Chapter 1Getting you mind warmed up

Part 1 of 2

About 500 years ago it began to be recognized that the universe behaves mathematically. This was a surprising result and still is when you think about it. Nonetheless, it is true: if you are to work with the simplest forms of nature, you must use the language of mathematics. This section is meant to review for you the concepts of algebra and introduce you to some of the vocabulary and tools of physics.

1. A car travels 486 miles in 8.7 hours. What is its average speed?

average speed = 486 mi/8.7 hraverage speed = 55.9 mi/hr

2. A snail slides 13 cm in 28 seconds. What is its average speed?

average speed = 13 cm/28 secaverage speed = .464 cm/sec

3. An electron travels an astonishing 1.63 meters in 5.7 x l0-8 seconds. How fast is it moving?

speed = 1.63 m/5.7 * 10-8 secspeed = 2.86 * 109 m/sec

4. Notebook paper measures 8.5 inches by 11 inches. What is its area? (That is, the area of one side of the paper.)

Area = Length * Width

A = 93.5 in2

5. A block of iron measures 3.6 cm by 8.4 cm by 4.4 cm. What is its volume?

Volume = l * w * hV = (3.6 cm)(8.4 cm)(4.4 cm)V = 133 cm3

Pay particular attention in these problems to the units. In problem 1 you divide miles by hours to get miles/hr (read this as "miles per hour). In problem 2 you divided centimeters by seconds to get cm/sec. In problem 4 you multiplied inches by inches to get inches2 or "square inches". Finally, in problem 5, by multiplying cm by cm by cm, you got cm3 or "cubic centimeters". Keep units in mind as you complete the next set of problems.

6. A car travels at 51.4 mi/hr for 1.6 hr. How far does it get in this time?

Distance (s) = speed * times = (51.4 mi/hr)(1.6 hr)s = 82.2 mi

7. A piece of wire has a mass of 6.4 gm/cm (grams per centimeter). What is the mass of 13.6 cm of this wire?

mass (m) = (6.4 gm/cm)(13.6 cm)m = 87.0 gm

8. Corn valley has a population of 40,200 and an area of 6.84 mi2. What is the population density?

population density = people/mi2

population density = 40,200 people/6.84 mi2

population density = 5.88 * 103 people/mi2

9. Lead has a density of 11.3 gm/cm3. What is the mass of 325 cm3 of lead?

Each cubic centimeter has a mass of 11.3 gm, so 325 cm3 has a mass:m = (11.3 gm/cm3)(325 cm3)m = 3.67 * 103 gm

Don't forget to be aware of units. They behave just as fractions did in basic math. Recall how you learned to cancel:

(3/5) * (5/8) = 3/8(5/16)/(9/16) = 5/99 * (5/9) = 5

This same approach may be used successfully on the units of quantities:

(2 ft/sec) * (60 sec/min) = 120 ft/min (30 gm/cm2)/(1000 gm/min) = 0.03 min/cm2

6 sec * (4 ft/sec) = 24 ft

Study solutions 6 through 9 until the manipulation of units in these problems comes easily to you. It's hard to overestimate the importance of being able to work with units. Here are yet more:

10. Anita Solution works steadily at making a rug. She completes 15 square inches in 37.2 minutes. What is her average rate of work?

average rate of work = 15 in2/37.2 min

average rate of work = .403 in2/min

11. C. D. Twain puts on his favorite album, which rotates at 33.3 revolutions/minute. How many times does the disk revolve for a song that lasts 4 minutes 22 seconds?

First express the time in minutes:4 min 22 sec = 4 + 22/60 = 4.37 min

revolutions = (33.3 rev/min)(4.37 min)revolutions = 146

12. Atmospheric pressure is 14.7 pounds per square inch. What is the total force on a table top 27 inches wide by 43 inches long?

Force = (14.7 lb/in2)(27 in)(43 in)Force = 1.71 * 104 lb

Be aware that you do not really need to know what the problem is all about. If you know what units you are supposed to end up with, you can solve the problem by making sure they cancel to give the correct final unit.

13. A block of wood measures 3.5 cm by 8.2 cm by 6.7 cm. Its mass is 131 grams. What is its density?

To find density we must determine how many grams are in 1 cubic centimeter. Thus we take the total mass and divide by the number of cm3.

density = 131 gm/[(3.5 cm)(8.2 cm)(6.7 cm)]density = .681 gm/cm3

14. The density of iron is 7.6 gm/cm3. What is the mass of an iron rod 0.44 cm in radius and 25 cm long?

Volume of cylinder = area of base * heightVolume of cylinder = (3.14)r2

Mass (m) = density * volumem = (7.6 gm/cm3)[(3.14)(.442 cm2)(25 cm)m = 116 gm

Check your units again. If these problems seem a bit too easy, bear with me a thorough familiarity with these concepts is too important to pass by quickly.

15. A bicyclist travels at 37 ft/sec. How long does it take him to travel 1.0 feet?

We know ft/sec, but we wish to know

sec/ft; that is, the number of seconds required to travel 1 foot:

time (t) = 1/(37 ft/sec)t = 1 sec/37 ft t = .0270 sec/ft

16. A sound wave vibrates at 13,540 cycles/sec. How long does it take for one vibration?

We know cycles per second, but we wish to know seconds per cycle:

t = 1/(13,540 cycle/sec)t = 7.39 * 10-5 sec/cycle

17. Water pressure on a surface is 128 pounds/in2. What surface area will feel the force of one pound?

As before:

surface area = 1/(128 lb/in2)surface area = .00781 in2/lb

18. A city has a population density of 18,757 people/mi2. How much area does this give each person?

Again:

Area = 1/(18,757 people/mi2)Area = 5.33 * 10-5 mi2/person

This is a square 39 ft on a side!

The problems you've just completed may look a bit trivial, but there's an important lesson. When the number is inverted, the units that go with the number are inverted as well. This will be useful in the following problems.

19. Rain drops fall on a tile surface at a density of 4,675 drops/ft2. There are 16 tiles on each square foot of floor space. How many drops fall on each tile?

We wish to end up with units of drops/tile. Thus ft2 must cancel out:

drops/tile = (4675 drops/ft2)/(16 tiles/ft2)drops/tile = 292

20. A stretch of desert highway outside Barstow, California still retains Burma-Shave signs from the early 1960's. There are 18 signs/mile, and a car speeds by them at 55 mi/hr. a) At what rate will the driver see the signs? (That is, how many signs/hour will pass by him?b) What time interval will pass between successive signs?

a)Miles will have to cancel out:

b)interval = 1/(990 sign/hr)

rate of signs = (18 sign/mi)(55 mi/hr)rate of signs = 990 sign/hr

note: this does not mean there are 990 signs on the road. Seeing 99 signs in 1/10 hr would do it.

interval = 1.01 * 10-3 hr/sign

note: this is 3.6 sec between signs

21. One quart of Slopiton paint covers 450 ft2. Doris Open wishes to paint baseballs, each of which has an area of 0.12 ft2. How many quarts of paint will be required to complete 3,486 baseballs?

We want an answer with quarts for a unit. Baseballs are ft2 will have to cancel out and a quart will need to end up in the numerator.

quarts of paint = [(.12 ft2/baseball)(3486 baseball)]/(450 ft2/qt)quarts of paint = .93

By now you should be aware of an important fact: Units can tell you how to solve a problem. If, on a test, you find you can't get your mind off the weekend's date, a little reasoning with units is a great substitute for understanding. This is most easily seen in the conversion of units from one system to another.

You will need the following English and metric conversions:

1 inch = 2.54 cm

1 foot = 12 inches

5280 feet = 1 mile

100 centimeters = 1 meter

1 kilometer = 1000 meters

1.057 quart = 1 liter

1 liter = 1000 cm3

1 pound = 4.45 newtons

Take the relationship that 1 inch = 2.54 cm. It follows, then, that:

1 in/2.54 cm = 12.54 cm/1 in = 1

Since we can multiply or divide any quantity by 1 without changing its value, the

following is mathematically permissible:

(26.8 cm)(1 in/2.54 cm) = 10.6 in

We have shown that 26.8 cm is equivalent to 10.6 inches. Now try to convert 16.0 inches to centimeters.

(16.0 in)((2.54 cm/1 in) = 40.6 cm

22. convert 18,722 ft into miles.

(18, 722 ft)(1 mi/5280 ft) = 3.55 mi

23. Convert 3.6 meters into centimeters.

(3.6 m)(100 cm/1 m) = 360 cm

24. Convert 8.6 x l0-4 cm into inches.

(8.6 * 10-4 cm)(1 in/2.54 cm) = 3.39 * 10-4 in

25. Convert 3.62 quarts (qt) into liters (l).

(8.6 * 10-4 cm)( 1 l/1.057 qt) = 3.42 l

26. Convert 3.86 x 103 cm3 into quarts.

(3.86 * 103 cm3)[(1.057 qt)/(103 cm3)] = 4.08 qt

In the next problems string together several factors to complete the conversions.

27. Convert 877 cm into miles.

convert cm-->inches-->feet-->miles

(877 cm)(1 mi/2.54 cm)(1 ft/12 in)(1 mi/5280 ft) = 5.45 * 10-3 mi

28. Convert 2.644 days into minutes.

convert days-->hours-->minutes

(2.644 days)(24 hr/1 day)(60 min/1 hr) = 3.81 * 103

29. Convert 15 in2 to cm2. (Remember, when you square a unit, you must also square the conversion factor.)

(15 in2)(2.54 cm/1 in)2 = (15)(6.45 cm2) = 96.8 cm2

30. Convert 3.7 x l0-3 ft3 into in3.

(3.7 * 10-3 ft3)(12 in/1 ft)3 = 6.39 in3

Its crucial that your place the power in the correct position. 12 in is equal to 1 ft, but 1728 in3 and equal to 1 ft3.

31. Convert 4.75 x 107 ft2 into mi2.

(4.75 * 107 ft3)(1 mi/5280 ft)2 = 1.70 mi2

32. Convert 3.6 m3 into in3.

covert meter3-->cm3-->inch3

(3.6 m3)(100 cm/1 m)3(1 in/2.54 cm)3 = 2.20 * 105 in3

The next problems get complex, but they are conceptually no more difficult than what you've been doing. Try to follow the steps in this example.

Convert 3.08 lb/in2 into nt/cm2

solution: First I must explain that nt is for newton, a metric unit of force. It converts directly to pounds by the factor 1 pound = 4.45 newtons. (The correct international system symbol for newton is N, but that can get confused with another N when you're first learning the subject.)

1) Start with your quantity: 3.08 ln/in2

2) Throw in some parentheses: (3.08 lb/in2)(______)(______)

3) Convert pounds to newtons: (3.08 lb/in2)(4.45 nt/1 lb)(______)

4) Convert in2 to cm2: (3.08 lb/in2)(4.45 nt/1 lb)(1 in/2.54 cm)2

5) Multiply it out: (3.08 lb/in2)(4.45 nt/1 lb)(1 in/2.54 cm)2 = 2.12 nt/cm2

As you can see, a systematic approach will substitute nicely for thinking or understanding on these problems, and will free your mind for more important matters.

33. Convert 5.62 nt-m/sec to lb-ft/sec.

(5.62 nt*m/sec)(1 lb/4.45 nt)(100 cm/1 m)(1 in/2.54 cm)(1 ft/12 in) = 4.14 lb*ft/sec

34. Convert 3.91 nt/liter into pounds/quart.

(3.91 nt/L)(1 lb/4.45 nt)(1 L/1.057 qt) = 0.831 lb/qt

35. Convert 4.6 x 104 nt/m2 to lb/in2.

(4.6 * 104 nt/m2)(1 lb/4.45 nt)(1 m/100 cm)2(2.54 cm/1 in) 2 = 6.67 lb/in2

Don't forget to square 100 and 2.54!

36. Convert 88 feet/second to mile/hour.

(88 ft/sec)(1 mi/5280 ft)(60 sec/1 min)(60 min/1 hr) = 60 mi/hr

A handy conversion: 88 ft/sec = 60 mi/hr

37. Convert 17.8 ft3/min2 to cm3/sec2.

(17.8 ft3/min2)(12 in/1 ft)3(2.54 cm/1 in)3(1 min/60 sec) 2 = 140 cm3/sec2

Chapter 1Getting you mind warmed up

Part 2 of 2

Here's another conversion to know, this time it's about angles. The most natural unit of angular measure is not the degree, but the radian. We can generate 1 radian in the following manner:

1) take a circle

2) remove its radius

3) squash the radius up against the perimeter of the circle

4) the angle thus formed is the radian

Notice that the radian measure is the same for large and small circles.

Identify the following central angles:

38. 39. 40.

41. 42. 43.

38.Ø = 2 m/2 mØ = 1 rad

39.Ø = 4 m/2 mØ = 2 rad

40.Ø = 1 m/2 mØ = .5 rad

41.Ø = 9 m/6 mØ = 1.5 rad

42.Ø = 5 m/25 mØ = .2 rad

43.Ø = s/r

The last relationship, Ø = s/r, should be memorized. (Actually, memorized is a poor term. Most relationships in physics are so reasonable that memory becomes automatic once the principle is used a little.)

Now we will convert from radians to degrees and vice-versa. Recall that the circumference of a circle is given by C = 2(3.14)r. That is, 2(3.14) radii can be wrapped neatly around a circle.

2(3.14) r's stretched out 2(3.14) r's wrapped around circle

Note that the interior angle defined by the end points of this wrapping is exactly one complete circle, or 360°. But in our new improved notation, it is 2(3.14) radians. Thus we may say that 360° = 2(3.14) radians.

44. Convert 1.6 radians to degrees.

(1.6 rad)(360°/2(3.14) rad) = 91.7°

45. Convert to radians or degrees, whichever is appropriate. a) 38°b) 0.22 radc) 3.14 radd) 422°

a)(38°)(2(3.14) rad/360°) = .66 rad

b)(.22 rad)(360°/2(3.14) rad) = 12.6°

c)(3.14 rad)(360°/2(3.14) rad) = 180°

d)(422°)(2(3.14) rad/360°) = 7.37 rad

46. Convert 2.0 complete revolutions to radians.

(2.0 rev)(2(3.14) rad/1 rev) = 4(3.14) rad

47. Convert 0.3 revolutions to radians and 1.12 x 103 rev to rad.

(.3 rev)(2(3.14) rad/1 rev) = 1.88 rev(1.12 * 103 rev)(2(3.14) rad/1 rev) =

7.04 * 103 rad

48. Convert 63.6 radians and 4.4 x 103 radians to revolutions.

(63.6 rad)(1 rev/2(3.14) rad) = 10.1 rev(4.4 * 10-3rad)(1 rev/2(3.14) rad) = 7.0 * 10-4 rev

49. A motor shaft turns at 28,000 rev/min. Convert this to (a) rad/sec and (b) deg/sec.

a)(28000 rev/min)(2(3.14) rad/1 rev)(1 min/60 sec) = 2.93 * 103 rad/sec

b)(28000 rev/min)(360°/1 rev)(1 min/60 sec) = 1.68 * 105 °/sec

50. Find the central angle in degrees.

a) b)

a)Ø = (4/6 rad)(360°/2(3.14) rad)Ø = 38.2°

b)Ø = [(12.1/3.6) rad](360°/2(3.14) rad)Ø = 193°

51. Earth's moon has a diameter of 3476 km, and is located 3.84 x 105 km from Earth. What is its apparent size in degrees, as seen from Earth?

Ø = [(3476/3.84 * 105) rad](360°/2(3.14) rad)Ø = .52°

52. The sun has the same apparent size as the moon. (That's why solar eclipses come out so nicely.) If the sun is 9.3 x 107 miles away, what is its diameter? You will need to use problem 51 in this.

Ø = s/rs = rØ

s = (9.3 * 107 mi)[(3476/3.84 * 105) rad]s = 8.4 * 105 mi

That is, the diameter of the sun is its distance times its apparent size in radius.

53. A merry-go-round rotates at 0.077 rad/sec. How long will it take to complete one full revolution?

We have rad/sec, but we want sec/rev. Thus (sec/rad) * (rad/rev) will give us our answer.

(1 sec/.077 rad)(2(3.14) rad/1 rev) = 81.6 sec/rev

54. A wheel of 80 cm diameter spins at 142 rad/sec.a) How far does a point on the rim travel in 1 second?b) How fast is a point 6 cm from the center traveling?

a)In 1 second it rotates:(142 rad/sec)(1 sec) = 142 rad

So the rim travels s = rØ:(80/2 cm)(142 rad) = 5680 cm

b)v = s/tv = rØ/tv = [(6 cm)(142)]/(1 sec)v = 852 cm/sec

55*. Find the rate of angular motion, in rad/sec, of Earth about the sun.

It takes 365 days for the earth to make one complete orbit.

angular motion = (1 rev/365 days)(2(3.14) rad/1 rev)(1 day/24 hr)(1 hr/60 min)(1 min/60 sec)angular motion = 2.0 * 10-7 rad/sec

56*. A beetle takes a joy ride on a pendulum. The string of the pendulum is 183 cm long. If the beetle rides through a swing of 40°, how far has he traveled along the path of the pendulum? (Your answer will be in centimeters.)

First we need the angle in radians through which the pendulum has swung:

(40°)(2(3.14) rad/360°) = .70 rad

s/r = Øs/r = .70 rads = (.70)rs = (.70)(183 cm)s = 128 cm

57*. If the aforementioned (problem 57) beetle swings through 0.15 radians, how far has he traveled?

s = rØs = (183 cm)(.15)s = 27.5 cm

58*. If the pendulum (problem 57) at some instant is swinging at 1.4 rad/sec, how fast is the beetle traveling in cm/sec?

In one second the beetle travels 1.4 rad, so:

Ø = s/r1.4 = s/183 cms = (1.4)(183 cm)s = 256 cmThus the speed is 256 cm/sec

59*. Convert the following into linear speed of the beetle (yes, still from problem 57).a) 0.88 rad/secb) 2.4 x 10-5 rad/secc) 38 °/secd) 127 °/sec

a)(.88 rad/sec)(183 cm) = 161 cm/sec

b)(2.4 * 10-5 rad/sec)(183 cm) = 4.39 * 10-3 cm/sec

c)(38 °/sec)(2(3.14) rad/360°)(183 cm) = 121 cm/sec

d)(127 °/sec)(2(3.14) rad/360°)(183 cm) = 406 cm/sec

Note that since a radian is dimensionless, it can be dropped in and out of any unit.

60*. A small wheel, with radius of 1.6 cm, drives a large wheel of 14.1 cm radius by their circumferences being pressed together. If the small wheel turns at 480 rad/sec, what does the big one turn at?

The rim of the small wheel moves at:

v = (480 rad/sec)(1.6 cm)v = 768 cm/sec

The rim of the large wheel is forced to move at the same speed, as the rims are in contact. Thus...

w = (768 cm/sec)/(14.1 cm)w = 54 rad/sec

61*. A bucket is pulled from a well by a hand-powered winch. The handle on the winch has a radius of 86 cm, the axle has a radius of 11 cm. If the handle is turned at 80 cm/sec, how fast does the bucket rise?

w = velocity/radiusw = (80 cm/sec)/(86 cm)w = .930 rad/sec

velocity = w * rv = (.930 rad/sec)(11 cm)v = 10.2 cm/sec

I have introduced the radians and rotational velocities because they are concepts that need some time to be thoroughly internalized. We will return to them periodically until they come out full blast in Circular and Rotational motion. Be sure you have made sense out of the last few problems, as they will help immensely later in the course.

Another concept that's good to be exposed to early is density. The Greek letter p (pronounced "ro") is used to represent density.

62. The density of a block of clay is 1.4 gm/cm3. What is the mass of 288 cm3 of clay? Hint: Let the units be your guide.

m = (1.4 gm/cm3)(288 cm3)m = 403 gm

63. A block of metal measures 3.6 cm x 4.2 cm x 6.7 cm. What is its density if the mass is 346 gm?

p = mass/volumep = 346 gm/[(3.6 cm)(4.2 cm)(6.7 cm)]p = 3.42 gm/cm3

64. A sphere whose radius is 5.7 cm has a mass of 426 gm. What is its density? (Recall that for a sphere, V = 4/3 r3.)

p = mass/volumep = m/[(4/3)(3.14)(r3)]p = (426 gm)/[(4/3)(3.14)(5.7)3]

p = 0.55 gm/cm3

65. A brass ball, density 5.6 gm/cm3, has a radius of 6.4 cm. What is its mass?

p = mass/volumem = p * Vm = (5.6 gm/cm3)[4/3(3.14)](6.4 cm)3

m = 6.15 * 103 gm

66. A 15 gm marble has a density of 1.6 gm/cm3. What is its radius?

p = m/VV = m/pV = (15 gm)/(1.6 gm/cm3)V = 9.38 cm3

But 9.38 cm3 = [4/3(3.14)r3]r = 1.31 cm

67. A cylinder, 5.0 cm in diameter, and 13 cm high, has a mass of 2.64 kg. What is its density in gm/cm3?

For a cylinder, V = (3.14)r2h (area of base times height)

p = m/Vp = [(2.64 kg)(1000 gm/1 kg)]/[(3.14)(2.5 cm)2(13 cm)]p = 10.3 gm/cm3

68*. A 422 gm cylinder is 2.2 inches high and has a density of 3.4 gm/cm3. What is its radius in cm? (Be careful of the mixed unit systems.)

First get everything into the same system:

(2.2 in)(2.54 cm/1 in) = 5.59 cm

Next:p = m/Vp = m/[(3.14)r2h]r = [m/[(3.14)hp]]1/2

r = [(422 gm)/[(3.14)(5.59 cm)(3.4 gm/cm3)]]1/2

r = 2.66 cm

69*. A piece of pipe has an outer radius of 4.7 cm and an inner radius of 2.6 cm. What is the mass of a 35 cm length of the pipe if its density is 8.4 gm/cm3?

To get the volume of the pipe, calculate the volume of the outer

cylinder and subtract the volume of the hole.

m = pVm = (8.4 gm/cm3)[(3.14)(4.7 cm)2(35 cm) - (3.14)(2.6 cm)2(35 cm)]m = 1.42 * 104 gm

70*. A hollow sphere has an inner radius of 7.6 cm, an outer radius of 11.4 cm, and a density of 3.8 gm/cm3. What is its mass?

As before:

m = pVm = (3.8 gm/cm3)[(4/3)(3.14)(11.4 cm)3

- (4/3)(3.14)(7.6 cm)3]m = 1.66 * 104 gm

I hope you noticed that in #70 and #71 that the terms could be simplified.

V = (3.14)r12h - (3.14)r2

2h = (3.14)h(r12 - r2

2) V = (4/3)(3.14)r1

3 - (4/3)(3.14)r23 = (4/3)(3.14)(r1

3 - r23)

71*. A cylinder, 15.0 cm long and 2.5 cm in radius, is made of two different metals bonded to make a single bar. If the densities are 4.1 gm/cm3 and 6.3 gm/cm3, what length of the lighter metal is needed? The total mass is 1500 gm.

m = p1V1 + p2V2

m = (4.1 gm/cm3)[x * (3.14) * (2.52 cm2)] + (6.3 gm/cm3)(15 - x)[(3.14) * (2.52 cm2)]m = 1500 gm

This messy algebraic equation can be simplified to:80.5x + 1855 - 123.70x = 1500x = 8.22 cm

72**. A hollow sphere has an outer radius of 21.6 cm and masses 6.8 x 104 gm. When the sphere is filled with liquid whose density is 4.8 gm/cm3 the total mass rises to 16.5 x 104 gm. What is the density of the material making up the hollow sphere?

For the inner sphere, m = pVThe mass increase was:

(16.5 * 104 - 6.8 * 104 gm) = (4.8 gm/cm3)(4/3(3.14)r3)r = 16.9 cm

So, for the outer sphere:

(6.8 * 104 gm) = p[4/3(3.14)(21.63 - 16.93) cm3]p = 3.09 gm/cm3

One last skill you will need for physics is the ability to manipulate expressions without using numbers. While the geometry of the next few problems is not used much in the rest of the course, the algebraic methods will be pervasive.

73. Express the perimeter of each figure in terms of the given dimensions.

a) b) c)

a)P = 2(s + 2s)P = 2(3s)P = 6s

b)P = 2s + s + [s2 + (2s)2]1/2

P = 3s + (5s2)1/2

P = [3 + 51/2]s

c)P = 2(3.14)r

74. Now for some slightly harder ones:

a) b) c)

a)For 1/2 circle:P1 = 1/2[2(3.14)r]P1 = (3.14)r

For base:P2 = 2r

Total:P = P1 + P2

P = 2r + (3.14)rP = [2 + (3.14)]r

b)Slant side:(s2 + s2)1/2 = (2)1/2s

Total:P = 2(2s) + 2(s) + (2)1/2sP = [6 + (2)1/2]s

c)P = 2r + 4r + 2r + r + (3.14)r + rP = [10 + (3.14)]r

75. Find the surface area of each object:

a) b) c)

a)Top and Bottom (sides #1):A1 = 2(s * 2s)A1 = 4s2

Front and Back (sides #2):A2 = 2(s * 2s)A2 = 4s2

Ends (sides #3):A3 = 2(s * s)A3 = 2s2

A = A1 + A2 + A3

A = 10s2

b)A1 = 2[(3.14)r2]A2 = [2(3.14)r]hA = A1 + A2

A = [2(3.14)r2 + 2(3.14)rh]A = [2(3.14)r(r + h)]

c)A = (2r)2 - 1/4(3.14)r2

A = [4 - (3.14/4)]r2

I hope you've noticed the units. For perimeter, which is a linear measure, the expressions are all in the first power of length. For area, they are in the second power of length, for volume, the third power. There are no exceptions to this rule.

Examples:4 (3.14)r2 must be an area.(3.14)r2h must be a volume.2L + 2W must be a perimeter.

76. Find the volumes of the figures.

a) b) c)

a)V = 1/2(b * h) * lV = (bhl)/2

b)V = [(3.14)r2]hV = (3.14)r2h

c)V = [s2 - (3.14)(s/4)2]tV = s2t[1 - (3.14/16)]

Remember that for cylinders, whether they are circular cylinders, triangular cylinders, or anything else, their volume is given by the area of the base times the height.

77*. Find the areas and volumes of the figures below.

a) b) c)

a)A1 = 2[1/2(3.14)r2]

A2 = 1/2[2(3.14)r](2r)A2 = 2(3.14)r2

A3 = (2r)(2r)A3 = 4r2

A = A1 + A2 + A3

A = [4 + 3(3.14)]r2

V = [1/2(3.14)r2](2r)V = (3.14)r3

b)A1 = 2[(2r)2 - 1/4[(3.14)r2]]A1 = [8 - (3.14/2))r2

A2 = 2(hr)A2 = 2hr

A3 = 1/4[2(3.14)r]hA3 = [(3.14)rh]/2

A4 = 2(2r * h)A4 = 4rh

A = [6 + (3.14/2)]rh + [8 - (3.14/2)]r2

V = [(2r)2 - 1/4(3.14)r2]hV = [4 - (3.14/4)]r2h

c)A1 = 1/2[4(3.14)r2

2]A1 = 2(3.14)r2

2

A2 = 1/2[4(3.14)r12]

A2 = 2(3.14)r12

A3 = (3.14)r22 - (3.14)r1

2

A3 = (3.14)(r22 - r1

2)

A = 2(3.14)(r22 + r1

2) + (3.14) (r2

2 - r12)

V = 1/2[(4/3)(3.14)r23] -

1/2[(4/3)(3.14)r13]

V = [2(3.14)/3][r23 - r1

3]

78. Karen Pheden can run around the unusual track shown in time t. What is her average velocity? (Express v in terms of r and t.)

velocity = distance/timevelocity = [1/2(2)(3.14)r + 2r]/tvelocity = [(3.14) + 2]r/t

79*. C. D. Twain can drive his car at a constant velocity v over any race track in the world. How long will it take him to complete the one at left?

velocity = distance/timetime = distance/velocitytime = [4r = 4[1/4(2)(3.14)r]]/vtime = [4 + 2(3.14)]r/v

79. The object shown at right has a density r. What is its mass?

p = m/Vm = pVm = p[(3.14)r2

2 - (3.14)r12]h

m = (3.14)ph(r22 - r1

2)

80*. If the quarter cylinder shown were to be melted down and poured into a spherical shape, what would the radius of the completed sphere be?

Let R = radius of final sphere:

4/3(3.14)R3 = 1/4(3.14)r2hR3 = 3/16r2hR = (3/16r2h)1/3

81** What size cube will have the same surface area as the hemisphere shown?

The surface area of the object is:A = 1/2[4(3.14)r2] + (3.14)r2 * 3(3.14)r2

The surface of a cube of edge S is:A = 6S6

Thus, 3(3.14)r2 = 6S6

S = [(3.14)/2]1/2r