geas mar2014 th2 - solutions
TRANSCRIPT
Excel Review Center ECE Refresher Course GEAS 2
Baguio: PiRC Building, Gapuz Centrum II, 80 T. Alonso St., New Lucban CP: 0928 934 1637 Manila: CMFFI Bldg. R. Papa St. Tel. 7365291
1. If the coefficient of sliding friction…
N = W = 150 lb
F = µN = 0.05( ) 150( )F = 7.5 lb
Fforward = F = 7.5 lb → Ans
2. A 3200 lb car traveling with a speed…
Fc = m v2
r
Fc =3200 lb
32.2 fts2
88 fts
⎛⎝⎜
⎞⎠⎟
2
484 ft
Fc = 1590 lb → Ans
3. A 400 kN block is resting on a rough…
N = WN = 400 kNF = µNF = 0.40( ) 400( )F = 160 kNP = F = 160 kN → Ans
4. A skier is on a 25° slope. His mass…
N = W cos25°N = 80( ) 9.81( )cos25°
N = 711.27 NF = µNF = 0.08( ) 711.27( )F = 56.9 N
∑F = ma80( ) 9.81( )sin25° − 56.90 = 80a
a = 3.43 m/s2 → Ans
5. A trapezoid with parallel bases…
y = 2a +b
a +bh3=
2 2( ) + 42+ 4
33= 1.33→ Ans
6. A 250-g stone is attached to a 50-cm…
F = mv2
r=
0.25 kg( ) 2 m/s( )2
0.50 m= 2N→ Ans
7. A 600 N block rests on a 30° plane…
Fx = 0∑
Pcos30° = F + 600sin30°Pcos30° = µN+ 600sin30° →Eq. 1
Fy = 0∑
N = 600cos30° +Psin30° → Eq. 2 Substitute Eq. 2 in Eq. 1:
Pcos30 = 0.2 600cos30 +Psin30⎡⎣ ⎤⎦ +
600sin300.866P = 103.923 + 0.1P + 300
P = 527.31 N → Ans
8. A pipeline crossing a river is…
w = wp + wc = 14 +1= 15 kg/m
H = wL2
8d= 15(100)2
8(2)= 9375 kg
T = wL2
⎛⎝⎜
⎞⎠⎟
2
+H2
T = 15(100)2
⎛⎝⎜
⎞⎠⎟
2
+ 93752
T = 9405 kg → Ans
9. A cable suspended from a level…
Using squared property of parabola:
1002
30= x2
10
x2 = 1002
3x = 57.8 → Ans
10. Determine the moment of inertia…
I= 2mr2
3
I= 2(10)(3)2
3I= 60 → Ans
11. An inverted cone (apex down) has….
I= 3mr2
10
I= 3(10)(2)2
10I= 12 → Ans
12. A car is traveling at 60 kph….
a = ΔvΔt
=40 km
hr1000 m
1 km⎛⎝⎜
⎞⎠⎟
1 hr3600 s
⎛⎝⎜
⎞⎠⎟
6 sa = 1.85 m/s2 → Ans
13. A man, standing on a cliff overlooking…
s = vot ±12
at2
−y = 0 − 12
9.81( ) 2( )2 ; y = 19.6 m → Ans
14. A tennis ball is dropped from the…
s = vot ±12
at2
−40 = 0 − 12
9.81( ) t2 ; t = 2.86 s → Ans
15. Two blocks weighing 8 kg and…
Block 1:
Block 2:
Solve for a in eq. 1 and eq. 2: a = 1.4
m/s2
16. A car travels at the constant speed… The average speed is equal to that of the constant speed which would be required for the object to travel the same distance d in the same time t.
W
F
25° N
P
600 N
a
F
N
30°
30°
30°
Excel Review Center ECE Refresher Course GEAS 2
Baguio: PiRC Building, Gapuz Centrum II, 80 T. Alonso St., New Lucban CP: 0928 934 1637 Manila: CMFFI Bldg. R. Papa St. Tel. 7365291
v =dtotal
ttotal
v = 20 + 20 + 202030
+ 2040
+ 2050
v = 38.2 mph → Ans
17. During the takeoff, an Airbus A380…
18. A dragster reaches a quarter mile…
v2 = vo2 ± 2as
802 = 0 + 2a 402( )a = 7.96 m/s2
v = vo ± at
80 m / s = 0 + 7.96 m/s2( ) tt = 10 s → Ans
19. A ball is released from rest at a…
20. An airplane lands on a carrier deck…
21. A ball is thrown with an initial velocity…
This is a problem in projectile motion. Analyze first the horizontal component of the motion. Note: The ball is not accelerating horizontally.
vx = vox ± axtvx = 160cos53° + 0vx = 96 ft/s → Ans
Analyze the vertical component of the motion. Note: The vertical acceleration is due to gravity (ay = g)
Get the resultant of the two components:
v = vx2 + vy
2
v = 962 + 642
v = 115.4 ft/s → Ans
22. A ball is thrown with an initial velocity…
Analyzing just the vertical component of the motion, at the highest point the velocity (vertical component only) is zero:
vy = voy ± ayt
0 = 160sin53° − 32.2( ) tt = 4 s → Ans
23. A ball is thrown with an initial velocity…
R =Vo
2 sin2θg
R =1602 sin2 53( )
32.2R = 764 ft → Ans
24. A 65-lb horizontal force is sufficient…
N = W = 65 lb
Fapplied = F = 1200 lb
F = µN1200 = µ 65( )
µ = 0.054 → Ans
25. A 1000-gram mass slides down an…
s = Vot ±12
at2
81= 0 + 12
a 0.6( )2
a = 450 cm/s2
F = maF = 1000 g( ) 450 cm/s2( )F = 450,000 dynes → Ans
26. A baseball pitcher throws a ball…
W = mg
m = Wg
=
13
lb
32.2 fts2
m = 0.01 slugs
F = maF = 0.01 slugs( ) 480 ft/s2( )F = 5 lb → Ans
27. The outside curve on a highway…
sinθ = 430
θ = 7.66°
tanθ = v2
Rg
tan7.66° = v2
150( ) 32.2( )v = 25.4 ft/s → Ans
28. A satellite is placed in a circular…
R = Rearth +100
R = 6.4 ×106m+1.6 ×105mR = 6.5 ×106m
Fgravity = Fc
mg = mv2
R
g = v2
R
v = 9.81ms2
⎛⎝⎜
⎞⎠⎟
6.5 ×106m( )v = 8000 m
s
T = 2πRv
T =2π 6.5 ×106m( )
8000 ms
T = 85 min → Ans
Excel Review Center ECE Refresher Course GEAS 2
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29. A cable 800 m long weighing 15.5…
30. From the cable in the previous…
x = cloge
s + yc
x = 300loge
400 + 500300
x = 330 m → Ans
31. On a certain stretch of the railroad…
32. Water drops from a faucet at the rate…
v = vo ± at
−3 = 0 − 9.81( ) tt = 0.306 s
s = Vot ±12
at2
−h = 0 − 12
9.81( ) 0.306( )2
h = 0.459 m → Ans
33. Suppose that you throw a ball…
v2 = vo2 ± 2as
0 = 102 − 2 9.81( )hh = 5.10 m
hmax = 5.10 + 2hmax = 7.10 m → Ans
34. A man weighing 70 kg is in an…
35. When a 3000 N boat is moving at…
v2 = vo2 ± 2as
0 = 32 − 2as
a = 4.5s
wg
a = R = 30V
wg
4.5s
= 30V
V =v1 + v2
2
V = 3 + 02
V = 1.5 m/s
30009.81
4.5s
= 30 1.5( )s = 30.6 m→ Ans
36. A projectile is fired with an initial…
y = x tanθ − gx2
2v2 cos2 θ
−80 = x tan30° −9.81( )x2
2 60( )2cos2 30°
x = 422 m → Ans
37. A rocket is released from a jet… Vertical component of the motion:
y = Voyt ±12
ayt2
−2400 = 0 − 12
9.81( ) t2
t = 22.12 s → Ans
Horizontal component of the motion:
x = Voxt ±12
axt2
x = 1200( )10003600
22.12( ) + 12
0.6( ) 9.81( ) 22.12( )2
x = 8813 m
38. An airplane makes a turn in a…
tanθ = v2
Rg
tanθ =215 m
s⎛⎝⎜
⎞⎠⎟
2
1750 m( ) 9.81 ms2
⎛⎝⎜
⎞⎠⎟
θ = 69.6° → Ans
39. A flywheel 6 ft in diameter accelerates…
α =ω f − ω i
t
α =4 rev
min× 2π rad
1 rev× 1 min
60 s⎡
⎣⎢
⎤
⎦⎥ − 0
10 sα = 0.42 rad/s2 → Ans
40. What is the velocity of a particle after…
v = x3 − 2x2 − 5x + 4
v = 5( )3 − 2 5( )2− 5 5( ) + 4
v = 54 m/s → Ans
41. What is the maximum speed at which…
Note: since the curve is “unbanked”, angle of elevation θ = 0; Φ is called angle of friction and is equal to Arctan µ where µ is the coefficient of friction
tan θ + φ( ) = vmax( )2
Rg
tan 0 + tan−10.80( ) = vmax( )2
60( ) 9.81( )vmax = 21.7 m/s → Ans
42. A man standing at the foot of a…
h = 12
gt1t2
h = 12
32.2( ) 1( ) 2( ) = 32.2 s → Ans
43. A car starting from rest accelerates…
v = vo ± at
120 = 0 + a 4( ); a = 30 ft/s2
s = vot ±12
at2
s = 0 + 12
30 ft/s2( ) 2( )2; s = 60 ft → Ans
44. A car is traveling at 40 kph…
v = vo ± at
0 = 40 kmhr
× 1000 m3600 s
⎛⎝⎜
⎞⎠⎟− 6 m
s2
⎛⎝⎜
⎞⎠⎟
t
t = 1.85 s → Ans
Excel Review Center ECE Refresher Course GEAS 2
Baguio: PiRC Building, Gapuz Centrum II, 80 T. Alonso St., New Lucban CP: 0928 934 1637 Manila: CMFFI Bldg. R. Papa St. Tel. 7365291
45. A projectile is fired from the top…
Vertical component of the motion:
sy = voyt ±12
ayt2
−200 = 1314sin45( ) t − 12
32.2( ) t2
t = 58 s
Horizontal component of the motion:
sx = voxt ±12
axt2
sx = 1314cos45( ) 58( ) + 0
sx = 53,890 ft → Ans
46. Two cars A and B are traveling in…
d = 100 m + 70( )10 − 60( )10⎡⎣ ⎤⎦d = 200 m → Ans
47. A block weighing 400 N is pulled…
∑Fhorizontal = 0Phorizontal − f = 0
220cos40° − µ 400( ) = 0
µ = 0.42 → Ans
48. A moon revolves around the earth…
aN = v2
r=
ωr( )2
r= ω2r
2.7 mms2 × 1 m
1000 mm= 2.7 ×10−6( )2
r
r = 3.7 ×106 m → Ans
49. A car enters a 350-m radius curve…
vo = 50 kmhr
= 13.89 ms
v2 = vo2 + 2ats
v2 = 13.892 + 2 2( ) 320( )v = 38.38 m
s
an =v2
r=
38.38 ms
⎛⎝⎜
⎞⎠⎟
2
320= 4.6 m
s2
atotal = an2 + at
2
atotal = 4.62 + 22 = 5 ms2 → Ans
50. A tennis ball is projected upward…
Upward motion:
v = vo ± at0 = 120 − 9.81t ; t = 12.23 s
It will also take 12.23 s to move
downwards. Therefore, total time is 25 s. 51. What is the resultant force on a body…
F = maF = 48 kg( ) 6 m/s2( )F = 288 N → Ans
52. Two horses on opposite banks…
ΣFx = −200cosθ − 240cos 60 − θ( )ΣFy = 0
ΣFy = 200sinθ − 240sin 60 − θ( )0 = 200sinθ − 240sin 60 − θ( )θ = 33°
R = ΣFx2 + ΣFy
2
R = −381.5( )2+ 0
R = 381.5 kN → Ans
53. What is the upward acceleration…
∑Fy = ma
R− W = ma1000 − 90( ) 9.81( ) = 90a
a = 1.3 m/s2 → Ans
54. A car (mass = 1800 kg) is traveling…
F = ma = 1800 kg( ) −20 cms2
⎛⎝⎜
⎞⎠⎟
1 m100 cm
F = −360 N → Ans
55. What is the weight of a person on…
Won moon =
16
120( ) = 20 lb → Ans
56. Two blocks of mass 6 kg and m kg…
W = mg174 = 6 +m( ) 9.67( ); m = 12 kg → Ans
57. What is the combined kinetic energy…
KE = 12
mv2
KE = 12
225 kg( ) 40 kmhr
× 1000 m3600 s
⎛⎝⎜
⎞⎠⎟
2
KE = 13.89 kJ → Ans
58. If a certain object with mass of 15 kg…
PE = mgh
PE = 15 kg( ) 13
9.81 ms2
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥ 100 ft × 1 m
3.3 ft⎡
⎣⎢
⎤
⎦⎥
PE = 1500 J → Ans
59. If a car with initial velocity, v…
KE1 =12
mv2 = 0.5mv2
KE2 =12
m 5v( )2= 1
2m 25v2( ) = 12.5mv2
ΔKE = 12.5mv2 − 0.5mv2
ΔKE = 12mv2 = 24 12
mv2⎛⎝⎜
⎞⎠⎟
ΔKE = 24x → Ans
60. Find the average power necessary…
P = Wt= mgh
t=
30( ) 9.81( ) 10( )60
P = 49 W → Ans
61. A charger delivers current of 6 A…
P = IV; Et= IV
E = IVt = 6 A( ) 12 V( ) 6 hr( )3600 s1 hr
E = 1.55 ×106 J → Ans
62. Find the volume (in cm3) of a piece…
ρ = mv
; v = mρ= 70 g
2.33 gcm3
= 30cm3 → Ans
63. A golf ball was hit with a velocity of…
y
x θ α
60°
200 kN
240 kN
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Ft = Δmv
F = Δmvt
=0.050 kg( ) 90 m
s⎛⎝⎜
⎞⎠⎟− 0
0.3 sF = 15 N → Ans
64. A toaster connected to a 220 V…
P = V2
R
Et= V2
R; E
60 s=
220 V( )2
15 ohmsE = 194 kJ → Ans
65. A 1-A lamp on a 220 V line operates…
E = Pt = IV( ) tE = 1 A( ) 220 V( ) 10 days( )24 hr
1 day= 52.8 kW ⋅hr
cost = 52.8 kW ⋅hr( ) P2.501 kW ⋅hr
= P132→ Ans
66. A 200 gram apple is thrown from…
ΔKE = 12
mvf2 − 1
2mvi
2
ΔKE = 12
m vf2 − vi
2( )ΔKE = 1
20.2 kg( ) 502 − 202( )
ΔKE = 210 J × 1 cal4.186 J
ΔKE = 50.17 cal → Ans
67. What is the energy of emitted light…
E = hf
E = 6.63×10−34 Js( ) 6.88×10−19s−1( )E = 4.56×10−19J → Ans
68. A body initially at rest is acted upon…
Ft = ΔmvFt = mv2 −mv1
18 5( )−12t = 0 − 0
t = 7.5 s
69. Three concurrent forces at the origin…
d = 32 + 52 + 72 = 9.11
20009.11
=Fx
3Fx = 660 N → Ans
70. A meteor is moving away from a…
v =v1 + v2
1+v1v2
c2
v = 0.75c + 0.48c
1+0.75c( ) 0.48c( )
c2
v = 0.904c → Ans
71. A 2-m long pendulum is pulled aside…
T = 2π L
g= 2π 2
9.81= 2.8 s → Ans
72. Two electrons (Q1 and Q2) of
electrical…
F = kQ1Q2
d2 = 9 ×109( ) 0.003( ) 0.005( )32
F = 15,000 N → Ans
73. A block, weighing 100 lbs is
suspended…
1keff
= 1k1
+ 1k2
= 120
+ 130
;keff = 12 lbin
T = 2π mkeff
= 2π
100lb
32.2 fts2
12 lbin
× 12 in1 ft
T = 0.92 s → Ans
74. How long will it take for sound to
travel…
Note: Solve first for the speed of sound in air at 20°C. The speed of sound in air at 0°C is approximately 330 m/s
v = 330 1+ 20273
= 341.9 ms
t = dv= 4000 m
341.9 ms
= 11.7 s → Ans
75. An electron at rest has a mass of…
E = mc2
E = 9.11×10−31( ) 3 ×108( )2
E = 8.198 ×10−14 J
E = mV2
8.198 ×10−14 J = 2 9.11×10−31( )v2
v = 2.12×108 m/s → Ans
76. A choir is composed of 39 singers…
I= 10log IIo−10log 39I
IoI= −16 dB → Ans
77. What is the intensity level at a point…
SIL = 10log
2W
4π 20m( )2
10−12 Wm2
= 86 dB → Ans
78. A sound has an intensity of 10-10…
I= 10log IIo
I= 10log10−10 W
cm2
⎛⎝⎜
⎞⎠⎟
100 cm1 m
⎛⎝⎜
⎞⎠⎟
2
10−12 Wm2
= 60 dB
79. A human ear can distinguish a…
Note: The time it travels one way is just 0.10 s.
d = vt = 1150 ft
s⎛⎝⎜
⎞⎠⎟
0.10s( ) = 115ft → Ans
80. A ray of light travels from air at an…
n = cv
1.33 = 3 ×108
v;v = 2.26 ×108 m
s→ Ans
81. Find the period of the 100-cm…
T = 2 1 s( ) = 2 s → Ans
82. The speed of light in a particular…
83. A 10-g mass attached to a spring…
Get the spring constant first using the period of the motion given in the problem:
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ω = 2π f1f= 2π
ω
T = 2πω
= 2π mk
2 = 2 10 0.010k
k = 0.10 N/m
Use the spring formula:
F = kxF = 0.10( ) 0.10( )F = 10−2 N
84. The star nearest Earth, Alpha
Centauri…
d = 4.3 LY × 3×108 m 1 s
× 31,536,000 s 1 year
d = 4.068×1016 m
d = vt
t = dv
= 4.068×1016 m
0.95 3×108 ms
⎛⎝⎜
⎞⎠⎟
t = 142.74×106 st = 4.53 yr
85. The star nearest Earth, Alpha
Centauri…
Time based on an Earth clock
d = 4.3 LY × 3×108 m 1 s
× 31,536,000 s 1 year
d = 4.068×1016 m
d = vt
t = dv
= 4.068×1016 m
0.95 3×108 ms
⎛⎝⎜
⎞⎠⎟
t = 142.74×106 st = 4.53 yr
Use time dilation (since clock is now moving):
t = t'
1− v2
c2
4.53 = t'
1−0.95c( )2
c2
t' = 1.41 years
86. An observer moves past a meter…
l' = l 1− v2
c2
l' = 1 m( ) 1−0.5c( )2
c2
l' = 0.866 m
87. An observer measures the length…
l' = l 1− v2
c2
0.5 m = 1 m( ) 1− v2
3×108 ms
⎛⎝⎜
⎞⎠⎟
2
v = 2.59×108 ms
88. A rocket is moving with a velocity…
l' = l 1− v2
c2
l' = 5 m( ) 1−0.385c( )2
c2
l' = 4.61 m
89. A spaceship moving away from the…
V = v1 + v2
1+v1v2
c2
V = 0.75c + 0.75c
1+0.75c( ) 0.75c( )
c2
V = 1.5c1+ 0.5625
V = 0.96c
90. An incident ray from water makes…
For reflection, angle of incidence is always equal to the angle of reflection. Therefore, the angle of reflection is also 55 degrees.
91. A diverging lens has a focal length of…
Use Thin-Lens Equation (it is assumed that the focal length is much greater than the thickness of the lens):
1f= 1
i+ 1
o1
−5 cm= 1
i+ 1
4 cmi = −2.22 cm (virtual)
Note:
f is + for a converging lens (thicker at the center than at the sides)
f is – for a diverging lens (thinner in the center than at the sides) o is + if object is on the same side of the lens as the incident light (real object) o is – if the object is on the other side i is + if the image is on the side of the lens opposite the incident light (real) i is – if the image is on the same side (virtual)
Use formula for Lateral Magnification:
m = − io
m = −−2.2( )
4m = 0.55 (erect, smaller)
Note: + value of m means image is erect - value of m means image is inverted m = 1 means object and image have same size m < 1 means image is smaller m > 1 means image is larger
92. A penny is placed 4.0 cm in front…
1f= 1
i+ 1
o1
15 cm= 1
i+ 1
4 cmi = −5.45 cm (virtual)
m = − io
m = −−5.45( )
4m = 1.4 (erect, larger)
93. A flea is located 3.0 cm from a… Spherical mirrors are curved mirrors used as image-forming devices. You can still use the thin-lens equation. Focal length of the spherical mirror is half its radius.
1f= 1
i+ 1
o1
−5 cm= 1
i+ 1
3 cmi = −1.9 cm (virtual) → Ans
94. A far-sighted classmate is unable…
Nearsightedness (myopia) can be corrected by using a diverging lens while Farsightedness (hyperopia) can be corrected by using a converging lens.
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D = 1f
1f= 1
i+ 1
o
D = 1i+ 1
o
D = 1−1.25
+ 10.3
D = 2.5 diopters → Ans
Note: i is – because the image and object are on the same side of the lens (image must be virtual)
95. A photographer has an 8x magnifier…
Magnifying glasses are converging lenses.
M= 25 cmf
8 = 25 cmf
f = 3.1 cm → Ans
96. A biology student wishes to use…
M= 25 cmf
M= 25 cm6 cm
M= 4.17 = 4x → Ans
97. For the biology student in the…
Mmax =25 cm
f
M= 25 cm6 cm
+1
M= 5x → Ans
98. A laboratory microscope has a 20X…
A typical microscope consists of a tube with a converging lens at both ends. The lens close to the object is called the objective lens. The lens through which one looks is called the eyepiece/ocular lens
Mobjective =16
fobjective
Meyepiece =25
feyepiece
Mtotal = MobjectiveMeyepiece
Mobjective =16 cmfobjective
20 = 16 cmfobjective
fobjective = 0.8 cm → Ans
99. From the previous problem…
Meyepiece =25 cmfeyepiece
10 = 25 cmfeyepiece
feyepiece = 2.5 cm → Ans
100. Given the microscope in…
Mtotal = MobjectiveMeyepiece
Mtotal = 20( ) 10( )Mtotal = 200x
200( ) 9 nm( ) = 1.8µm → Ans
END