Answer Key

Chapter 1

Problems & Exercises

1.

(a) 27.8 m/s

(b) 62.1 mph

3.

$$\frac{\text{1.0 m}}{s} = \frac{1\text{.0
m}}{s} \times \frac{\text{3600 s}}{\text{1 hr}} \times \frac{1\
km}{\text{1000 m}}$$

= 3.6 km/h.

5.

length: 377 ft; 4.53 × 10³ in. width: 280 ft; 3.3 × 10³ in.

7.

8.847 km

9.

(a) 1.3 × 10⁻⁹ m

(b) 40 km/My

11.

2 kg

13.

(a) 85  to 95  km/h

(b) 53  to 59  mi/h

15.

(a) 7.6 × 10⁷ beats

(b) 7.57 × 10⁷ beats

(c) 7.57 × 10⁷ beats

17.

(a) 3

(b) 3

(c) 3

19.

(a) 2.2%

(b) 59 to 61 km/h

21.

80 ± 3 beats/min

23.

2.8 h

25.

11.4 ± 0.6 cm³

27.

12.06 ± 0.04 m²

29.

Sample answer: 2 × 10⁹
heartbeats

31.

Sample answer: 2 × 10³¹
if an average human lifetime is taken to be about 70 years.

33.

Sample answer: 50 atoms

35.

Sample answers:

(a) 10¹²
cells/hummingbird

(b) 10¹⁶ cells/human

Chapter 2

Problems & Exercises

1.

(a) 7 m

(b) 7 m

(c) +7 m

3.

a. [latex]8\text{~m} + 2\text{~m} + 3\text{~m} = {13\text{~m}}[/latex]

b. [latex]{9\text{~m}}[/latex]

c. [latex]\Delta x = 11\text{~m} - 2\text{~m} = {9\text{~m}}[/latex]

5.

(a) 3.0 × 10⁴ m/s

(b) 0 m/s

7.

2 × 10⁷ years

9.

34.689 m/s = 124.88 km/h

11.

(a) 40.0 km/h

(b) 34.3 km/h, 25º S of E.

(c)average speed = 3.20
km/h

13.

384,000 km

15.

(a) 6.61 × 10¹⁵ rev/s

(b) 0 m/s

16.

4.29 m/s²

18.

(a) 1.43 s

(b) −2.50 m/s²

20.

(a) 10.8 m/s

(b)

Image Description

The image is a graph titled “Position vs. Time.” It depicts a curve showing how position changes over time. The horizontal axis is labeled “Time (s)” and spans from 0 to 6 seconds, with intervals marked at each second. The vertical axis is labeled “Position (m)” and ranges from 0 to 60 meters, with intervals marked every 10 meters.

The graph is a curve that begins at the origin (0, 0) and rises steadily, illustrating an acceleration. Data points marked by diamonds are plotted along the curve at various points, indicating that the object’s position increases more rapidly as time progresses.

21.

38.9 m/s (about 87 miles per hour)

23.

(a) 16.5 s

(b) 13.5 s

(c) −2.68 m/s²

25.

(a) 20.0 m

(b) −1.00 m/s

(c) This result does not really make sense. If the runner starts at
9.00 m/s and decelerates at 2.00
m/s², then she will have stopped after 4.50 s. If she
continues to decelerate, she will be running backwards.

27.

0.799 m

29.

(a) 28.0 m/s

(b) 50.9 s

(c) 7.68 km to accelerate and 713 m to decelerate

31.

(a) 51.4 m

(b) 17.1 s

33.

(a) −80.4 m/s²

(b) 9.33 × 10⁻² s

35.

(a) 7.7 m/s

(b) −15 × 10² m/s². This is
about 3 times the deceleration of the pilots, who were falling from
thousands of meters high!

37.

(a) 32.6 m/s²

(b) 162 m/s

(c) v > v_max,
because the assumption of constant acceleration is not valid for a
dragster. A dragster changes gears, and would have a greater
acceleration in first gear than second gear than third gear, etc. The
acceleration would be greatest at the beginning, so it would not be
accelerating at 32.6 m/s²
during the last few meters, but substantially less, and the final
velocity would be less than 162 m/s.

39.

104 s

40.

(a) v = 12.2 m/s; a = 4.07 m/s²

(b) v = 11.2 m/s

41.

(a) y₁ = 6.28 m;
v₁ = 10.1 m/s

(b) y₂ = 10.1 m;
v₂ = 5.20 m/s

(c) y₃ = 11.5  m;
v₃ = 0.300 m/s

(d) y₄ = 10.4 m;
v₄ = −4.60 m/s

43.

v₀ = 4.95
m/s

45.

(a) a = −9.80
m/s²; v₀ = 13.0 m/s; y₀ = 0 m

(b) v = 0m/s. Unknown is
distance y to top of
trajectory, where velocity is zero. Use equation v² = v₀² + 2a(y − y₀)
because it contains all known values except for y, so we can solve for y. Solving for y gives

[latex]\begin{align*}v^2-v^2_0&=2a\left(y-y_0\right)\\[1.5ex]\frac{v^2-v^2_0}{2a}&=y-y_0\\[1.5ex]y&=y_0+\frac{v^2-v^2_0}{2a}=0\text{m}+\frac{\left(0\text{m/s}\right)^2-\left(13.0\text{m/s}\right)^2}{2\left(-9.80\text{m/s}\right)^2}=8.62\text{m}\end{align*}[/latex]

Dolphins measure about 2 meters long and can jump several times their
length out of the water, so this is a reasonable result.

(c) 2.65 s

Image Description

The image is a diagram illustrating a projectile motion scenario. It shows a side view of a surface with two distinct levels. The left side is a raised platform with a vertical drop to the lower level on the right. An arrow, labeled with an “S”, begins from the top of the platform, arcs upward, and then downward to the lower level, indicating the trajectory of a projectile. There are two labels: “y₀ = ?” near the top of the platform, prompting the initial vertical position, and “y = 0 m” on the lower level, indicating the final vertical position is at ground level.

47.

(a) 8.26 m

(b) 0.717 s

49.

1.91 s

51.

(a) 94.0 m

(b) 3.13 s

53.

(a) -70.0 m/s (downward)

(b) 6.10 s

55.

(a) 19.6 m

(b) 18.5 m

57.

(a) 305 m

(b) 262 m, -29.2 m/s

(c) 8.91 s

59.

(a) 115 m/s

(b) 5.0 m/s²

61.

[latex]v = \frac{\left( \text{11.7} - 6.95 \right) \times \text{10}^{3}\ \text{m}}{\left( 40\text{.0 – 20}.0 \right)\ \text{s}} = \text{238 m/s}[/latex]

63.

Image Description

The image is a graph titled “Position vs. Time” showing the relationship between time in seconds (x-axis) and position in kilometers (y-axis). The x-axis ranges from 0 to 25 seconds, while the y-axis ranges from 4.68 to 4.82 kilometers. A red line with two data points illustrates the position increasing over time. Initially, the line is relatively flat from 0 to about 9 seconds, indicating a slower change in position. From 9 to 21 seconds, the line curves upward, suggesting an acceleration in the change of position.

65.

(a) 6 m/s

(b) 12 m/s

(c) 3 m/s²

(d) 10 s

67.

(a) Car A is traveling faster at the checkpoint because it must go
past the speed of car B to reach the same distance.

(b) i. Yes, the equation is consistent with the answer because the
speed of car A is only a constant away from the correct answer. ii. Yes,
the equation makes sense because V = 2V₀.

(c)
gn: center;”>

Image Description

The image is a graph with two lines representing different functions. The graph has an X-axis labeled as x and a Y-axis labeled as V. There are two important lines on the graph:

• A horizontal blue line labeled as V_B, which represents the constant value V₀ and runs parallel to the X-axis.
• A slanted maroon line labeled as V_A, which starts at the origin (0,0) and increases linearly to reach 2V₀ at t_A.

The Y-axis is marked at two points: V₀ and 2V₀. There is also a dashed line extending horizontally from the point where the slanted line reaches 2V₀ down to the X-axis at t_A. The background is a light beige color.

Chapter 3

Problems & Exercises

1.

(a) 480 m

(b) 379 m, 18.4º east of north

3.

north component 3.21 km, east component 3.83 km

5.

19.5 m, 4.65º south of west

7.

(a) 26.6 m, 65.1º north of east

(b) 26.6 m, 65.1º south of west

9.

52.9 m, 90.1º with respect to the
x-axis.

11.

x-component 4.41 m/s

y-component 5.07 m/s

13.

(a) 1.56 km

(b) 120 m east

15.

North-component 87.0 km, east-component 87.0 km

17.

30.8 m, 35.8 west of north

19.

(a) 30.8 m, 54.2º south of west

(b) 30.8 m, 54.2º north of east

21.

18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at
45.0º west of north

23.

7.34 km, 63.5º south of east

25.

[latex]\begin{align*}x&=1.30\text{m}\times 10^2\\[1.5ex]y=30.9\text{m}\end{align*}[/latex]

27.

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, 50.2º below
horizontal

29.

(a) 18.4º

(b) The arrow will go over the branch.

[latex]R=\frac{v^2_0}{\sin2\theta_0g}[/latex]

31.

[latex]\text{For}\;\theta=45\degree,\;\;R=\frac{v^2_0}{g}[/latex]

R = 91.8 m for v₀ = 30 m/s; R = 163 m for v₀ = 40 m/s; R = 255 m for v₀ = 50 m/s.

33.

(a) 560 m/s

(b) 8.00 × 10³ m

(c) 80.0 m. This error is not significant because it is only 1% of
the answer in part (b).

35.

1.50 m, assuming launch angle of 45º

37.

θ = 6.1º

yes, the ball lands at 5.3 m from the net

39.

(a) −0.486 m

(b) The larger the muzzle velocity, the smaller the deviation in the
vertical direction, because the time of flight would be smaller. Air
resistance would have the effect of decreasing the time of flight,
therefore increasing the vertical deviation.

41.

4.23 m. No, the owl is not lucky; he misses the nest.

43.

No, the maximum range (neglecting air resistance) is about 92 m.

45.

15.0 m/s

47.

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of
wind.

49.

[latex]y - y_{0} = 0 = v_{0y}t - \frac{1}{2}{gt}^{2} = (v_{0}\ \text{sin}\ \theta)t - \frac{1}{2}{gt}^{2}[/latex],

so that [latex]t = \frac{2\left( v_{0}\ \text{sin}\ \theta \right)}{g}[/latex]

x − x₀ = v_0xt = (v₀ cos θ)t = R,
and substituting for t
gives:

$$R = v_{0}\ \text{cos}\ \theta\left(
\frac{{2v}_{0}\ \text{sin}\ \theta}{g} \right) = \frac{{2v}_{0}^{2}\
\text{sin}\ \theta\ \text{cos}\ \theta}{g}$$

since 2 sin θ cos θ = sin 2θ,
the range is:

[latex]R = \frac{{v_{0}}^{2}\ \text{sin}\ 2\theta}{g}[/latex].

52.

(a) 35.8 km, 45º south of east

(b) 5.53 m/s, 45º south of east

(c) 56.1 km, 45º south of east

54.

(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m

56.

17.0 m/s, 22.1º

58.

(a) 230 m/s, 8.0º south of west

(b) The wind should make the plane travel slower and more to the
south, which is what was calculated.

60.

(a) 63.5 m/s

(b) 29.6 m/s

62.

6.68 m/s, 53.3º south of west

64.

(a) [latex]H_{\text{average}} = \text{14.9}\frac{\text{km/s}}{\text{Mly}}[/latex]

(b) 20.2 billion years

66.

1.72 m/s, 42.3º north of east

71.

(a) Since sin²50 > sin²40,
then B reaches the greatest height.

(b) i. Yes, it is consistent because sin 50 > sin40.
ii. No, it does not make sense because y is proportional to
sin²θ.

(c)

Image Description

The image is a graph illustrating two lines labeled A and B on a coordinate system. The horizontal axis is labeled “y,” and the vertical axis is labeled “V.”

The line labeled “A” is shown in a purple color and starts at a point labeled “V_0A” on the vertical axis, sloping downwards towards the horizontal axis.

The line labeled “B” is shown in a blue color and starts at a higher point labeled “V_0B” on the vertical axis, also sloping downwards towards the horizontal axis. It intersects the horizontal axis farther along than line A.

 

Chapter 4

Problems & Exercises

1.

265 N

3.

13.3 m/s²

7.

(a) 12 m/s².

(b) The acceleration is not one-fourth of what it was with all
rockets burning because the frictional force is still as large as it was
with all rockets burning.

9.

(a) The system is the child in the wagon plus the wagon.

(b)

Image Description

The image shows a diagram of forces acting on an object represented by a small circle labeled “m” at the center. There are four arrows originating from this circle in different directions:

• An upward arrow labeled “N” indicating a normal force.
• A downward arrow labeled “w” depicting weight.
• A rightward arrow labeled “F₂” showing an applied force.
• A leftward arrow labeled “F₁” indicating another applied force, with an additional smaller arrow labeled “f” pointing in the left direction representing friction.

(c) a = 0.130 m/s² in the
direction of the second child’s push.

(d) a = 0.00 m/s²

11.

(a) 3.68 × 10³ N . This
force is 5.00 times greater than his weight.

(b) 3750 N; 11.3º above
horizontal

13.

1.5 × 10³ N, 150 kg, 150
kg

15.

Force on shell: 2.64 × 10⁷ N

Force exerted on ship = −2.64 × 10⁷ N, by Newton’s third
law

17.

(a) 0.11 m/s²
(b) 1.2 × 10⁴ N

19.

(a) 7.84 × 10⁻⁴ N

(b) 1.89 × 10⁻³ N. This
is 2.41 times the tension in the vertical strand.

21.

Newton’s second law applied in vertical direction gives

F_y = F − 2T sin θ = 0

F = 2T sin θ

$$T = \frac{F}{\text{2 sin}\
\theta}.$$

Image Description

The image is a physics diagram representing forces acting on a mass. In the center, there is a circle labeled with the letter “m” to denote mass. There are three arrows indicating forces acting on this mass:

• An upward arrow labeled “T” representing tension.
• A downward arrow labeled “f” representing force, likely friction or another opposing force.
• A downward arrow labeled “mg” representing the gravitational force, where “m” is mass and “g” is the acceleration due to gravity.

23.

Using the free-body diagram:

F_net = T − f − mg = ma,

so that

[latex]a = \frac{T - f - \text{mg}}{m} = \frac{1\text{.250} \times \text{10}^{7}\ \text{N} - 4.50 \times \text{10}^{\text{6}}\ N - (5.00 \times \text{10}^{5}\ \text{kg})\left( 9.\text{80 m/s}^{2} \right)}{5.00 \times \text{10}^{5}\ \text{kg}} = \text{6.20}\ \text{m/s}^{2}[/latex].

25.

1. Use Newton’s laws of motion.

   

   Image Description

   The image is a physics diagram illustrating forces acting on a mass. In the center, there is a circle labeled with the letter “m,” representing mass. There are two arrows extending vertically from this circle:

   • An upward arrow on the left side is labeled “F,” indicating a force applied upwards.
   • A downward arrow on the left side is labeled “w,” representing the weight acting downwards.

   To the right of the circle, there is another upward arrow labeled “a,” representing acceleration in the upward direction.

2. Given : a = 4.00g = (4.00)(9.80
   m/s²) = 39.2 m/s²; m = 70.0
   kg,

• Find: F.

3. ∑F=+F − w = ma, so
   that F = ma + w = ma + mg = m(a + g).

• F = (70.0 kg)[(39.2
  m/s²) + (9.80
  m/s²)] = 3.43 × 10³N. The force exerted by
  the high-jumper is actually down on the ground, but F is up from the ground and makes
  him jump.

4. This result is reasonable, since it is quite possible for a
   person to exert a force of the magnitude of 10³ N.

27.

(a) 4.41 × 10⁵ N

(b) 1.50 × 10⁵ N

29.

(a) 910 N

(b) 1.11 × 10³ N

31.

a = 0.139 m/s, θ = 12.4º north of east

33.

1. Use Newton’s laws since we are looking for forces.
2. Draw a free-body diagram:
   
   Image Description

   The image depicts a diagram with three arrows forming angles with a horizontal dashed line. The central arrow points vertically downward, creating a vertical line which intersects the horizontal dashed line. Two additional arrows extend from the intersection point, pointing downward and outward at symmetrical angles. These two arrows are each labeled with “15°”, indicating that they form a 15-degree angle with the horizontal dashed line. The overall formation resembles an inverted and narrow ‘V’ shape with the central vertical arrow splitting the ‘V’.

3. The tension is given as T = 25.0 N. Find F_app. Using Newton’s laws
   gives: Σ F_y = 0, so
   that applied force is due to the y-components of the two
   tensions: F_app = 2 T sinθ = 2(25.0
   N)sin(15º) = 12.9 N

• The x-components of the tension cancel. ∑F_x = 0.

4. This seems reasonable, since the applied tensions should be
   greater than the force applied to the tooth.

40.

10.2 m/s², 4.67º from
vertical

Image Description

The image depicts a free body diagram of an object with mass labeled as “m” at the center. Three forces are acting on the object:

• Tension T1: Represented by an upward and leftward pointing dashed arrow inclined at 15 degrees from the vertical axis.
• Tension T2: Represented by a rightwards pointing dashed arrow inclined at 10 degrees from the horizontal axis.
• Weight w: Represented by a downward pointing solid arrow along the vertical axis.

 

42.

T₁ = 736 N

T₂ = 194 N

44.

(a) 7.43 m/s

(b) 2.97 m

46.

(a) 4.20 m/s

(b) 29.4 m/s²

(c) 4.31 × 10³ N

48.

(a) 47.1 m/s

(b) 2.47 × 10³ m/s²

(c) 6.18 × 10³ N . The
average force is 252 times the shell’s weight.

52.

(a) 1 × 10⁻¹³

(b) 1 × 10⁻¹¹

54.

10²

55.

(a) Box A travels faster at the finishing distance since a greater
force with equal mass results in a greater acceleration. Also, a greater
acceleration over the same distance results in a greater final
speed.

(b) i. Yes, it is consistent because a greater force results in a
greater final speed. ii. It does not make sense because [latex]V = \sqrt{2(f/m)x} = \text{K}\sqrt{(F)}[/latex].

(c)

Image Description

The image shows a plot with a downward sloping curve on a two-dimensional graph. The x-axis is labeled with the letter “F” and the y-axis is labeled with the letter “x”. The curve starts high on the y-axis and slopes downward as it moves to the right along the x-axis, indicating an inverse relationship. The equation x ∝ 1/F is written below the graph, suggesting that x is inversely proportional to F.

56.

5.00 N

59.

(a) 588 N

(b) 1.96 m/s²

61.

(a) 3.29 m/s²

(b) 3.52 m/s²

(c) 980 N; 945 N

65.

1.83 m/s²

69.

(a) 4.20 m/s²

(b) 2.74 m/s²

(c) –0.195 m/s²

71.

(a) 1.03 × 10⁶ N

(b) 3.48 × 10⁵ N

73.

(a) 51.0 N

(b) 0.720 m/s²

Chapter 5

Problems & Exercises

1.

723 km

3.

5 × 10⁷ rotations

5.

117 rad/s

7.

76.2 rad/s

728 rpm

8.

(a) 33.3 rad/s

(b) 500 N

(c) 40.8 m

10.

12.9 rev/min

12.

4 × 10²¹ m

14.

a) 3.47 × 10⁴ m/s², 3.55 × 10³ g

b) 51.1 m/s

16.

a) 31.4 rad/s

b) 118 m/s

c) 384 m/s

d)The centripetal acceleration felt by Olympic skaters is 12 times
larger than the acceleration due to gravity. That’s quite a lot of
acceleration in itself. The centripetal acceleration felt by Button’s
nose was 39.2 times larger than the acceleration due to gravity. It is
no wonder that he ruptured small blood vessels in his spins.

18.

a) 0.524 km/s

b) 29.7 km/s

20.

(a) 1.35 × 10³ rpm

(b) 8.47 × 10³ m/s²

(c) 8.47 × 10^–12 N

(d) 865

21.

(a) 16.6 m/s

(b) 19.6 m/s²

(c)

Image Description

The image depicts a simple diagram featuring a rectangle with a vertical line passing through its center. The line extends beyond the rectangle’s top and bottom edges. At the top end of the line is an arrow pointing upward labeled “N” for North. At the bottom end, another arrow points downward labeled “W” for West. The rectangle seems to represent some form of directional layout.

(d) 1.76 × 10³ N or
3.00 w , that is, the normal force (upward) is three
times her weight.

(e) This answer seems reasonable, since she feels like she’s being
forced into the chair MUCH stronger than just by gravity.

22.

a) 40.5 m/s²

b) 905 N

c) The force in part (b) is very large. The acceleration in part (a)
is too much, about 4 g.

d) The speed of the swing is too large. At the given velocity at the
bottom of the swing, there is enough kinetic energy to send the child
all the way over the top, ignoring friction.

23.

a) 483 N

b) 17.4 N

c) 2.24 times her weight, 0.0807 times her weight

25.

4.14º

27.

a) 24.6 m

b) 36.6 m/s²

c) a_c = 3.73 g.
This does not seem too large, but it is clear that bobsledders feel a
lot of force on them going through sharply banked turns.

29.

a) 2.56 rad/s

b) 5.71º

30.

a) 16.2 m/s

b) 0.234

32.

a) 1.84

b) A coefficient of friction this much greater than 1 is unreasonable
.

c) The assumed speed is too great for the tight curve.

33.

a) [latex]g=\frac{GM}{r^{2}} \rightarrow M=\frac{r^{2}g}{G}=5.95296\times 10^{24}\text{kg}[/latex].

b) This is slightly, but measurably, smaller than the accepted value,
differing in the third significant digit. The equation we used to find
g assumes a sphere, but we already noted that the radius at the
pole is smaller than at the equator. The calculation gives the mass for
a sphere of the radius we used—and ignores the additional mass outside
that radius.

35.

a) 1.62 m/s²

b) 3.75 m/s²

37.

a) 3.42 × 10^–5 m/s²

b) 3.34 × 10^–5 m/s²

The values are nearly identical. One would expect the gravitational
force to be the same as the centripetal force at the core of the
system.

39.

a) 7.01 × 10^–7 N

b) 1.35 × 10^–6 N, 0.521

41.

a) 1.66 × 10^–10 m/s²

b) 2.17 × 10⁵ m/s

42.

a) 2.937 × 10¹⁷ kg

b) 4.91 × 10^–8

of the Earth’s mass.

c) The mass of the mountain and its fraction of the Earth’s mass are
too great.

d) The gravitational force assumed to be exerted by the mountain is
too great.

Chapter 6

Problems & Exercises

1.

3.00 J = 7.17 × 10⁻⁴ kcal

3.

(a) 5.92 × 10⁵ J

(b) −5.88 × 10⁵ J

(c) The net force is zero.

5.

3.14 × 10³ J

7.

(a) −700 J

(b) 0

(c) 700 J

(d) 38.6 N

(e) 0

9.

1/250

11.

1.1 × 10¹⁰ J

13.

2.8 × 10³ N

15.

102 N

16.

(a) 1.96 × 10¹⁶ J

(b) The ratio of gravitational potential energy in the lake to the
energy stored in the bomb is 0.52. That is, the energy stored in the
lake is approximately half that in a 9-megaton fusion bomb.

18.

(a) 1.8 J

(b) 8.6 J

20.

$$v_{f} = \sqrt{2\text{gh} + {v_{0}}^{2}}
= \sqrt{2\left( \text{9.80 m}\text{/s}^{2} \right)( – 0\text{.180 m}) +
(2\text{.00 m/s})^{2}} = 0\text{.687 m/s}$$

22.

7.81 × 10⁵ N/m

24.

9.46 m/s

26.

4 × 10⁴ molecules

27.

Equating ΔPE_g
and ΔKE, we obtain [latex]v = \sqrt{2\text{gh} + {v_{0}}^{2}} = \sqrt{2\left( \text{9.80 m}\text{/s}^{2} \right)\left( \text{20.0 m}) + (\text{15.0 m/s} \right)^{2}} = \text{24.8 m/s}[/latex]

29.

(a) 25 × 10⁶ years

(b) This is much, much longer than human time scales.

30.

2 × 10⁻¹⁰

32.

(a) 40

(b) 8 million

34.

$149

36.

(a) 208 W

(b) 141 s

38.

(a) 3.20 s

(b) 4.04 s

40.

(a) 9.46 × 10⁷ J

(b) 2.54 y

42.

Identify knowns: m = 950
kg, slope
angle θ = 2.00º, v = 3.00 m/s,
f = 600 N

Identify unknowns: power P
of the car, force F that car
applies to road

Solve for unknown:

$$P = \frac{W}{t} = \frac{\text{Fd}}{t} =
F\left( \frac{d}{t} \right) = \text{Fv},$$

where F is parallel to the
incline and must oppose the resistive forces and the force of
gravity:

F = f + w = 600 N + mg sin θ

Insert this into the expression for power and solve:

[latex]\begin{align*}P&=\left(f+mg\sin\theta\right)v\\[1.5ex]&=\left[600\text{N}+\left(950\text{kg}\right)\left(9.80\text{m/s}^2\right)\sin2\degree\right]\left(30.0\text{m/s}\right)\\[1.5ex]&=2.77\times 10^4\text{W}\end{align*}[/latex]

About 28 kW (or about 37 hp) is reasonable for a car to climb a
gentle incline.

44.

(a) 9.5 min

(b) 69 flights of stairs

46.

641 W, 0.860 hp

48.

31 g

50.

14.3%

52.

(a) 3.21 × 10⁴ N

(b) 2.35 × 10³ N

(c) Ratio of net force to weight of person is 41.0 in part (a); 3.00
in part (b)

54.

(a) 108 kJ

(b) 599 W

56.

(a) 144 J

(b) 288 W

58.

(a) 2.50 × 10¹² J

(b) 2.52%

(c) 1.4 × 10⁴ kg (14
metric tons)

60.

(a) 294 N

(b) 118 J

(c) 49.0 W

62.

(a) 0.500 m/s²

(b) 62.5 N

(c) Assuming the acceleration of the swimmer decreases linearly with
time over the 5.00 s interval, the frictional force must therefore be
increasing linearly with time, since f = F − ma. If the
acceleration decreases linearly with time, the velocity will contain a
term dependent on time squared (t²). Therefore, the water
resistance will not depend linearly on the velocity.

64.

(a) 16.1 × 10³ N

(b) 3.22 × 10⁵ J

(c) 5.66 m/s

(d) 4.00 kJ

66.

(a) 4.65 × 10³ kcal

(b) 38.8 kcal/min

(c) This power output is higher than the highest value on Table
7.5, which is about 35 kcal/min (corresponding to 2415 watts) for
sprinting.

(d) It would be impossible to maintain this power output for 2 hours
(imagine sprinting for 2 hours!).

69.

(a) 4.32 m/s

(b) 3.47 × 10³ N

(c) 8.93 kW

70.

(a) Box A has the greatest speed since it has the greatest work done;
work equals change in kinetic energy and thus fastest speed.

(b) i. Yes, this equation is consistent. An increased force or
distance increases the work done, and thus increases the change in
kinesthetic energy. ii. No, this equation does not make sense.

(c)

Image Description

The image is a graph on a light beige background illustrating the equation w = kx². The graph features two axes: the horizontal axis labeled as x and the vertical axis labeled as w. The curve on the graph starts at the origin (0,0) and curves upward, showing an exponential growth pattern consistent with a quadratic function. To the right of the curve, the equation w = kx² is displayed, emphasizing the relationship between w and x in the context of the graph.

Chapter 7

Problems & Exercises

1.

a) 46.8 N·m

b) It does not matter at what height you push. The torque depends on
only the magnitude of the force applied and the perpendicular distance
of the force’s application from the hinges. (Children don’t have a
tougher time opening a door because they push lower than adults, they
have a tougher time because they don’t push far enough from the
hinges.)

3.

23.3 N

5.

Given:

[latex]\begin{align*}m_1&=26.0\text{kg,}\;m_2=32.0\text{kg,}\;m_{\text{s}}=12.0\text{kg,}\\[1.5ex]r_1&=1.60\text{m,}\;r_{\text{s}}=0.160\text{m, find (a)}\;r_2\text{, (b)}\;F_{\text{p}}\end{align*}[/latex]

a) Since children are balancing:

[latex]\begin{align*}\text{net}\;\tau_{\textrm{cw}}=-\text{net}\;\tau_{\textrm{ccw}}\\[1.5ex]\Rightarrow w_1r_1+m_{\text{s}}gr_{\text{s}}=w_2r_2\end{align*}[/latex]

So, solving for r₂ gives:

[latex]\begin{align*}r_2&=\frac{w_1r_1+m_{\text{s}}gr_{\text{s}}}{w_2}=\frac{m_1gr_1+m_{\text{s}}gr_{\text{s}}}{m_2g}=\frac{m_1r_1+m_{\text{s}}r_{\text{s}}}{m_2}\\[1.5ex]&=\frac{\left(26.0\text{kg}\right)\left(1.60\text{m}\right)+\left(12.0\text{kg}\right)\left(0.160\text{m}\right)}{32.0\text{kg}}\\[1.5ex]&=1.36\text{m}\end{align*}[/latex]

b) Since the children are not moving:

[latex]\begin{align*}\text{net}\;F=0=F_{\text{p}}-w_1-w_2-w_{\text{s}}\\[1.5ex]\Rightarrow F_{\text{p}}=w_1+w_2+w_{\text{s}}\end{align*}[/latex]

So that

[latex]\begin{align*}F_{/text{p}}&=\left(26.0\text{kg}+32.0\text{kg}+12.0\text{kg}\right)\left(9.80\text{m/s}^2\right)\\[1.5ex]&=686\text{N}\end{align*}[/latex]

6.

F_wall = 1.43 × 10³ N

8.

a) 2.55 × 10³ N, 16.3º to the
left of vertical (i.e., toward the wall)

b) 0.292

10.

F_B = 2.12 × 10⁴ N

12.

a) 0.167, or about one-sixth of the weight is supported by the
opposite shore.

b) F = 2.0 × 10⁴ N, straight
up.

14.

a) 21.6 N

b) 21.6 N

16.

350 N directly upwards

19.

25

50 N

21.

a) MA = 18.5

b) F_i = 29.1
N

c) 510 N downward

23.

1.3 × 10³ N

25.

a) T = 299 N

b) 897 N upward

26.

[latex]\begin{align*}F_{\text{B}}&=470\text{N;}\;r_1=4.00\text{cm;}\;w_{\text{a}}=2.50\text{kg;}\;r_2=16.0\text{cm;}\;w_{\text{b}}=4.00\text{kg;}\;r_3=38.0\text{cm}\\[1.5ex]F_{\text{E}}&=w_{\text{a}}\left(\frac{r_2}{r_1}-1\right)+w_{\text{b}}\left(\frac{r_3}{r_1}-1\right)\\[1.5ex]&=\left(2.50\text{kg}\right)\left(9.80\text{m/s}^2\right)\left(\frac{16.0\text{cm}}{4.0\text{cm}}-1\right)+\left(4.00\text{kg}\right)\left(9.80/text{m/s}^2\right)\left(\frac{38.0\text{cm}}{4.00\text{cm}}-1\right)\\[1.5ex]&=407\text{N}\\[1.5ex]1.1\times 10^3\text{N}\end{align*}[/latex]

28.

[latex]\theta=190\degree\;\text{ccw from positive}\;x\;\text{axis}[/latex]

30.

F_V = 97 N, θ = 59º

32.

(a) 25 N downward

(b) 75 N upward

33.

(a) F_A = 2.21 × 10³ N
upward

(b) F_B = 2.94 × 10³ N
downward

35.

(a)  F_{teeth on
bullet} = 1.2 × 10² N  upward

(b) F_J = 84 N
downward

37.

(a) 147 N downward

(b) 1680 N, 3.4 times her weight

(c) 118 J

(d) 49.0 W

39.

a) [latex]{\overset{-}{x}}_{2} = \text{2.33 m}[/latex]

b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board
on the other side of the pivot. The second child is off the board.

c) The position of the first child must be shortened, i.e. brought
closer to the pivot.

Chapter 8

Problems & Exercises

1.

ω = 0.737 rev/s

3.

(a) −0.26 rad/s²

(b) 27 rev

5.

(a) 80 rad/s²

(b) 1.0 rev

7.

(a) 45.7 s

(b) 116 rev

9.

a) 600 rad/s²

b) 450 rad/s^s

c) 21.0 m/s^s

10.

(a) 0.338 s

(b) 0.0403 rev

(c) 0.313 s

12.

0.50 kg ⋅ m²

14.

(a) 50.4 N ⋅ m

(b) 17.1 rad/s²

(c) 17.0 rad/s²

16.

3.96 × 10¹⁸ s

or 1.26 × 10¹¹ y

18.

[latex]\begin{align*}I_{end}&=I_{center}+m\left(\frac{1}{2}\right)^2\\[1.5ex]\text{Thus,}\;I_{center}&=I_{end}-\frac{1}{4}ml^2=\frac{1}{3}ml^2-\frac{1}{4}ml^2=\frac{1}{12}ml^2\end{align*}[/latex]

19.

(a) 2.0 ms

(b) The time interval is too short.

(c) The moment of inertia is much too small, by one to two orders of
magnitude. A torque of 500 N ⋅ m is
reasonable.

20.

(a) 17,500 rpm

(b) This angular velocity is very high for a disk of this size and
mass. The radial acceleration at the edge of the disk is > 50,000
gs.

(c) Flywheel mass and radius should both be much greater, allowing
for a lower spin rate (angular velocity).

21.

(a) 185 J

(b) 0.0785 rev

(c) W = 9.81 N

23.

(a) 2.57 × 10²⁹ J

(b) KE_rot = 2.65 × 10³³ J

25.

KE_rot = 434 J

27.

(a) 128 rad/s

(b) 19.9 m

29.

(a) 10.4 rad/s²

(b) net W = 6.11 J

34.

(a) 1.49 kJ

(b) 2.52 × 10⁴ N

36.

(a) 2.66 × 10⁴⁰ kg ⋅ m²/s

(b) 7.07 × 10³³ kg ⋅ m²/s

The angular momentum of the Earth in its orbit around the Sun is
3.77 × 10⁶ times larger than
the angular momentum of the Earth around its axis.

38.

22.5 kg ⋅ m²/s

40.

25.3 rpm

Chapter 9

Problems & Exercises

1.

1.610 cm³

3.

(a) 2.58 g

(b) The volume of your body increases by the volume of air you
inhale. The average density of your body decreases when you take a deep
breath, because the density of air is substantially smaller than the
average density of the body before you took the deep breath.

4.

2.70 g/cm³

6.

(a) 0.163 m

(b) Equivalent to 19.4 gallons, which is reasonable

8.

7.9 × 10² kg/m³

9.

15.8 g/cm³

10.

(a) 10¹⁸ kg/m³

(b) 2 × 10⁴ m

11.

3.59 × 10⁶ Pa; or 521 lb/in²

13.

2.36 × 10³ N

14.

0.760 m

16.

[latex]\begin{align*}\left(hpg\right)_{\text{units}}&=\left(\text{m}\right)\left(\text{kg/m}^3\right)\left(\text{m/s}^2\right)=\left(\text{kg}\cdot\text{m}^2\right)/\left(\text{m}^3\cdot\text{2}^2\right)\\[1.5ex]&=\left(\text{kg}\cdot\text{m/s}^2\right)\left(1/\text{m}^2\right)\\[1.5ex]&=\text{N/m}^2\end{align*}[/latex]

18.

(a) 20.5 mm Hg

(b) The range of pressures in the eye is 12–24 mm Hg, so the result
in part (a) is within that range

20.

1.09 × 10³ N/m²

22.

24.0 N

24.

2.55 × 10⁷ Pa; or 251
atm

26.

5.76 × 10³ N extra
force

28.

(a) [latex]V = d_{\text{i}}A_{\text{i}} = d_{\text{o}}A_{\text{o}} \Rightarrow d_{\text{o}} = d_{\text{i}}\left( \frac{A_{\text{i}}}{A_{\text{o}}} \right)\text{.}[/latex]

Now, using equation:

$$\frac{F_{1}}{A_{1}} =
\frac{F_{2}}{A_{2}} \Rightarrow F_{\text{o}} = F_{\text{i}}\left(
\frac{A_{\text{o}}}{A_{\text{i}}} \right)\text{.}$$

Finally,

$$W_{\text{o}} = F_{\text{o}}d_{\text{o}}
= \left( \frac{F_{\text{i}}A_{\text{o}}}{A_{\text{i}}} \right)\left(
\frac{d_{\text{i}}A_{\text{i}}}{A_{\text{o}}} \right) =
F_{\text{i}}d_{\text{i}} = W_{\text{i}}.$$

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With
friction, we know there are losses, so that W_out = W_in − W_f;
therefore, the work output is less than the work input. In other words,
with friction, you need to push harder on the input piston than was
calculated for the nonfriction case.

29.

Balloon:

[latex]\begin{align*}P_{\text{g}}&=5.00\text{cm H}_2\text{O},\\[1.5ex]P_{\text{abs}}&=1.035\times 10^3\text{cm H}_2\text{O}.\end{align*}[/latex]

Jar:

[latex]\begin{align*}P_{\text{g}}&=-50.0\text{mm Hg},\\[1.5ex]P_{\text{abs}}&=710\text{mm Hg}.\end{align*}[/latex]

31.

4.08 m

33.

[latex]\begin{align*}\Delta P=38.7\text{mm Hg,}\\[1.5ex]\text{Leg blood pressure}=\frac{159}{119}\end{align*}[/latex]

35.

22.4 cm²

36.

91.7%

38.

815 kg/m³

40.

(a) 41.4 g

(b) 41.4 cm³

(c) 1.09 g/cm³

42.

(a) 39.5 g

(b) 50 cm³

(c) 0.79 g/cm³

It is ethyl alcohol.

44.

8.21 N

46.

(a) 960 kg/m³

(b) 6.34%

She indeed floats more in seawater.

48.

(a) 0.24

(b) 0.68

(c) Yes, the cork will float because ρ_obj < ρ_{ethyl
alcohol}(0.678 g/cm³ < 0.79 g/cm³)

50.

The difference is 0.006%.

52.

F_net = F₂ − F₁ = P₂A − P₁A = (P₂ − P₁)A

= (h₂ρ_flg − h₁ρ_flg)A

= (h₂ − h₁)ρ_flgA

where ρ_fl =
density of fluid. Therefore,

F_net = (h₂ − h₁)Aρ_flg = V_flρ_flg = m_flg = w_fl

where is w_fl the
weight of the fluid displaced.

54.

592 N/m²

56.

2.23 × 10⁻² mm Hg

58.

(a) 1.65 × 10⁻³ m

(b) 3.71 × 10⁻⁴ m

60.

6.32 × 10⁻² N/m

Based on the values in table, the fluid is probably glycerin.

62.

[latex]\begin{align*}P_{\text{w}}&=14.6\;\text{N/m}^2,\\[1.5ex]P_{\text{a}}&=4.46\;\text{N/m}^2,\\[1.5ex]P_{\text{sw}}&=7.40\;\text{N/m}^2.\end{align*}[/latex]

Alcohol forms the most stable bubble, since the absolute pressure
inside is closest to atmospheric pressure.

64.

5.1º

This is near the value of θ = 0º for most organic
liquids.

66.

−2.78

The ratio is negative because water is raised whereas mercury is
lowered.

68.

479 N

70.

1.96 N

71.

−63.0 cm H₂O

73.

(a) 3.81 × 10³ N/m²

(b) 28.7 mm Hg, which is sufficient
to trigger micturition reflex

75.

(a) 13.6 m water

(b) 76.5 cm water

77.

(a) 3.98 × 10⁶ Pa

(b) 2.1 × 10⁻³ cm

79.

(a) 2.97 cm

(b) 3.39 × 10⁻⁶ J

(c) Work is done by the surface tension force through an effective
distance h/2 to raise the
column of water.

81.

(a) 2.01 × 10⁴ N

(b) 1.17 × 10⁻³ m

(c) 2.56 × 10¹⁰ N/m²

83.

(a) 1.38 × 10⁴ N

(b) 2.81 × 10⁷ N/m²

(c) 283 N

85.

(a) 867 N

(b) This is too much force to exert with a hand pump.

(c) The assumed radius of the pump is too large; it would be nearly
two inches in diameter—too large for a pump or even a master cylinder.
The pressure is reasonable for bicycle tires.

Chapter 10

Problems & Exercises

1.

2.78 cm³/s

3.

27 cm/s

5.

(a) 0.75 m/s

(b) 0.13 m/s

7.

(a) 40.0 cm²

(b) 5.09 × 10⁷

9.

(a) 22 h

(b) 0.016 s

11.

(a) 12.6 m/s

(b) 0.0800 m³/s

(c) No, independent of density.

13.

(a) 0.402 L/s

(b) 0.584 cm

15.

(a) 127 cm³/s

(b) 0.890 cm

17.

[latex]\begin{align*}P&=\frac{\text{Force}}{\text{Area}},\\[1.5ex]\left(P\right)_{\text{units}}&=\text{N/m}^2=\text{N}\cdot\text{m/m}^3=\text{J/m}^3\\[1.5ex]&=\text{energy/volume}\end{align*}[/latex]

19.

184 mm Hg

21.

2.54 × 10⁵ N

23.

(a) 1.58 × 10⁶ N/m²

(b) 163 m

25.

(a) 9.56 × 10⁸ W

(b) 1.41

27.

1.26 W

29.

(a) 3.02 × 10⁻³ N

(b) 1.03 × 10⁻³

31.

1.60 cm³/min

33.

8.7 × 10⁻¹¹ m³/s

35.

0.316

37.

(a) 1.52

(b) Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure.

39.

225 mPa ⋅ s

41.

0.138 Pa ⋅ s,

or

Olive oil.

43.

(a) 1.62 × 10⁴ N/m²

(b) 0.111 cm³/s

(c)10.6 cm

45.

1.59

47.

2.95 × 10⁶ N/m²(gauge
pressure)

Chapter 11

Problems & Exercises

1.

102ºF

3.

20.0ºC and 25.6ºC

5.

9890ºF

7.

(a) 22.2ºC

(b) [latex]\begin{align*}\Delta T\left(\degree\text{F}\right)&=T_2\left(\degree\text{F}\right)-T_1\left(\degree\text{F}\right)\\[1.5ex]&=\frac{9}{2}T_2\left(\degree\text{C}\right)+32.0\degree-\left(\frac{9}{5}T_1\left(\degree\text{C}\right)+32.0\degree\right)\\[1.5ex]&=\frac{9}{5}\left(T_2\left(\degree\text{C}\right)-T_1\left(\degree\text{C}\right)\right)=\frac{9}{5}\Delta T\left(\degree\text{C}\right)\end{align*}[/latex]

9.

169.98 m

11.

5.4 × 10⁻⁶ m

13.

Because the area gets smaller, the price of the land DECREASES by
~$17, 000.

15.

[latex]\begin{align*}V&=V_0+\Delta V=V_0\left(1+\beta\Delta T\right)\\[1.5ex]&=\left(60.00\text{L}\right)\left[1+\left(950\times 10^{-6}/\degree\text{C}\right)\left(35.0\degree\text{C}\right)\left(35.0\degree\text{C}-15.0\degree\text{C}\right)\right]\\[1.5ex]&=61.1\text{L}\end{align*}[/latex]

17.

(a) 9.35 mL

(b) 7.56 mL

19.

0.832 mm

21.

We know how the length changes with temperature: ΔL = αL₀ΔT.
Also we know that the volume of a cube is related to its length by V = L³, so the
final volume is then V = V₀ + ΔV = (L₀ + ΔL)³.
Substituting for ΔL gives

V = (L₀ + αL₀ΔT)³ = L₀³(1 + αΔT)³.

Now, because αΔT
is small, we can use the binomial expansion:

V ≈ L₀³(1 + 3αΔT) = L₀³ + 3αL₀³ΔT.

So writing the length terms in terms of volumes gives V = V₀ + ΔV ≈ V₀ + 3αV₀ΔT,
and so

ΔV = βV₀ΔT ≈ 3αV₀ΔT, or β ≈ 3α.

Chapter 12

Problems & Exercises

1.

5.02 × 10⁸ J

3.

3.07 × 10³ J

5.

0.171ºC

7.

10.8

9.

617 W

11.

35.9 kcal

13.

(a) 591 kcal

(b) 4.94 × 10³ s

15.

13.5 W

17.

(a) 148 kcal

(b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s

19.

33.0 g

20.

(a) 9.67 L

(b) Crude oil is less dense than water, so it floats on top of the
water, thereby exposing it to the oxygen in the air, which it uses to
burn. Also, if the water is under the oil, it is less efficient in
absorbing the heat generated by the oil.

22.

a) 319 kcal

b) 2.00ºC

24.

20.6ºC

26.

4.38 kg

28.

(a) 1.57 × 10⁴ kcal

(b) 18.3 kW ⋅ h

(c) 1.29 × 10⁴ kcal

30.

(a) 1.01 × 10³ W

(b) One

32.

84.0 W

34.

2.59 kg

36.

(a) 39.7 W

(b) 820 kcal

38.

35 to 1, window to wall

40.

1.05 × 10³ K

42.

(a) 83 W

(b) 24 times that of a double pane window.

44.

20.0 W, 17.2% of 2400 kcal per day

45.

10 m/s

47.

85.7ºC

49.

1.48 kg

51.

2 × 10⁴ MW

53.

(a) 97.2 J

(b) 29.2 W

(c) 9.49 W

(d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W.
While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person.

55.

−21.7 kW Note that the negative
answer implies heat loss to the surroundings.

57.

−266 kW

59.

−36.0 W

61.

(a) 1.31%

(b) 20.5%

63.

(a) −15.0 kW

(b) 4.2 cm

65.

(a) 48.5ºC

(b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than 48.5ºC, and the rate of radiant heat transferred to the rider would be less than 20.0 W.

67.

(a) 3 × 10¹⁷ J

(b) 1 × 10¹³ kg

(c) When a large meteor hits the ocean, it causes great tidal waves,
dissipating large amount of its energy in the form of kinetic energy of
the water.

69.

(a) 3.44 × 10⁵ m³/s

(b) This is equivalent to 12 million cubic feet of air per second.
That is tremendous. This is too large to be dissipated by heating the air by only 5ºC. Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because evaporation removes larger quantities of heat than was considered in part (a).

71.

20.9 min

73.

(a) 3.96×10^-2 g

(b) 96.2 J

(c) 16.0 W

75.

(a) 1.102

(b) 2.79 × 10⁴ J

(c) 12.6 J. This will not cause a significant cooling of the air
because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0ºC to 50.0ºC.

76.

(a) 36ºC

(b) Any temperature increase greater than about 3ºC would be unreasonably large. In this case the final temperature of the person would rise to 73ºC (163ºF).

(c) The assumption of 95% heat retention is unreasonable.

78.

(a) 1.46 kW

(b) Very high power loss through a window. An electric heater of this power can keep an entire room warm.

(c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler.

Chapter 13

Problems & Exercises

1.

(a) 1.23 × 10³ N/m

(b) 6.88 kg

(c) 4.00 mm

3.

(a) 889 N/m

(b) 133 N

5.

(a) 6.53 × 10³ N/m

(b) Yes

7.

16.7 ms

8.

0.400 s/beats

9.

400 Hz

10.

12,500 Hz

11.

1.50 kHz

12.

(a) 93.8 m/s

(b) 11.3 × 10³ rev/min

13.

2.37 N/m

15.

0.389 kg

18.

94.7 kg

21.

1.94 s

22.

6.21 cm

24.

2.01 s

26.

2.23 Hz

28.

(a) 2.99541 s

(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period
changes by (0.01)² = 0.01% so it is necessary to have at least 4 digits after the decimal to see the changes.

30.

(a) Period increases by a factor of 1.41 ($\sqrt{2}$)

(b) Period decreases to 97.5% of old period

32.

Slow by a factor of 2.45

34.

length must increase by 0.0116%.

35.

(a) 1.99 Hz

(b) 50.2 cm

(c) 1.40 Hz, 71.0 cm

36.

(a) 3.95 × 10⁶ N/m

(b) 7.90 × 10⁶ J

37.

a). 0.266 m/s

b). 3.00 J

39.

$$\pm \frac{\sqrt{3}}{2}$$

42.

384 J

44.

(a). 0.123 m

(b). −0.600 J

(c). 0.300 J. The rest of the energy may go into heat caused by
friction and other damping forces.

46.

(a) 5.00 × 10⁵ J

(b) 1.20×10³ s

47.

t = 9.26 d

49.

f = 40.0 Hz

51.

v_w = 16.0
m/s

53.

λ = 700 m

55.

d = 34.0 cm

57.

f = 4 Hz

59.

462 Hz,

4 Hz

61.

(a) 3.33 m/s

(b) 1.25 Hz

63.

0.225 W

65.

7.07

67.

16.0 d

68.

2.50 kW

70.

3.38 × 10^–5 W/m²

Chapter 14

Problems & Exercises

1.

0.288 m

3.

332 m/s

5.

[latex]\begin{align*}v_{\text{w}}&=\left(331\text{m/s}\right)\sqrt{\frac{T}{273\text{K}}}=\left(331\text{m/s}\right)\sqrt{\frac{293\text{K}}{273\text{K}}}\\[1.5ex]&=343\text{m/s}\end{align*}[/latex]

7.

0.223

9.

(a) 7.70 m

(b) This means that sonar is good for spotting and locating large
objects, but it isn’t able to resolve smaller objects, or detect the
detailed shapes of objects. Objects like ships or large pieces of
airplanes can be found by sonar, while smaller pieces must be found by
other means.

11.

(a) 18.0 ms, 17.1 ms

(b) 5.00%

(c) This uncertainty could definitely cause difficulties for the bat,
if it didn’t continue to use sound as it closed in on its prey. A 5%
uncertainty could be the difference between catching the prey around the
neck or around the chest, which means that it could miss grabbing its
prey.

12.

3.16 × 10⁻⁴ W/m²

14.

3.04 × 10⁻⁴ W/m²

16.

106 dB

18.

(a) 93 dB

(b) 83 dB

20.

(a) 50.1

(b) 5.01 × 10⁻³ or $\frac{1}{200}$

22.

70.0 dB

24.

100

26.

1.45 × 10⁻³ J

28.

28.2 dB

30.

(a) 878 Hz

(b) 735 Hz

32.

3.79 × 10³ Hz

34.

(a) 12.9 m/s

(b) 193 Hz

36.

First eagle hears 4.23 × 10³ Hz

Second eagle hears 3.56 × 10³ Hz

38.

0.7 Hz

40.

0.3 Hz, 0.2 Hz, 0.5 Hz

42.

(a) 256 Hz

(b) 512 Hz

44.

180 Hz, 270 Hz, 360 Hz

46.

1.56 m

48.

(a) 0.334 m

(b) 259 Hz

50.

3.39 to 4.90 kHz

52.

(a) 367 Hz

(b) 1.07 kHz

54.

(a) f_n = n(47.6 Hz), n = 1, 3, 5, …, 419

(b) f_n = n(95.3 Hz), n = 1, 2, 3, …, 210

55.

1 × 10⁶ km

57.

498.5 or 501.5 Hz

59.

82 dB

61.

approximately 48, 9, 0, –7, and 20 dB, respectively

63.

(a) 23 dB

(b) 70 dB

65.

Five factors of 10

67.

(a) 2 × 10⁻¹⁰ W/m²

(b) 2 × 10⁻¹³ W/m²

69.

2.5

71.

1.26

72.

170 dB

74.

103 dB

76.

(a) 1.00

(b) 0.823

(c) Gel is used to facilitate the transmission of the ultrasound
between the transducer and the patient’s body.

78.

(a) 77.0 μm

(b) Effective penetration depth = 3.85 cm, which is enough to examine the eye.

(c) 16.6 μm

80.

(a) 5.78 × 10⁻⁴ m

(b) 2.67 × 10⁶ Hz

82.

(a) [latex]\left. \ v_{w} = 1540\ m/s = f\lambda \Rightarrow \lambda = \frac{1540\ m/s}{100 \times 10^{3}\ \text{Hz}} = 0.0154\ m\ < \ 3.50\ m. \right.[/latex] Because the wavelength is much shorter than the distance in question, the wavelength is not the limiting factor. (b) 4.55 ms 84. 974 Hz (Note: extra digits were retained in order to show the difference.)

Chapter 15

Problems & Exercises

1.

(a) 1.25 × 10¹⁰

(b) 3.13 × 10¹²

3.

-600 C

5.

1.03 × 10¹²

7.

9.09 × 10⁻¹³

9.

1.48 × 10⁸ C

11.

(a) 0.263 N

(b) If the charges are distributed over some area, there will be a
concentration of charge along the side closest to the oppositely charged
object. This effect will increase the net force.

13.

The separation decreased by a factor of 5.

17.

[latex]\begin{align*}F&=k\frac{|q_1q_2|}{r^2}=ma\Rightarrow a=\frac{kq^2}{mr^2}\\[1.5ex]&=\frac{\left(9.00\times 10^9\;\text{N}\cdot\text{m}^2/\text{C}^2\right)\left(1.60\times 10^{-19}\text{m}\right)^2}{\left(1.67\times 10^{-27}\text{kg}\right)\left(2.00\times 10^{-9}\text{m}\right)^2}\\[1.5ex]&=3.45\times 10^{16}\text{m/s}^2\end{align*}[/latex]

18.

(a) 3.2

(b) If the distance increases by 3.2, then the force will decrease by
a factor of 10 ; if the distance decreases by 3.2, then the force will
increase by a factor of 10. Either way, the force changes by a factor of
10.

20.

(a) 1.04 × 10⁻⁹ C

(b) This charge is approximately 1 nC, which is consistent with the
magnitude of charge typical for static electricity

23.

1.02 × 10⁻¹¹

25.

1. 0.859 m beyond negative charge on line connecting two charges
2. 0.109 m from lesser charge on line connecting two charges

28.

8.75 × 10⁻⁴ N

30.

(a) 6.94 × 10⁻⁸ C

(b) 6.25 N/C

32.

(a) 300 N/C (east)

(b) 4.80 × 10⁻¹⁷ N (east)

42.

(a) 2.12 × 10⁵ N/C

(b) one charge of +q

44.

(a) 0.252 N to the left

(b) x = 6.07 cm

46.

(a)The electric field at the center of the square will be straight
up, since q_a and q_b are positive and q_c and q_d are negative and all have the same magnitude.

(b) 2.04 × 10⁷ N/C (upward)

48.

0.102 N, in the −y direction

50.

(a) [latex]\overset{\rightarrow}{E} = 4.36 \times \text{10}^{3}\ \text{N/C},\ 35.0º[/latex], below the horizontal.

(b) No

52.

(a) 5.58 × 10⁻¹¹ N/C

(b)the coulomb force is extraordinarily stronger than gravity

54.

(a) −6.76 × 10⁵ C

(b) 2.63 × 10¹³ m/s² (upward)

(c) 2.45 × 10⁻¹⁸ kg

56.

The charge q₂ is
9 times greater than q₁.

69.

(a) The forces are balanced in the x-direction, so the net
force is vertical. It is composed of the sum of the vertical components of the Coulomb force and the gravitational force.

[latex]\begin{align*}\text{F}_y&=-2\left(\frac{kq_1q_2}{r^2}+G\frac{m_1m_2}{r^2}\right)\cos45\degree\\[1.5ex]&=-2\left(\frac{8.99\times 10^9\left(1.00\times 10^{-9}\right)\left(2.00\times 10^{-9}\right)}{8}+6.67\times 10^{-11}\frac{\left(10.0\right)\left(10.0\right)}{8}\right)\cos45\degree\\[1.5ex]&=-4.36\times 10^{-9}\text{N}\end{align*}[/latex]

(b) No, it is in a metastable position. Since it cannot move
horizontally, it cannot traverse any part of the track.

(c) [latex]\begin{align*}\text{F}_x&=-\left(\frac{kq_1q_2}{4}+G\frac{m_1m_2}{4}\right)\frac{1}{2}+\left(\frac{kq_1q_2}{12}+G\frac{m_1m_2}{12}\right)\frac{\sqrt{2}}{2}\\[1.5ex]&=-\left(\frac{8.99\times 10^9\left(-1.00\times 10^{-9}\right)\left(2.00\times 10^{-9}\right)}{4}\frac{1}{2}-6.67\times 10^{-11}\frac{\left(10.0\right)\left(10.0\right)}{4}\frac{1}{2}\right)\\[1.5ex]&\;\;\;\;\;+\left(\frac{8.99\times 10^9\left(-1.00\times 10^{-9}\right)\left(2.00\times 10^{-9}\right)}{12}\frac{\sqrt{3}}{2}-6.67\times 10^{-11}\frac{\left(10.0\right)\left(10.0\right)}{12}\frac{\sqrt{3}}{2}\right)\text{N}\\[1.5ex]&=2.60\times 10^{-9}\text{N}\end{align*}[/latex]

[latex]\begin{align*}\text{F}_y&=-\left(\frac{kq_1q_2}{4}+G\frac{m_1m_2}{4}\right)\frac{\sqrt{3}}{2}-\left(\frac{kq_1q_2}{12}+G\frac{m_1m_2}{12}\right)\frac{1}{2}\\[1.5ex]&=-2.08\times 10^{-9}\text{N}-\left(\frac{8.99\times 10^9\left(1.00\times 10^{-9}\right)\left(2.00\times 10^{-9}\right)}{12}\frac{\sqrt{3}}{2}-6.67\times 10^{-11}\frac{\left(10.0\right)\left(10.0\right)}{4}\frac{\sqrt{3}}{2}\right)\\[1.5ex]&\;\;\;\;-\left(\frac{8.99\times 10^9\left(10.0^{-9}\right)}{12}\frac{1}{2}+6.67\times 10^{-11}\frac{\left(10.0\right)\left(10.0\right)}{12}\frac{1}{2}\right)\text{N}\\[1.5ex]&=-6.36\times 10^{-9}\text{N}\end{align*}[/latex]

(d) Yes.

(e) It will reach [latex](1, - \sqrt{3)}[/latex]
before it will change direction.

(f) The metastable positions where there is no component in one
direction are (0, 0), (4, 0), (2, 2), and (2, −2). They number 4.

Chapter 16

Problems & Exercises

1.

0.278 mA

3.

0.250 A

5.

1.50ms

7.

(a) 1.67kΩ

(b) If a 50 times larger resistance existed, keeping the current
about the same, the power would be increased by a factor of about 50 (based on the equation P = I²R), causing much more energy to be transferred to the skin, which could cause serious burns. The gel used reduces the resistance, and therefore reduces the power transferred to the skin.

9.

(a) 0.120 C

(b) 7.50 × 10¹⁷electrons

11.

96.3 s

13.

(a) 7.81 × 10¹⁴ He⁺⁺ nuclei/s

(b) 4.00 × 10³ s

(c) 7.71 × 10⁸ s

15.

−1.13 × 10⁻⁴m/s

17.

9.42 × 10¹³electrons

18.

0.833 A

20.

7.33 × 10⁻² Ω

22.

(a) 0.300 V

(b) 1.50 V

(c) The voltage supplied to whatever appliance is being used is
reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly.

24.

0.104 Ω

26.

2.81 × 10⁻⁴ m

28.

1.10 × 10⁻³ A

30.

−5ºC to 45ºC

32.

1.03

34.

0.06%

36.

−17ºC

38.

(a) 4.7 Ω (total)

(b) 3.0% decrease

40.

2.00 × 10¹² W

44.

(a) 1.50 W

(b) 7.50 W

46.

$$\frac{V^{2}}{\Omega} =
\frac{V^{2}}{\text{V/A}} = \text{AV} = \left( \frac{C}{s} \right)\left(
\frac{J}{C} \right) = \frac{J}{s} = 1\ \text{W}$$

48.

$$1\ \text{kW} \cdot \text{h=}\left(
\frac{1 \times \text{10}^{3}\ \text{J}}{\text{1 s}} \right)(1\ h)\left(
\frac{\text{3600 s}}{\text{1 h}} \right) = 3\text{.60} \times
\text{10}^{6}\ \text{J}$$

50.

$438/y

52.

$6.25

54.

1.58 h

56.

$3.94 billion/year

58.

25.5 W

60.

(a) 2.00 × 10⁹ J

(b) 769 kg

62.

45.0 s

64.

(a) 343 A

(b) 2.17 × 10³ A

(c) 1.10 × 10³ A

66.

(a) 1.23 × 10³ kg

(b) 2.64 × 10³ kg

69.

(a) 2.08 × 10⁵ A

(b) 4.33 × 10⁴ MW

(c) The transmission lines dissipate more power than they are
supposed to transmit.

(d) A voltage of 480 V is unreasonably low for a transmission
voltage. Long-distance transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses.

73.

480 V

75.

2.50 ms

77.

(a) 4.00 kA

(b) 16.0 MW

(c) 16.0%

79.

2.40 kW

81.

(a) 4.0

(b) 0.50

(c) 4.0

83.

(a) 1.39 ms

(b) 4.17 ms

(c) 8.33 ms

85.

(a) 194 kW

(b) 880 A

87.

(a) 0.400 mA, no effect

(b) 26.7 mA, muscular contraction for duration of the shock (can’t
let go)

89.

1.20 × 10⁵ Ω

91.

(a) 1.00 Ω

(b) 14.4 kW

93.

Temperature increases 860º C. It is
very likely to be damaging.

95.

80 beats/minute

97.

(a) [latex]\begin{align*}\frac{3.25\times 10^{-3}}{1}\text{j/s}&=Power=I^2R=\frac{1}{r^2}\frac{\left(2.50\right)^2\left(1.72\times 10^{-8}\right)\left(1.25\right)}{\pi}\text{jm}^2/\text{s}\\[1.5ex]r&=\sqrt{\frac{\left(2.50\right)^2\left(1.72\times 10^{-8}\right)\left(1.25\right)}{0.00325\pi}}=3.63\;\text{mm}\end{align*}[/latex]

(b) [latex]Power = \frac{(2.50)^{2}\left( 1.74 \times 10^{- 8} \right)(1.25)}{\left( 3.63 \times 10^{- 3} \right)^{2}\pi}\text{j/s}[/latex], so 3.28 j of energy is expended in 1
s. Since α is specified in the
chart to only two significant figures, round to 3.3 j.

(c) Since the energy dissipation is directly proportional to the
resistivity, the increase of 0.01% in energy dissipation is when the
resistivity increases 1%. Or, 1.01 = 1 + 0.0039ΔT. T = 23^∘C.

98.

(a) 2.75 kΩ

(b) 27.5 Ω

100.

(a) 786 Ω

(b) 20.3 Ω

102.

29.6 W

104.

(a) 0.74 A

(b) 0.742 A

106.

(a) 60.8 W

(b) 3.18 kW

108.

(a) [latex]R_{\text{s}}=R_1+R_2\Rightarrow R_{\text{s}}\approx R_1\left(R_1>>R_2\right)[/latex]

(b) [latex]\frac{1}{R_{p}} = \frac{1}{R_{1}}+\frac{1}{R_{2}} = \frac{R_{1} + R_{2}}{R_{1}R_{2}}[/latex],

so that

[latex]R_{\text{p}}=\frac{R_1R_2}{R_1+R_2}\approx\frac{R_1R_2}{R_1}=R_2\left(R_1>>R_2\right)[/latex]

110.

(a) −400 kΩ

(b) Resistance cannot be negative.

(c) Series resistance is said to be less than one of the resistors,
but it must be greater than any of the resistors.

Chapter 17

Problems & Exercises

1.

(a) Left (West)

(b) Into the page

(c) Up (North)

(d) No force

(e) Right (East)

(f) Down (South)

3.

(a) East (right)

(b) Into page

(c) South (down)

5.

(a) Into page

(b) West (left)

(c) Out of page

7.

7.50 × 10⁻⁷ N
perpendicular to both the magnetic field lines and the velocity

9.

(a) 3.01 × 10⁻⁵ T

(b) This is slightly less then the magnetic field strength of 5 × 10⁻⁵ T at the surface of the
Earth, so it is consistent.

11.

(a) 6.67 × 10⁻¹⁰ C
(taking the Earth’s field to be 5.00 × 10⁻⁵ T)

(b) Less than typical static, therefore difficult

12.

4.27 m

14.

(a) 0.261 T

(b) This strength is definitely obtainable with today’s technology.
Magnetic field strengths of 0.500 T are obtainable with permanent
magnets.

16.

4.36 × 10⁻⁴ m

18.

(a) 3.00 kV/m

(b) 30.0 V

20.

0.173 m

22.

7.50 × 10⁻⁴ V

24.

(a) 1.18 × 10 ³ m/s

(b) Once established, the Hall emf pushes charges one direction and
the magnetic force acts in the opposite direction resulting in no net
force on the charges. Therefore, no current flows in the direction of
the Hall emf. This is the same as in a current-carrying
conductor—current does not flow in the direction of the Hall emf.

26.

11.3 mV

28.

1.16 μV

30.

2.00 T

31.

(a) west (left)

(b) into page

(c) north (up)

(d) no force

(e) east (right)

(f) south (down)

33.

(a) into page

(b) west (left)

(c) out of page

35.

(a) 2.50 N

(b) This is about half a pound of force per 100 m of wire, which is
much less than the weight of the wire itself. Therefore, it does not
cause any special concerns.

37.

1.80 T

39.

(a) 30º

(b) 4.80 N

41.

(a) τ decreases by 5.00% if B
decreases by 5.00%

(b) 5.26% increase

43.

10.0 A

45.

[latex]A \cdot m^{2} \cdot T = A \cdot m^{2}\left( \frac{N}{A \cdot m} \right) = N \cdot m[/latex].

47.

3.48 × 10⁻²⁶ N ⋅ m

49.

(a) 0.666 N ⋅ m west

(b) This is not a very significant torque, so practical use would be
limited. Also, the current would need to be alternated to make the loop
rotate (otherwise it would oscillate).

50.

(a) 8.53 N, repulsive

(b) This force is repulsive and therefore there is never a risk that
the two wires will touch and short circuit.

52.

400 A in the opposite direction

54.

(a) 1.67 × 10⁻³ N/m

(b) 3.33 × 10⁻³ N/m

(c) Repulsive

(d) No, these are very small forces

56.

(a) Top wire: 2.65 × 10⁻⁴ N/m s, 10.9º to left of up

(b) Lower left wire: 3.61 × 10⁻⁴ N/m, 13.9º down from right

(c) Lower right wire: 3.46 × 10⁻⁴ N/m, 30.0º down from left

58.

(a) right-into page, left-out of page

(b) right-out of page, left-into page

(c) right-out of page, left-into page

60.

(a) clockwise

(b) clockwise as seen from the left

(c) clockwise as seen from the right

61.

1.01 × 10¹³ T

63.

(a) 4.80 × 10⁻⁴ T

(b) Zero

(c) If the wires are not paired, the field is about 10 times stronger
than Earth’s magnetic field and so could severely disrupt the use of a
compass.

65.

39.8 A

67.

(a) 3.14 × 10⁻⁵ T

(b) 0.314 T

69.

7.55 × 10⁻⁵ T,
23.4º

71.

10.0 A

73.

(a) 9.09 × 10⁻⁷ N upward

(b) 3.03 × 10⁻⁵ m/s²

75.

60.2 cm

77.

(a) 1.02 × 10³ N/m²

(b) Not a significant fraction of an atmosphere

79.

17.0 × 10⁻⁴%/ºC

81.

18.3 MHz

83.

(a) Straight up

(b) 6.00 × 10⁻⁴ N/m

(c) 94.1 μm

(d)2.47 Ω/m, 49.4 V/m

85.

(a) 571 C

(b) Impossible to have such a large separated charge on such a small
object.

(c) The 1.00-N force is much too great to be realistic in the Earth’s
field.

87.

(a) 2.40 × 10⁶ m/s

(b) The speed is too high to be practical ≤ 1% speed of light

(c) The assumption that you could reasonably generate such a voltage
with a single wire in the Earth’s field is unreasonable

89.

(a) 25.0 kA

(b) This current is unreasonably high. It implies a total power
delivery in the line of 50.0×10^9 W, which is much too high for standard
transmission lines.

(c)100 meters is a long distance to obtain the required field
strength. Also coaxial cables are used for transmission lines so that
there is virtually no field for DC power lines, because of cancellation
from opposing currents. The surveyor’s concerns are not a problem for
his magnetic field measurements.

92.

(a) F = qvBsin θ
Since sin θ = 0, F = 0 for the
first electron.
For the second and third electrons, sin θ = 1.
So, F = (1.60 × 10⁻¹⁹C)(5.00 × 10⁷m/s)(2.00 × 10⁻⁷T) = 1.60 × 10⁻¹⁸N.

(b) The first electron moves in a straight line. The other two
electrons take a circular path.

(c) The first electron will never return to the origin. The other two
will return.

[latex]\begin{align*}r&=\frac{\left(9.11\times 10^{-11}\right)\left(5.00\times 10^7\right)}{\left(1.60\times 10^{-19}\right)\left(2.00\times 10^{-7}\right)}\text{m}\\[1.5ex]v&=\frac{r}{t}\\[1.5ex]t&=\frac{\left(9.11\times 10^{-11}\right)}{\left(1.60\times 10^{-19}\right)\left(2.00\times 10^{-7}\right)}\text{s}=2.85\times 10^{15}\text{s}\end{align*}[/latex]