Physics Textbook Answer Key PDF

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Answers Answers to Stop to Think Questions and Odd-Numbered Exercises and Problems Chapter 1

u

11. a.

a

u

Stop to Think Questions

u

v3

1. B. The images of B are farther apart, so it travels a larger distance than does A during the same intervals of time. 2. a. Dropped ball. b. Dust particle. c. Descending rocket. 3. e. The average velocity vector is found by connecting one dot in the motion diagram to the next. u u u u u 4. b. v 2 = v 1 + ∆v , and ∆v points in the direction ofa.

4

v2 3

2

b.

u

a

u

u

v3

v2

4

3

2

13.

u

v1

u u

u u

u

∆v

v2

u

u

a

a

Stops 5. d + c + b = a.

u

a=0

u

a Brakes

u

a=0

y

15.

Highest point

Exercises and Problems 1.

Skid begins

Stops

x Wad loses contact with rubber band

3.

t=0s

Reaches 70 mph t = 10 s

t = 20 s

y

5.

Ball released

u

v0

Wad released

u

v1

17. u

v

Starts u

v2

u

a u

a

u

v

u

a

u

a

u

v3 u

a

u

u

u

a

Ground u

7. v

Stop

9.

u

u

v0

u

v

u

v1

0

1 a1

v3 3

u

a

u

v2 2

a

4

u

a

u u

u

a2 = 0

a3

Same point shown twice

v

A-10Answers

39. Pictorial representation

21. u

a0 x x0, v0x, t0

x0, v0, t0

x1, v1x, t1

u

a

Known v0x = 0 m/s t0 = 0 s x0 = 0 m a 0x = 1.5 m/s2 v1x = 7.5 m/s

Known x0 = 0 m x1 = 10 m/s t0 = 0 s v0x = 0 m/s

30° x1, v1, t1

Find x1

Find v1x

Motion diagram u

23. a. 1 b. 3 c. 2 d. 5 25. a. 1.9 m b. 1.09 * 1014 s c. 2.2 * 10-4 m/s 27. a. 7 m d. 100,000 m c. 30 m/s d. 0.2 m 29. a. 32,590 b. 9.0 c. 0.237 d. 4.78 31. a. 15 m 33. 32 ms 35. Pictorial representation

v d. 5.7 * 1010 m2

u

a

41. Pictorial representation

u

a

u

a

x0 , v0 x , t0

x1, v1x , t1

10°

x1, v1x , t1 x0 , v0 x , t0

Motion diagram u

a u

v

Motion diagram Known x0 = 0 m v0x = 300 m/s

u

a

x1 = 4 km t0 = 0 s

Stops u

v

Find ax

Known

Pictorial representation y

37.

x0 = 0 m v1x = 0 m/s t0 = 0 s v0x = 30 m/s

Motion diagram

Find x1

Ceiling

y1, v1y, t1

43. Pictorial representation

Known y0 = 0 m v0y = 10 m/s t0 = 0 s y1 = 3.0 m a 0y 6 0

u

u

aD = 0

David xD0 , tD0 , vD0 x u

a0

xD1, tD1, vD1x

u

a0

u

aT Tina

Find t1

xT0 , tT0 , vT0 x

xT1, tT1, vT1x

Motion diagram u

u

y0, v0y, t0

aD = 0

u

Start u

vD

v

u

aT u

vT Known xD0 = 0 m xT0 = 0 m tD0 = 0 s t T0 = 0 s vD0 x = 30 m/s vT0 x = 0 m/s a D0 x = 0 m/s2

Find xT1

Answers to Stop to Think Questions and Odd-Numbered Exercises and ProblemsA-11

49. a.

9. a x (m/s2 )

u

a u

v

1

Coasting begins

Stop

0

51. a.

t (s) 2

4

-1 u

a 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43.

u

a u

v

u

v 3

2

53. Smallest: 6.4 * 10 m , largest: 8.3 * 103 m2 55. 2.9 * 10-4 m3 57. a. 83.3 kg/m3 b. 810 kg/m3

Chapter 2 Stop to Think Questions 1. d. The particle starts with positive x and moves to negative x. 2. c. The velocity is the slope of the position graph. The slope is positive and constant until the position graph crosses the axis, then positive but decreasing, and finally zero when the position graph is horizontal. 3. b. A constant positive vx corresponds to a linearly increasing x, starting fromxi = - 10 m. The constant negativevx then corresponds to a linearly decreasing x. 4. a and b. The velocity is constant while a= 0; it decreases linearly while a is negative. Graphs a, b, and c all have the same acceleration, but only graphs a and b have a positive initial velocity that represents a particle moving to the right. 5. d. The acceleration vector points downhill (negative s-direction) and has the constant value- g sin u throughout the motion. 6. c. Acceleration is the slope of the graph. The slope is zero at B. Although the graph is steepest at A, the slope at that point is negative, and so a A 6 a B . Only C has a positive slope, soa C 7 a B .

a. 8.0 m b. 2.0 m/s c. –2.0 m/s2 a. 2.7 m/s2 b. 1.3 * 102 m = 4.3 * 102 feet a. 360 d b. 4.6 * 1015 m c. 0.49 2.8 m/s2 216 m 3.2 s 5.2 cm 73 m 265 m a. 16 m/s b. 31 m 16 m/s a. 21 m b. 26 m/s c. 24 m/s2 a. 0 s and 3 s b. 12 m and –18 m/s2; –15 m and 18 m/s2 a. –10 m/s b. –20 m/s c. 95 m/s a. 2 s, 5 s b. –3 m/s2, 3 m/s2 a. 20 s b. 667 m s

t

0 vs

0 as

t

0

t

Exercises and Problems 1. a. Beth b. 20 min 3. a. 48 mph b. 50 mph 5. a. vx (m/s)

b. None

10 45. 0

t (s) 1

-10 7. 8.0 cm

2

3

4 47. 49. 51. 53. 55. 57. 59.

a. Yes b. 35 s c. No a. 5 m b. 22 m/s 5.7 m/s Yes, 10 m a. 54.8 km b. 228 s 19.7 m 55 cm

A-12Answers 61. a. 2.0 h b. 73 m c. x (mi)

3. 5. 7. 9. 11. 13.

73

a. E sin u, - E cos u b. E cos f, - E sin f 8m a. 3.8 m/s, 6.5 m/s b. –1.3 m/s2 , 0.80 m/s2 c. –30 N, 40 N 100 m, west a. 7.6, 67° b. 4.9 m/s2, 66° c. 18 m/s, 38 ° d. 4.0 m, 34° a. 1in + 3jn b. y 6

l

4

An n

Ca

ro

Known u A = 4dn - 2en u B = - 3dn + 5en

2.4

t (h) 0

63. 65. 67. 69. 71. 73. 77. 79. 81. 83. 85.

0.5

1.0

1.5

vf = 22gh 17 cm 14 m/s 0.36 m 4.4 m/s2 gh/d c. 17.2 m/s c. 750 m a. 10 s b. 3.8 m/s2 c. 5.6% 12.5 m/s 4500 m/s2

2.0

u

u

B

2

u

Find u

C

u

u

C=A+B C and u

-4

-2

x

0

2

4

u

A -2

-4 15. 17. 19. 21. 23. 25.

Chapter 3

27.

Stop to Think Questions u

u

u

1. c. The graphical construction ofA 1 + A2 + A 3 is shown in the figure. u

u

A1 + A2

u

A1 u

A3 u

A2 u

Parallelogram u u of A1 and A2 u

u

A1 + A2 + A3 u

u

29. 31. 33. 35. 37. 39. 41. 43. 45.

c. 3.2, 72° above the+ x-axis a. - 1in + 11jn c. 11, 5.2 ° ccw from the+ y-axis a. 2.8 b. 4.1 c. 6.1 vx = - 50 m /s, vy = - 87 m /s B = 2.2 T, u = 27° u a. 0 m, 26 m, 160 m b. v 1t2 = 110i n+ 8.0jn2 t m/s c. 0 m/s, 26 m/s, and 6 m/s u C = 0.8 ni - 4.5 jn 1 1 n u B= in + j 22 22 u u a. ∆r = 11.7 cm2in - 13.0 cm2jn b. ∆ r = 0.0 cm 90 m, 46° south of west vx = 3.7 m/s, vy = 2.6 m/s a. –3.4 m/s b. –9.4 m/s 280 N, 350 N 570 N, –380 N 5.2 mm/s at 27 ° below the positive x-axis 4.4 units at 83° below the negative x-axis 7.3 N at 79° below the negative x-axis

u

Parallelogram of (A1 + A2) and A3 u

u

2. a. The graphical construction of2A - B is shown in the figure.

Chapter 4 Stop to Think Questions

u

-B u

u

2A - B u

2A

3. Cx = − 4 cm, Cy = 2 cm. u 4. c. VectorC points to the left and down, so bothC x and C y are negative. f. C x is in the numerator because it is the side opposite

1. c. v = 0 requires both vx = 0 and vy = 0. Neither x nor y can be changing. u u 2. d. The parallel component ofa is opposite v and will cause the particle u to slow down. The perpendicular component aofwill cause the particle to change direction downward. u a = - gj does not depend on its mass. The 3. d. A projectile’s acceleration second marble has the same initial velocity and the same acceleration, so it follows the same trajectory and lands at the same position. u v PH = vuPG + uv GH = 4. f. The plane’s velocity relative to the helicopter is u u u v PH v PG - vHG, where G is the ground. The vector addition shows that is to the right and down with a magnitude greater than the 100 m/s u of v PG . u

vPG of plane relative to ground

Exercises and Problems

u

u

u

1. a. A + B

B

u

b.

u

A

u

-B u

A

u

u

A-B

u

u

u

vPH = vPG - vHG of plane relative to helicopter

- vHG

Answers to Stop to Think Questions and Odd-Numbered Exercises and ProblemsA-13 5. b. An initial cw rotation causes the particle’s angular position to become increasingly negative. The speed drops to half after reversing direction, so the slope becomes positive and is half as steep as the initial slope. = 0°. Turning through the same angle returns the particleu to 6. ab + ae + aa = ac + ad. Centripetal acceleration isv 2/r. Doubling r decreasesa r by a factor of 2. Doubling v increasesa r by a factor of 4. a r. Reversing direction doesn’t change v is negative but 7. c. v is negative because the rotation is cw. Because becoming less negative, the change∆v is positive. So a is positive.

75. 77. 79. 81. 83. 85.

1940 rpm 550 rpm b. 30 m west 34.3° 3.8 m 10°

Chapter 5 Stop to Think Questions

Exercises and Problems 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25.

1. c. u

C E 2.2 m/s2 12 ni - 8 jn2 m/s2 19.6 m 680 m r = 16.4 m vx = 27.7 m/s 30 s a. 42° west of north b. 45 s 10 rev u (rad) 80

F1

u

F2

u

F3

u

The y-component of F3 u cancels the y-component of F1. u

The x-component of F3 is to the left and larger uthan the x-component of F2. 2. a, b, and d. Friction and the normal force are the only contact forc Nothing is touching the rock to provide a “force of the kick.” 3. b. Acceleration is proportional to force, so doubling the number rubber bands doubles the acceleration of the original object fr 2 m/s2 to 4 m/s 2 . But acceleration is also inversely proportional to mas Doubling the mass cuts the acceleration in half, back to2 m/s2. 4. d.

60

u

u

u

F1

F1 + F2

u

u

F3

F3

u

u

F2

u

u

u

a is in the same u direction as Fnet. u

t (s) 0

2

4

6

8

a. 4.7 rad/s b. 1.3 s 680 km/h, 1040 mph 43 m a. 3.0 * 104 m/s b. 2.0 * 10-7 rad/s c. 6.0 * 10-3 m/s2 v = 5.7 m/s, a r = 108 m/s2 a. 3.75 rad/s b. 5.0 rad/s c. 5.0 rad/s 98 rpm 47 rad/s2 38 rev 1 1bt 2 + v0x 2 ni + 1e -ct + v0y2jn 2

v 20 sin2 u b. 14.4 m, 28.8 m, 43.2 m 2g a. 12 m/s b. 0.90 m Clears by 1.0 m 7.4 m/s a. 13 m/s b. 48° 470 m/s2 4.8 m/s a. 39 mi b. 20 mph No. The angular acceleration is negative. a. 1.75 * 104 m/s 2 b. 4.4 * 103 m/s2 69 m/s at 21° with the vertical a. - 100 rad/s 2 b. 50 rev 0.75 rad/s 2 2 ∆u R 22 ∆uR b

Exercises and Problems 1. 3. 5. 7. 9. 11. 13. 15. 17.

Gravity, normal force, static friction Gravity, normal force, kinetic friction Gravity, drag a. 2.4 m/s2 b. 0.60 m/s2 m 1 = 0.080 kg, m 3 = 0.50 kg 1.5 J a. 4 m/s2 b. 2 m/s2 25 kg u

47. a. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73

a

F net 5. c. The acceleration vector points downward as the elevator slows. u u points in the same direction asa , so F net also points down. This will be T 6 FG . true if the tension is less than the gravitational force:

0 27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

Then add F3. This is Fnet.

First addu u F1 and F2.

20

u

Fnet

40

F3 u

u

F2

F2

u u

F1

u

F1 + F2

u

F1

u

19. F1 u

F3 u

A-14Answers 21.

33. a. 16 m/s2 b. 4.0 m/s2 c. 8.0 m/s2 d. 32 m/s2 y 35.

A rocket accelerates upward

y u

Fthrust u

u

Fnet

n

u

u

u

D

FG

D

23. Force identification

u

Fthrust

Free-body diagram

u

x

u

FG

u

Fnet = 0

y

u

u

a=0 u

n

u

u

v

u

Fnet = 0 x u

FG

u

Normal force n

y

37.

u

Gravity FG Force identification

25.

Free-body diagram x

y u

FG

u

n

u

a

x

u

fk

u

Gravity FG

u

Fnet

u

u

FG

u

Normal force n u Kinetic friction fk

Fnet

39.

y u

n

27. Force identification

x

u

fk

Free-body diagram

u

Fnet

y u

u

Drag Fdrag

Thrust Fthrust

u

a

u

FG

u

n u

u

Fdrag

F thrust

u

v x

u

FG

u

Gravity FG

Normal force n

u

Fnet

41. a.

Fx (N)

0.8

6 4

0.6 0.4

2

0.2

0

0.0 0.0

t (s) 1

2

3

4

-2 31.

1.0

Force (N)

29.

u

F = (0.57 { 0.03 kg) a 0.5 1.0 Acceleration (m/s2)

1.5

b. Yes. 0 m/s2, 0 N c. 57 kg

a x (m/s2) 43. 1.5

u

u

v

Air resistance (drag) F drag

y u

1.0

F thrust u

Fnet x

t (s)

0.0 1 - 0.5

u

a

0.5

2

3

u

Propulsion Fthrust

4 u

Gravity FG

u

Fdrag

u

FG

Answers to Stop to Think Questions and Odd-Numbered Exercises and ProblemsA-15 47.

b.

c.

u

y

v

u

Gravity FG

u

a

y

u

Normal force n

u

n u

Fnet

u

n u

Fnet

x

x u u

fk

Normal force un u Kinetic friction fk

Gravity FG

u

FG

u

FG

u

49.

v

Chapter 6 Stop to Think Questions y u

a

u

u

Tension T

Fnet

u

T

x u

FG

u

Gravity FG

51.

y

1. a. The lander isu descending and slowing. The acceleration vector poi upward, and soF net points upward. This can be true only if the thrust ha a larger magnitude than the weight. 2. a. You are descending and slowing, so your acceleration vector points ward and there is a net upward force on you. The floor pushes up agai your feet harder than gravity pulls down. 3. f b + f c = f d = f e + f a Situations c, d, and e are all kinetic friction, which does not depend on either velocity or acceleration. Kinetic friction is smaller than the maximum static friction that is exerted in b. fa = 0 because no friction is needed to keep the object at rest. 4. d. The ball is shot down at 30 m/s, so v0y = - 30 m/s. This exceeds the terminal speed, so the upward drag force is larger than the downward weight force. Thus the ball slows down even though it is “falling.” It will slow until vy = - 15 m/s, the terminal velocity, then maintain that velocity.

u

Spring force F Sp

u

n

u

Exercises and Problems

u

u

fk

FSp

v

x

u u

Gravity FG

u

a

Friction f k

53.

u u

FG Fnet

u

Normal force n

y u

n

x u

FG

55. u

v

y u

u

v

u

v

u

n u fs

u

a

Fnet

u

Gravity F G

u

Normal force n u Static friction f s

u

FG

x

1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

57. a.

First contact

Loses contact u

u

u

a

a u

v u

a

Expanding

Magnified view of ball in contact with ground

Compressing

v

47. 49. 51. 53. 55. 57. 59. 61.

94 N, 58° below the horizontal 510 N 160 N a. 0.0036 N b. 0.010 N a x = 1.5 m/s2, a y = 0 m/s2 0 m/s2, 4 m/s a. 490 N b. 490 N c. 740 N d. 240 N 40 s a. 540 N b. m = 55 kg; mg = 21...