of the image? b. How tall is the image?

12. What happens when light travels from air and enters into another medium such as

water?

13. Find the angle of refraction for light as it leaves air and enters water at an angle of 52.3°.

14. What happens to the light if it leaves water and attempts to enter the air at an angle of 52.3°?

15. What is a critical angle?



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16. Explain how fiber optic cables make use of the critical angle.

17. Why do diamonds appear to emit so much light from the top of the diamond? (Hint: Use your knowledge of critical angles and total internal reflection.)

18. Find the critical angle of incidence of a fluid that has an index of refraction of 1.495.

19. Find the index of refraction of a medium for which the angle on incidence of a light beam is 33.5° and the angle of refraction is 29.2°.

20. Find the speed of light in a liquid with an index of refraction of 1.46.



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CHAPTER 22: COLOR

1. Describe how Isaac Newton discovered that white light consists of all the colors of the visible light spectrum.

2. Explain why a blue car looks blue.

3. The visible light spectrum ranges from red (low frequency) to violet (high

frequency). What other light is not visible, but is emitted as heat and the other is emitted from sources such as the sun.

4. Explain the difference between primary colors and primary pigments.

5. Explain why sunsets appear red.

6. Explain diffraction.

7. Provide an everyday example of diffraction.

8. Sun glare is sunlight that is reflected horizontally off of objects on a bright, sunny

day. If provided with polarized lenses, would you prefer to have vertically aligned polarized filters or horizontally aligned filters?

9. What happens to the light when one polarized filter is placed horizontally and the

other is placed vertically?

10. Besides sunglasses, what are other uses of polarized filters?



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CHAPTER 23: SURVY OF MODERN PHYSICS

1. Briefly explain quantum mechanics. 2. What is an energy level? 3. What is the force responsible for keeping the nucleus of an atom together? 4. Explain the difference between atomic number and the atomic mass number. 5. Find the mass (in kg) of the 92

230U atom if its mass in atomic mass units is 232.0372 u. 6. If the mass of the neutral 4 He atom is 6.6463 ×10−27 kg, the mass of a proton is

1.67262 ×10−27 kg, and the mass of the neutron is 1.67493 ×10−27 kg, find the binding energy of

a. a 24He nucleus (in kg).

b. a 24He nucleus (in MeV).

7. Find the binding energy of a 40

228Th nucleus in MeV if the mass of the neutral Th atom is 228.02716 u.

8. What is radioactive decay? 9. What is half-life? 10. What type of radioactive particle is also an electromagnetic wave? (Alpha, Beta, or

Gamma) 11. Find the probability of decay of a single atom during the next 5.00 years if the decay

rate for this isotope is 1.82 ×10−5 per year. 12. If there is one chance in 545 that a single atom will decay in the next year, find

a. the probability that the atom will decay in the next two years. b. the decay rate for this atom.

13. Find the amount of radioactive radium remaining after 235 years if the original

amount of radium is 1.65 ×1023 atoms. The half-life of radium is 1620 years. 14. If the quantity of a radioactive isotope remaining after 13.0 years is one-fourth of the

original amount, what is the half-life of the isotope? 15. What is nuclear fission? 16. What is nuclear fusion?



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17. Find the activity of a 3.42-Ci sample of 7

13N 35 minutes after certification. Its half-life is 1 0.0 min.

18. If the activity of an isotope after 34 days is 0.15 of the original activity, what is the

half-life of the isotope? 19. If the activity of an isotope after 6.00 years is 0.28 of the original activity, what is the

half-life of the isotope? 20. Provide an example of a beneficial use of radiation.



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CHAPTER 24: SPECIAL AND GENERAL RELATIVITY

1. Briefly describe Einstein’s Special Theory of Relativity.

2. Briefly describe Einstein’s General Theory of Relativity.

3. What is Space-Time? Why is space-time important to Einstein’s theories?

4. What is the physical significance of E = mc2?

5. What causes a warp in space-time?

6. How can light be effected by the warp in space-time?

7. Describe the eclipse experiment that proved Einstein’s theory of General Relativity correct.

8. Calculate the mass of an object that contains 1.50 x 10-10 J of energy.

9. Calculate the amount of energy that is contained in one electron. (m = 9.10 x 10-31 kg)

10. What theory did Albert Einstein not complete before his death in 1955?



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CHAPTER 1: THE PHYSICS TOOLKIT

1. centi 2. m 3. 4.50 x 104 4. 7.42 x 10-5 5. 0.000328 6. 265,000,000 7. 1200 8. 6900 9. 0.75 10. 6.7 11. 0.0065 12. 166 13. 2.67 14. 0.425 15. 3 16. 4 17. 2 18. 0.1 m 19. 10 ft 20. 1000 s 21. 40.2 m 22. 890 kg m/s2 23. 104 m2 24. 970 cm3 25. 1.042 x 10-4 mi 26. 6.8 x 106 in. 27. 4.81 x 10-3 L 28. 1.5 x 10-9 gal 29. 1.275 x 105 cm 30. 8.207 x 10-5 km 31. 8.96 x 107 g 32. 1.8 x 104 in. 33. 355 gal = 1340 L 34. 2380 ft2 = 22, 200,000cm 22, 200,000cm 35. 1210 yd3; 243,000 gal 36. 208 L; 7.36 ft3



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CHAPTER 2: PROBLEM SOLVING

1. An exponent is a mathematical operation. A subscript is used to define a variable a specific feature or component of a variable.

2. A formula is a basic equation, usually expressed in letters and numbers. A working equation is created when the desired variable is isolated on one side of the equation.

3. Estimating the expected answer in problem solving can serve as a check to make sure the answer is correct.

4. m =Fa

5. t =2s

vf − vi

6. vf =2st

− vi

7. h =PEmg

8. A = 61.0 cm2 9. h = 5.35 cm 10. 84.7 mm 11. A = 1520 cm2 12. 12.0 cm 13. r = 6.32 m 14. h = 52.9 m 15. A = 5810 m2



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CHAPTER 3: VECTORS

1. Scalar is a quantity that specifies magnitude. Vector is a quantity that specifies both magnitude and direction.

2. Distance is a scalar that specifies the magnitude or the amount that an object has changed its position. Displacement is a vector that specifies the direct distance and direction that an object has moved.

3. 18 cm 4. 100 km at 27° north of east 5. 130 km at 38° south of east 6. 0.317 7. 2.385 8. 0.9058 9. 27.6° 10. 9.5° 11. 32.2° 12. 43.3 cm 13. 64.4 m 14. 55.5° 15. a = 20.4 m; b = 24.7 m; B = 50.5° 16. A = 33.6°; B = 56.4°; a = 26.5 km 17. 58.0° 18. 20.0 m 19. Rx = -8; Ry = -3 20. vx = -40.8 m/s; vy = -19.0 m/s



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CHAPTER 4: MOTION

1. 94 km 2. 2.7 h 3. 50 mi/h 4. 26.4 m/s 5. 59.0 mi/h 6. 86.6 ft/s 7. (a) 21 ft/s2 (b) 230 ft 8. (a) 4.40 ft/s2 (b) 385 ft 9. 1600 mi 10. –15,000 m/s2 11. (a) 31 m/s (b) 49 m 12. (a) 2580 m (b) 23.0 s (c) 46.0 s 13. (a) 0.621 s (b) 6.21 ft (c) 48.3 ft/s (d) 2.12 s 14. (a) 6.60 s (b) 212 ft/s 15. (a) 627 ft (b) 81.8 ft/s = 55.7 mi/h 16. (a) 295 ft/s = 201 mi/h (b) 32.9 ft/s 17. (a) 61.5 m/s; (b) 193 m 18. The vertical velocity decreases on the way up, reaches zero at the top of its flight,

and increases as it falls back down. The horizontal velocity remains constant throughout the flight.

19. There is no horizontal acceleration during the flight. The vertical acceleration (the acceleration due to gravity) remains a constant downward 9.80 m/s2 throughout the flight.

20. 64.5 m 21. 71.1 m 22. range increases; range decreases.



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CHAPTER 5: FORCE 1. Inertia is the property of a body that causes it to resist a change in its motion.

Inertia is defined by the objects mass. 2. 36.7 m/s 2 3. 19,300 N 4. Starting friction is the force that attempts to prevent a stationary object from sliding.

Sliding friction is the force that resists the continuation of the object’s movement. Starting friction tends to have a higher coefficient of friction.

5. Lubricate the area between the two surfaces. Place rollers or ball bearings between the two surfaces. Rolling friction can be significantly lower than sliding friction.

6. 1500 lb 7. 0.36 8. 37N to the left 9. 23.5 N resulting in a net force of 0 N 10. 0.143 m/s2 11. 9 12. 1.15 m/s 2 13. Mass is the amount of matter an object has, while weight is the amount of

gravitational force that is exerted on the mass. 14. There is a normal force from the floor and scale supporting the person with as much

force upward as gravity is exerting downward. The net force is therefore 0 N. 15. 666 N 16. 24.1 slugs 17. 103 N 18. For every force applied by object A to object B, there is an equal but opposite force

exerted by object B on object A. 19. –34.0 N 20. The less massive, 64.5 kg person will have a greater acceleration than the more

massive rowboat.



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CHAPTER 6: MOMENTUM 1. Momentum is a measure of the amount of inertia and motion an object has or the

difficulty in bringing a moving object to rest. 2. The tossed tennis ball has greater momentum because it has both mass and motion. 3. 3.9 kg m/s 4. The tennis racket applies a force for a period of time. The impulse alters the ball’s

velocity, thereby also changing its momentum. 5. 53.8 N; 3.90 Ns 6. 171 g 7. “Crumple Zones” are used to increase the amount of time of a collision, thus

reducing the amount of force experienced by the automobile and the passengers. 8. A bouncing motion results in a larger change in velocity, thereby resulting in a

greater change in momentum and impulse. A ball that simply strikes a surface and stops experiences a much smaller change in velocity and therefore smaller change in momentum and impulse.

9. Inelastic collisions result in two or more objects being coupled together after the collision. Elastic collisions result in two or more objects bouncing off one another and returning to their original shapes.

10. Elastic collision – two billiard balls, bowling ball and pins, tennis ball and racket. Inelastic collision – automobiles locking together, train cars coupling together, baseball player catching a ball in a glove.

11. 0.00252 s 12. 14,200 N 13. 35.7 kg m/s 14. 35.7 N s 15. 15.8 kg m/s 16. 30.0 kg m/s 17. 3.33 kg m/s



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CHAPTER 7: CONCURRENT AND PARALLEL FORCES 1. 356 N (right) 2. (a) 643 N (b) 53 N 3. 391 N at 56.7 ° 4. 567 km/h at 310.2° 5. 2260 N at 53.5 ° from F1 6. 3500 N at 45.0° from F1 7. 7 8. F1 = 1380 N; F2 = 2380 N 9. F1 = 36,800 lb; F2 = 31,900 lb 10. T = 3230 N; C = 2650 N 11. T1 = 24,900 N; T2 = 10,300 N 12. (a) FR = 133 lb @ -160° FE = 133 lb @ 20° (b) FR = 168 N @ 142° FE = 168 N @ -38° (c) FR = 76.1 N @ -179° FE = 76.1 N @ 1° 13. T = 3759 N C = 5064 N 14. (a) 15. (b) 16. Yes 17. Yes 18. 980 N 19. 160 lb 20. 32.0 N 21. 200 N m 22. 125 kg 23. 10.9 lb 24. 51.0 cm 25. 600 0kg 26. 5220 kg, 6580 kg



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CHAPTER 8: WORK AND ENERGY 1. 756 ft lb 2. 130,000 ft lb 3. 6370 N m 4. 102,000 N m 5. 4310 ft lb/s 6. 6620 W or 6.62 kW 7. 6.95 s 8. 51.5 s 9. 204 kg 10. 1200 L/min 11. 56,300 ft lb 12. 2500 J or 2.5 kJ 13. 140,000 ft lb 14. (a) 3370 kW (b) 4520 hp (c) 202 MJ (d) 4210 kW 15. 53.8 ft/s 16. (a) 0 (b) 797 m (c) 125 m/s 17. (a) 56.7 ft/s (b) 1.76 s (c) 99.8 ft lb 18. (a) 24.3 m/s (b) 50.5 m/s (c) 325 J 19. W = 420 0 ft lb 20. (a) 70 ,000 ft lb (b) P = 1.1 hp 21. (a) P = 960 w (b) P = 1.29 hp



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CHAPTER 9: ROTATIONAL MOTION 1. (a) 30π or 94.3 rad (b) 5400° 2. 6.98 rad/s 3. 110 cm/s 4. 26.5 N m 5. 26.3 lb 6. 6.25 N 7. 8.59 ft/s 8. Inertia is an object’s resistance to change its linear motion and is determine by the

object’s mass. Moment of Inertia, or rotational inertia, is an object’s resistance the change its rotational motion and is determined by its mass and the distance the mass is located from the axis of rotation.

9. Law of Conservation of Angular Momentum 10. 5.67 ft lb/s 11. 10.0 rad/s 12. 6.00 rpm 13. 30 teeth 14. 57.5 rpm 15. Yes 16. 150 rev 17. 2.00 cm 18. An idler changes the direction of rotation of the driven gear. 19. Opposite 20. Since the gears are connected, they both rotate together.



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CHAPTER 10: SIMPLE MACHINES 1. Simple machines result in useful work. Simple machines are typically used to

reduce the amount of effort force that is needed to do work. 2. Pencil sharpener, bicycle, snow blower, can opener, automobile. 3. The ratio of work output of a machine to the work input of a machine. 4. No. The work input to a machine is partially used to overcome internal friction

within the machine. 5. Third Class Lever; MA < 1.00 6. An increased radius provides the wheel and axle with a greater mechanical

advantage. Such an advantage is needed for a massive truck. 7. 32 : 1 8. Increase the length of the handle or increase the effort force 9. 340 lb; 4.00:1 10. 84.4 lb 11. 20 lb 12. 496 N 13. 5 : 1 14. 6.00 : 1; 1590 N 15. 62 ft 16. 29.7 in 17. The MA would decrease 18. 80 0 : 1 19. 15.0 : 1 20. 20 lb



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CHAPTER 11: UNIVERSAL GRAVITATION AND SATELLITE MOTION

1. Mass of two or more objects and the distance between them. 2. ¼ the original gravitational force 3. 2 x the original gravitational force. 4. The gravitational constant – it never changes. 5. 7.05 x 10-8 N 6. 1.08 x 10-9 m/s2 (much smaller than the earth’s 9.80 m/s2) 7. 3.52 x 1022 N 8. 1.98 x 1020 N 9. The mass of the sun is significantly more than that of the moon. 10. An orbit is the path taken by an object during its revolution around another object.

A centripetal force, in this case a gravitational force, is responsible for the orbit. 11. Mass of the orbited object and the distance from the center of the orbited object to

the satellite. 12. 2.98 x 104 m/s 13. 1.02 x 103 m/s 14. The velocity would decrease 15. Mass of the satellite plays no role 16. 3.17 x 107 s = 1.01 yr 17. 2.37 x 106 s = 27.4 d 18. The period would increase 19. Mass of the satellite plays no role 20. The gravitational force is stronger



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CHAPTER 12: MATTER 1. Stress caused by two forces acting directly opposite each other. Pulling on a rope. 2. Stress caused by two forces acting directly toward each other. Supporting column. 3. Stress caused by two forces applied in opposite, parallel directions. Knife cutting. 4. Stress caused by two torques applied in opposite direction. Screw tightening. 5. Stress caused by both a tension and a compression stress. Bending bookshelf. 6. 155 N 7. 25.2 cm 8. (a) 1.92×106 lb/in (b) 0.0313 in 9. 1.37 lb/in 3 10. 2.00 N 11. 900 0 kg/m 3 12. 0.0158 lb/in 3 13. 58.8 cm 3 14. 22.9m 3 15. 1430 kg 16. 62,400 lb 17. 0.00970 lb/in 3 18. 119 lb/ft 3 19. 0.0930 ft 3 20. 0.700 21. 90.5 lb



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CHAPTER 13: FLUIDS 1. External pressure applied to a fluid is transmitted to all inner surfaces of the liquid’s

container. 2. The top of the wing is curved more than the bottom so that the velocity of air

rushing past the top must be larger than that going past the bottom. This creates low pressure area on the top of the wing.

3. Gauge pressure is the pressure measured relative to atmospheric pressure. 4. The pressures are identical. The forces are different. 5. The pressure is greater at the bottom that at the top. 6. 39.2 kPa 7. 692 ft 8. 30.3 m 9. 11,200 lb 10. 343 kPa 11. 3.45 ×107 N 12. 8.62 ×106N 13. 2.85 ×104 N 14. 3.88 ×106 N / m 15. 4930 kPa 16. 119 lb 17. 8.40 18. 31.3 N 19. 301 kPa 20. 554 kPa 21. 7.4 N



252

CHAPTER 14: TEMPERATURE AND HEAT TRANSFER 1. Heat is the amount of energy. Temperature is a measure of the average kinetic

energy of atoms and molecules in matter. 2. 10 kg of ice at 0° C; the heat of fusion must be added to the ice to turn it to water. 3. Coolants boil at a higher temperature at high pressure. Therefore the engine coolant

can be at a higher temperature and transfer more heat to the atmosphere because of the higher temperature difference.

4. 117° F 5. 297 K 6. 500 °C 7. 100 0°F 8. Aluminum is a better conductor of heat than wood and therefore has a more

difficult time retaining heat. 9. 23.9 cal 10. 1.67 cal 11. 7.78 ×106 ft lb 12. 1.8 ×105 Btu 13. 4.97 ×104 Btu 14. 3.03 kcal 15. 77.3 C 16. 0.751 cm 17. 2.64 L 18. More heat is released when water is converted to steam. 19. 13,500 kcal 20. 130 0Btu



253

CHAPTER 15: PROPERTIES OF GASES

1. decrease 2. decrease 3. increase 4. decrease 5. 16.5 ft 3 6. 591 ° F 7. 333 ° C 8. 138 kPa 9. 29.5 ft 3 10. 481 in 3 11. 2240 psi 12. -22° F 13. 405 kPa 14. 56.2 kPa 15. 0.848 m 3



254

CHAPTER 16: WAVE MOTION AND SOUND 1. wavelength 2. period 3. frequency 4. water waves 5. Interference is the result of two or more waves traveling through the same region at

the same time. Diffraction is the bending of a single wave passing near an obstacle with an opening nearly the same size as its wavelength.

6. Two or more waves add together to form a larger wave in constructive interference. Destructive interference occurs when two or more waves form a smaller wave.

7. The velocity of sound is higher in water than in air. The higher density of water provides a higher velocity of sound.

8. Motion toward an oncoming sound wave increases the frequency at which the maximum pressure regions in the sound wave strike an observer, thereby producing a higher frequency sound. Motion away from an oncoming sound wave decreases the frequency at which the maximum pressure regions strike an observer, thereby producing a lower frequency sound.

9. Resonance occurs when an object vibrates at its natural frequency in response to the vibration at the same frequency of a nearby object.

10. 2.5 ×10−6 s 11. 3.13 Hz 12. 6.00 ×1014Hz 13. 1.67 ×10−15 s 14. 5.87 m/s 15. 50.0 Hz 16. 4.29 m 17. 377 m/s 18. 310 m/s 19. 1.00 s 20. 655 Hz 21. 5520 Hz or 5.52 kHz 22. 4570 Hz or 4.57 hKz 23. 1.27 s 24. 35.7 cm



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CHAPTER 17: BASIC ELECTRICITY 1. 3.35 ×10−5C 2. (a) 7290 N (b) Force is attractive 3. 3.66 ×10−11C 4. (a) 6.88 N (in negative direction) (b) 0.393 N (in negative direction) 5. 0.0247 m 6. 7.33μC 7. 1.86μC;3.72μC 8. (a) 9.82 N (in negative direction) (b) 34.8 N (in negative direction) (c) 34.8 N (in positive direction) 9. 22.1 Ω 10. 0.526 Ω 11. 1.79 Ω 12. 66.8 ft 13. 0.561 A 14. 20.8 V 15. 13.8 Ω 16. 26.3 V 17. (a) 5.29 A (b) 6.80 Ω (c) 17.2 V 18. 41.1 V 19. (a) 11.4 Ω (b) 1.58 A (c) 14.6 V (d) 3.40 V (e) 0.986 A 20. (a) 3.97 V (b) 1.12 A (c) 17.3 V (d) 21.3 V 21. (a) I = 33.3 mA (b) V = 8.33 (c) V = 16.7 V 22. I = 33.0 mA I = 10.0 mA I = 15.0 mA I = 5.0 mA I = 3.0 mA 23.

R1 = 8.0Ω R2 =12Ω R3 = 24Ω V1 = 24V V2 = 24V V3 = 24V I1 = 3.0A I2 = 2.0A I3 = 1.0A



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24. 429 Ω 25. 0.360 Ω 26. (a) 0.430 A (b) 2.93 V (c) 2.93 V 27. (a) 0.0517 A (b) 1.45 V 28. 250 W 29. 0.50 A 30. $0.38 31. 476 h 32. $8.02



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CHAPTER 18: MAGNETISM 1. Magnetic Field lines should come out of the north pole and progress towards the

south pole of the other magnet. 2. Ampere’s Rule 3. Perpendicular 4. The magnetic field is reversed. 5. The magnetic field surrounding the coiled wire induces a strong magnetic field in

the electromagnet’s iron core. 6. Tesla 7. 1.18 ×10−8 T 8. 3280 A 9. (a) 5.92 ×10−7 T (b) 0 10. 4. 7.52 ×10−4 T 11. 5. 1.71×10−3 T 12. 0.112 A 13. Hold the solenoid in the right hand so the fingers represent the direction of the

current. The thumb points to the north pole of the solenoid. 14. The magnetic field is cut in half. 15. The strength of the magnetic field is cut in half. 16. Motor converts electrical energy to mechanical energy while a generator converts

mechanical energy into electrical energy.



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CHAPTER 19: ALTERNATING CURRENT ELECTRICITY 1. 6.81 A 2. 18.7 v 3. 138 V 4. 131 V 5. Amount of current that produces the same amount of heat in a resistance as an equal

number of amperes of a steady direct current. 6. The current at any time in the circuit. 7. primary 8. 327 turns 9. 134 W 10. It is not possible to have a 100% efficient transformer. Copper losses, magnetic

losses, and eddy current losses account for some energy loss within the transformer. 11. High voltage and lower currents reduce the heat in the wires, which in turn

conserves more of the electric current. 12. An inductor is a coil of wire, in which an induced emf opposes any change in the

current. 13. (a) 61.8 Ω (b) 48.4 ° (c) 1.86 A 14. (a) 40.2 Ω (b) 61.0° (c) 2.74 A 15. (a) 199 (b) 0.553 A 16. (a) 152 Ω (b) 0.726 A 17. The inductive reactance must equal the capacitive reactance. 18. Diode allows current to only flow in one direction. 19. 6.10 ×105Hz or610 kHz 20. 3050 kW



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CHAPTER 20: LIGHT 1. A wave packet or bundle of energy that carries electromagnetic radiation. 2. Both. Known as dual nature of light. 3. 10.0 m 4. 140 km 5. 2.96 μs 6. The distance that light can travel in one earth year. 7. 1.09 ×1016Hz 8. 3.00 x 1010 Hz 9. Frequency 10. 1.03 ×10−24 J 11. 7.92 ×1011Hz 12. Double the original intensity 13. ¼ the original intensity 14. 13.3 lux 15. 6730 ft candles



260

CHAPTER 21: REFLECTION AND REFRACTION 1. An opaque object absorbs or reflects almost all light, while translucent objects

allow some, but not all light to pass through. 2. 15.5 cm behind the mirror. 3. A real image is formed by rays of light and can be projected on a screen, while a

virtual image is only appears to be formed by rays of light, but can not be projected on a screen.

4. Concave mirrors are used to magnify or concentrate waves. Concave mirrors in telescopes, close-up mirrors. Satellite dishes for other electromagnetic waves.

5. Convex mirrors are used to gain a wide field of viewing. Convex mirrors are often used as security mirrors or in the passenger side mirror in an automobile.

6. (a) 19.3 cm (b) 32.7 cm 7. 1.57 m 8. Converging lens 9. (a) 4.97 cm (b) 1.02 cm 10. 11.4 cm 11. 49.7 cm 12. Light slows down and refracts (if the light does not strike perpendicular to the

surface of the medium). 13. 36.5° 14. The light internally reflects at 52.3° 15. The smallest angle of incidence at which all light striking a surface is totally

internally reflected. 16. Fiber optic cables are made of a material that has a very small critical angle. This

allows the light to totally internally reflect back into the fiber optic cable when the light strikes the side of the cable.

17. The angles of the diamond’s cuts are made so after light enters the diamond, the light always strikes the edges of the diamond at angles beyond its critical angle. The light emerges from the top of the diamond creating the brilliance.

18. 42° 19. 1.13 20. 2.72 ×105mi / s



261

CHAPTER 22: COLOR

1. Newton sent a beam of white light through a prism causing the light to disperse into the various colors of the rainbow.

2. All the colors of the visible light spectrum are absorbed, except for blue, which is reflected back to the observer’s eye, thus making the car appear blue.

3. Infrared; Ultraviolet 4. Primary colors consist of lights and when combined form white light. Primary

pigments consist of dyes and when combined form black. 5. Since the sun is low on the horizon, the high frequency blue, indigo, and violet light

is scattered before the light reaches the observer. Therefore, the only light still transmitted through that section of the sky is the orange and reddish light.

6. Diffraction is the ability for light or any wave to bend around obstacles in its path. 7. Light entering through a doorway. Light bends around the opening of the doorway

and lights up an entire room. 8. Vertically aligned filters would block out the horizontal sun glare. 9. Perpendicular filters block out all light. 10. Liquid crystal displays consist of polarized filters. The windshield of airplanes are

polarized. 3-D glasses use polarized filters.



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CHAPTER 23: SURVEY OF MODERN PHYSICS 1. Quantum theory states that matter, just like light, behaves sometimes as a particle

and sometimes as a wave. 2. An energy level translates into the orbit of an electron. The lower the energy level,

the closer the orbit is to the nucleus of the atom. This is represented by “n”. 3. Strong Force 4. The atomic number is the number of protons in a nucleus. The atomic mass number

is the number of protons and neutrons in a nucleus. 5. 3.853×10−25 kg 6. (a) 4.88 ×10−29kg (b) 4.39 ×10−12J 7. 1700 MeV 8. Radioactive decay occurs when an unstable nucleus of an atom is transformed into a

new element through the spontaneous disintegration of its atomic nuclei. 9. Half-life is the time required for one half of the radioactive atoms in a sample to

radioactively decay. 10. Gamma 11. 9.10 ×10−5 12. (a) 33.67 10−× (b) 1.84 ×10−3 /yr 13. 1.49 ×1023 14. 6.48 yr 15. Nuclear fission is a nuclear reaction in which an atomic nucleus splits into

fragments and releases energy. 16. Nuclear fusion is a nuclear reaction in which light nuclei interact to form heavier

nuclei with the release of energy. 17. 0.428 Ci 18. 12.4 days 19. 3.27 yr 20. Food irradiation, identification and treatment of cancer cells and other defective

cells.



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CHAPTER 24: SPECIAL AND GENERAL RELATIVITY 1. The speed of light is constant regardless of the speed of the observer. 2. Gravity and acceleration are equivalent. In addition, light has mass and therefore

can be warped by gravity. 3. Space-Time is the combination of the three dimensions of space (x, y, and z) and

the fourth dimension time. Space and time dimensions are needed to locate one’s place in the universe.

4. This formula proved to the world that light can be represented as mass, and that all objects of mass have a tremendous amount of energy.

5. Any object of mass creates a warp in the fabric of space and time. 6. Since light has mass, its movement throughout the universe is determined by the

warping of the universe. A massive object such as a star or a planet could cause the light to change direction.

7. The eclipse of the sun showed that light, emitted from two starts, seemed to change position, when the light from the stars traveled close to the sun. The sun produced a warp in space-time, thereby shifting the apparent position of the stars.

8. 1.67 x 10-27 kg 9. 8.19 x 10-14 J 10. The Grand Unified Theory



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