Physics: Principles and Problems
Physics: Principles and Problems
9th Edition
Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz
ISBN: 9780078458132
Table of contents
Textbook solutions

All Solutions

Page 150: Practice Problems

Exercise 1
Step 1
1 of 5
$$
78.4 = dfrac{1}{2}9.8t^2
$$
$Vo$ = (Zero) so
we will use newotns law of motion

$$
s = V_o t+dfrac{1}{2}at^2
$$

Step 2
2 of 5
$$
4.9t^2=78.4
$$
$$
t^2 = 16
$$

then t = 4

Step 3
3 of 5
distance = 5 x 4
5 x 4 = 20 m
since velocity = $dfrac{distance}{time}$
then distance = velocity x time
Step 4
4 of 5
$$
V_f = 0-9.8 times 4
$$
vertical component meant that he wants final velocity that can be calculated from this law

$$
V_f = V_o +at
$$

Result
5 of 5
a) time = 4 sec
b) distance = 20 m
c) vertical component = -39.2 m/sec
Exercise 2
Solution 1
Solution 2
Step 1
1 of 2
It takes t seconds the giraffes to reach the bottom of the belt:

$t = sqrt{dfrac{-2y}{g}} = sqrt{dfrac{-2*(-0.6)}{9.8}} = 0.35 s$

The velocity of the giraffes must be:

$$
v_x = dfrac{x}{t} = dfrac{0.4}{0.35} = 1.1 m/s
$$

Result
2 of 2
$$
1.1 m/s
$$
Step 1
1 of 5
**Given:**
– Height: $h = – 0.6 mathrm{~m}$;
– Distance: $s = 0.4 mathrm{~m}$;

**Required:**
– The velocity of the giraffes at the end of the belt $v$;

Step 2
2 of 5
The conveyor belt is moving at a constant velocity. After reaching the end of the belt, the giraffes are subject to only the acceleration of gravity, and we call that projectile motion. In projectile motion, the horizontal speed remains constant, and the vertical speed is changing due to gravity. The time of the fall is given by the second relation.
$$begin{align*}
v &= frac{s}{t} &&(1) \
t &= sqrt{ frac{-2y}{g} } &&(2)
end{align*}$$
Step 3
3 of 5
Equal amount of time is needed for the giraffes to fall vertically into the box as it is needed for them to travel the horizontal distance to the box. That time can be calculated using the second equation where $y$ is the height the giraffes fall through.
$$begin{align*}
t &= sqrt{ frac{-2h}{g} } \
&= sqrt{ frac{-2 cdot (- 0.6 mathrm{~m})}{9.8 ,frac{text{m}}{text{s}^2}} } \
&= 0.35 mathrm{~s}
end{align*}$$
Step 4
4 of 5
That time is also the time needed for the horizontal distance to be crossed. Since the horizontal velocity is constant we can use the first equation with the time calculated above to obtain the velocity of the giraffes at the end of the belt.
$$begin{align*}
v &= frac{s}{t} \
&= frac{ 0.4 mathrm{~m}}{0.35 mathrm{~s}} \
&= 1.1 ,frac{text{m}}{text{s}} \
&approx 1 ,frac{text{m}}{text{s}}
end{align*}$$
$$boxed{ v = 1 ,frac{text{m}}{text{s}} }$$
Result
5 of 5
$$v = 1 ,frac{text{m}}{text{s}}$$
Exercise 3
Solution 1
Solution 2
Step 1
1 of 2
It takes t seconds to fall:

$t = dfrac{x}{v_x} = dfrac{0.070}{2.0} = 0.035 s$

They will fall a distance of y at the t=0.035s:

$y = (-1/2) g t^2 = (-1/2) (9.8) (0.035)^2 = -6.0e-3 m$

Thus they fall 6.0 mm.

Result
2 of 2
6.0 mm
Step 1
1 of 5
**Given:**
– Speed: $v = 2 ,frac{text{m}}{text{s}}$;
– Distance: $s = 7 mathrm{~cm}$;

**Required:**
– The distance of the sundaes from the end of the counter $s$;

Step 2
2 of 5
The sundaes are sliding at a constant velocity across the counter. After reaching the end of it, the sundaes are subject to only the acceleration of gravity, and we call that projectile motion. In projectile motion, the horizontal speed remains constant, and the vertical speed is changing due to gravity. The time of the fall is given by the second relation.
$$begin{align*}
v &= frac{s}{t} &&(1) \
t &= sqrt{ frac{-2y}{g} } &&(2)
end{align*}$$
Step 3
3 of 5
Equal amount of time is needed for the sundaes to travel the vertical distance as it is needed for them to travel the horizontal distance to the server. That time can be calculated multiplying the first equation by $t$ and dividing it by $v$ on both sides:
$$begin{align*}
t &= frac{s}{v} \
&= frac{ 7 mathrm{~cm}}{ 2 ,frac{text{m}}{text{s}} } \
&= frac{ 0.07 mathrm{~m}}{ 2 ,frac{text{m}}{text{s}} } \
&= 0.035 mathrm{~s}
end{align*}$$
Step 4
4 of 5
The second equation describes the change in time for constant acceleration motion. We are interested in the height at which the server catches the sundaes. We can calculate it if we solve the second equation for $y$ by first squaring it and then multiplying it by $g$ and dividing it by $-2$:
$$begin{align*}
y &= – frac{t^2 g }{2} \
&= – dfrac{(0.035 mathrm{~s})^2 cdot 9.8 ,frac{text{m}}{text{s}^2} }{2} \
&= 0.006 mathrm{~m}
end{align*}$$
$$boxed{ y = 0.006 mathrm{~m} }$$
Result
5 of 5
$$ y = 0.006 mathrm{~m} $$
Exercise 4
Solution 1
Solution 2
Step 1
1 of 5
First, let us examine the given values. We have the initial velocity $v_i$ and the angle $theta$ in which the ball was kicked.

$$
begin{align*}
v_i &= 27.0 ,mathrm{frac{m}{s}}\
theta &= 30.0^{circ}
end{align*}
$$

Step 2
2 of 5
textbf{a.)} We will now calculate for the ball’s hang time.\
Given the equation
begin{align}
v_y = v_i sin{theta}
end{align}
where $v_y$ is the velocity of the ball with respect to the $y$-axis. Now through the $3^{rd}$ kinematic equation, we also know that when the ball lands, its $y$-component is $0$
begin{align}
y = v_{y}t – frac{1}{2}gt^{2} = 0
end{align}
Therefore, we can calculate for the hang time by first, isolating time $t$ from the equation above.
begin{align*}
v_{y}t – frac{1}{2}gt^{2} &= 0\
frac{1}{2}gt cdot cancel{t} &= v_{y}cancel{t}\
gt &= 2v_{y}\
t &= frac{2v_{y}}{g}\
intertext{We, then, substitute $v_i sin{theta}$ to $v_y$}
t &= frac{2(v_i sin{theta})}{g}
intertext{and now, we substitute the values,}
t &= frac{2(27.0,mathrm{frac{m}{s}})sin{(30.0^{circ})}}{9.8 ,mathrm{frac{m}{s^2}}}\
&= boxed{2.76 ,mathrm{s}}
end{align*}
Step 3
3 of 5
$textbf{b.)}$ Next, we solve for the ball’s maximum height. Since we know that the ball’s trajectory is similar to that of a parabola, we can conclude that the maximum height is reached at half of the ball’s hang time which is at $t = 1.38,mathrm{s}$. Still using the same equation

$$
begin{align}
y = v_{y}t – frac{1}{2}gt^{2}
end{align}
$$

We will now solve for the maximum height $y$ by substituting the values.

$$
begin{align*}
y &= v_{y}t – frac{1}{2}gt^{2}\
&= (v_i sin{theta})t – frac{1}{2}gt^{2}\
&= (27.0,mathrm{frac{m}{s}})(1.38,mathrm{s}) sin{(30.0^{circ})} – frac{1}{2}(9.8 ,mathrm{frac{m}{s^2}})(1.38,mathrm{s})^2\
&= boxed{9.30 ,mathrm{m}}
end{align*}
$$

Step 4
4 of 5
$textbf{c.}$ And lastly, we solve for the ball’s range.

Given the equation

$$
begin{align}
v_x = v_i cos{theta}
end{align}
$$

where $v_x$ is the ball’s velocity with respect to the $x$-axis.
We can solve for the range using the formula

$$
begin{align}
x = v_{x}t
end{align}
$$

We will substitute $v_x$ with the $v_i cos{theta}$ and then substitute the values to obtain the value of the ball’s range.

$$
begin{align*}
x &= v_{x}t\
&= v_i cos{theta} cdot t\
&= (27.0 ,mathrm{frac{m}{s}})(2.76 ,mathrm{s}) cos{(30.0^{circ})}\
&= boxed{64.5 ,mathrm{m}}
end{align*}
$$

Result
5 of 5
$textbf{a.)}$ $t = 2.76 ,mathrm{s}$

$textbf{b.)}$ $y = 9.30 ,mathrm{m}$

$textbf{c.)}$ $x = 64.5 ,mathrm{m}$

Step 1
1 of 4
begin{align*}
intertext{Data known:}
&v_i=27hspace{1mm}frac{m}{s}hspace{1mm}text{initial velocity}\
&theta=30^{circ}
intertext{Part a:}
intertext{We are looking for ball’s hang time:}
intertext{We known that horizontal velocity in this kind of motion remains constant, so we can find time using formula:}
&v_x=frac{x}{t}hspace{2cm}rightarrowhspace{2cm}
&t=frac{x}{v_x}\
intertext{We can find horizontal component of velocity if we use initial velocity and angle. And we can find horizontal distance $x$ by looking at a graph, and read the value:}
intertext{As we can see $x$ is between 50 and 60, so:}
&x=55hspace{1mm}m
intertext{Horizontal velocity:}
&v_x=v_{0x}=const\
intertext{As we can see $theta$ is angle between velocity (total velocity, sum of horizontal and vertical) and horizontal line, so horizontal component of velocity is:}
&v_x=v_icos 30\
&v_x=27cdot 0.866\
&v_x=23.382hspace{1mm}frac{m}{s}
intertext{Now we can find time:}
&t=frac{x}{v_x}\
&t=frac{55}{23.382}\
&t=2.35hspace{1mm}s
end{align*}
Step 2
2 of 4
begin{align*}
intertext{Part b:}
intertext{We are looking for the ball’s maximum height}
intertext{When ball is on maximum height, it’s vertical velocity is equal to zero, because gravity force is pushing the ball down.}
intertext{We can find it using formula:}
&y_{max}=v_{yi}t_m-frac{1}{2}gt_m^2
intertext{- is in formula because g gas different direction from y axis, and $v_{yi}$ is initial vertical velocity.}
intertext{IT takes $frac{1}{2}$ of time from part a for the ball to get to the maximum height, because when we look at x axis, maximum height is at half of total x, so time is: }
&t_m=frac{2.35}{2}\
&t_m=1.175hspace{1mm}s
intertext{We can find initial vertical velocity by using total initial velocity and angle: }
&v_{yi}=v_isin 30\
&v_{yi}=27cdot0.5\
&v_{yi}=13.5hspace{1mm}frac{m}{s}\
intertext{Now we can find $y_{max}$}
&y_{max}=13.5cdot1.175-frac{1}{2}9.8cdot1.175^2\
&y_{max}=9hspace{1mm}m
end{align*}
Step 3
3 of 4
begin{align*}
intertext{Part c:}
intertext{We are looking for the ball’s range.}
intertext{Range is value of x, it is how far ball get on horizontal axis, which is x axis, so:}
&R=55hspace{1mm}m
end{align*}
Result
4 of 4
begin{align*}
intertext{Part a:}
&t=2.35hspace{1mm}s
intertext{Part b:}
&y_{max}=9hspace{1mm}m
intertext{Part c:}
&R=55hspace{1mm}m
end{align*}
Exercise 5
Step 1
1 of 5
Let us examine the given values, we know that the initial velocity $v_i$ is the same from the previous problem but the angle $theta$ is changed to $60.0^{circ}$

$$
begin{align*}
v_i &= 27.0 ,mathrm{frac{m}{s}}\
theta &= 60.0^{circ}
end{align*}
$$

Using the same method from the previous problem, we calculate the following: hang time, maximum height, and range or distance.

Step 2
2 of 5
textbf{a.) Hang time}
We know that $v_y = v_{i} sin{theta}$ such that $v_y$ is the velocity with respect to the $y$-axis. Using the equation
begin{align}
y = v_{y}t – frac{1}{2}gt^{2} = 0
end{align}
We isolate time $t$ and substitute the given values to obtain the hang time.
begin{align*}
v_{y}t – frac{1}{2}gt^{2} &= 0\
frac{1}{2}gt^2 &= v_{y}t\
gt &= 2v_{y}\
t &= frac{2v_{y}}{g}\
intertext{We substitute the given values.}
t &= frac{2(27.0,mathrm{frac{m}{s}})}{9.8,mathrm{frac{m}{s^2}}}\
t &= boxed{4.77 ,mathrm{s}}
end{align*}
Step 3
3 of 5
textbf{b.) Maximum height}
Using the equation
begin{align}
y = v_{y}t – frac{1}{2}gt^{2}
end{align}
We can compute for the maximum height $y$, knowing that the ball will reach maximum height at $t_{max height} = frac{1}{2}t_{hang time} = 2.38,mathrm{s}$.
begin{align*}
y &= v_{y}t – frac{1}{2}gt^{2}
intertext{We substitute $v_y$ with $v_{i} sin{theta}$}
y &= (v_{i} sin{theta})t – frac{1}{2}gt^{2}\
&= (27.0,mathrm{frac{m}{s}})(2.38 ,mathrm{s})sin{(60.0^{circ})} – frac{1}{2}(9.8,mathrm{frac{m}{s^2}})(2.38,mathrm{s})^2\
&= boxed{27.9 ,mathrm{m}}
end{align*}
Step 4
4 of 5
textbf{c.) Range}
We can calculate the range $x$ of the ball given that $v_x = v_{i} cos{theta}$, such that $v_x$ is the initial velocity with respect to the $x$-axis. We will use the time $t$ which is the hang time of the ball.
begin{align}
x = v_{x}t
end{align}
begin{align*}
intertext{We substitute $v_x$ with $v_{i} cos{theta}$ and plug in the given values.}
x &= (v_{i} cos{theta})t \
&= (27.0,mathrm{frac{m}{s}})(4.77 ,mathrm{s})cos{(60.0^{circ})}\
&= boxed{64.4 ,mathrm{m}}
end{align*}
Result
5 of 5
$textbf{a.)}$ Hang time $t = 4.77 ,mathrm{s}$

$textbf{b.)}$ Maximum height $y = 27.9 ,mathrm{m}$

$textbf{c.)}$ Range $x = 64.4 ,mathrm{m}$

Exercise 6
Solution 1
Solution 2
Step 1
1 of 5
begin{align*}
intertext{Data known:}
&v_i=7hspace{1mm}frac{m}{s}hspace{1mm}text{initial velocity}\
&theta=53^{circ}hspace{1mm}text{° above the horizontal}\
&y_i=50hspace{1mm}m
end{align*}
Step 2
2 of 5
begin{align*}
intertext{We are looking for velocity vector, when rock hits the ground. }
intertext{Velocity has two components, horizontal component $v_x$ and vertical component $v_y$, total velocity is:}
&v^2=v_x^2+v_y^2\
&v=sqrt{v_x^2+v_y^2}\
intertext{We can solve this problem by finding those components first. We can say that motion like this can be observed as combination of two motions, motion due horizontal axis, and due to vertical. }
intertext{Let’s find horizontal component firs. We known that horizontal component of motion remains the same all the time in motion like this, so:}
&v_x=v_{xi}=const\
intertext{As we can see $theta$ is angle between initial velocity (total velocity, sum of horizontal and vertical) and horizontal line, so horizontal component of velocity is:}
&v_x=v_icos 53\
&v_x=7cdot 0.6\
&v_x=4.2hspace{1mm}frac{m}{s}
end{align*}
Step 3
3 of 5
begin{align*}
intertext{Now we should find vertical component of velocity. We can use formula:} &v_y^2=v_{yi}^2-2g(y-y_i)
intertext{- is in formula because g has opposite direction comparing to y axis.}
intertext{y is equal to zero, because it hits the ground, and it is 0 m.And we can fin initial vertical velocity by using angle and initial total velocity}
&v_{yi}=v_isin 53\
&v_{yi}=7cdot0.8\
&v_{yi}=5.6hspace{1mm}frac{m}{s}\
\
&v_y^2=5.6^2-2cdot9.8(0-50)\
&v_y^2=31.36+980\
&v_y^2=1011.36
intertext{value of vertical component can be positive or negative, we use negative because direction of that component when rock hits the ground is opposite comparing to direction of y axis.}
&v_y^2=sqrt{1011.36}\
&v_y=-31.8hspace{1mm}frac{m}{s}
end{align*}
Step 4
4 of 5
begin{align*}
intertext{Now we can find velocity:}
&v=sqrt{v_x^2+v_y^2}\
&v=sqrt{4.2^2+(-31.8)^2}\
&v=sqrt{17.64+1011.36}\
&v=32.07hspace{1mm}frac{m}{s}
end{align*}
Result
5 of 5
$$
begin{align*}
&v_x=4.2hspace{1mm}frac{m}{s}\
&v_y=-31.8hspace{1mm}frac{m}{s}\
&v=32.07hspace{1mm}frac{m}{s}
end{align*}
$$
Step 1
1 of 8
We have the following values:

$$
begin{align*}
y &= 50.0 ,mathrm{m}\
v_i &= 7.0 ,mathrm{frac{m}{s}}\
theta &= 53.0^{circ}
end{align*}
$$

such that $y$ is the height of the cliff, $v_i$ is the initial velocity, and $theta$ is the angle of the projectile.

Step 2
2 of 8
We know that the velocity $v = sqrt{v_{ix}^{2} + v_{iy}^{2}}$. Given the equations:

$$
begin{align}
v_{ix} &= v_i cos{theta}\
v_{iy} &= v_i sin{theta} + gt
end{align}
$$

such that $v_{ix}$ is the initial horizontal velocity and $v_{iy}$ is the initial vertical velocity.

Step 3
3 of 8
Since we do not know time $t$, we can derive it from the equation:

$$
begin{align}
y = v_{y}t – frac{1}{2}gt^2
end{align}
$$

We know that at $y = -50.0 ,mathrm{m}$, which is the height of the rock as it reaches the bottom of the cliff, $v_{y} = 0$, such that $v_{y}$ is the vertical velocity. therefore

$$
begin{align*}
y &= -frac{1}{2}gt^2\
-2y &= gt^{2}\
-frac{2y}{g} &= t^2\
sqrt{-frac{2y}{g}} &= t
end{align*}
$$

Step 4
4 of 8
Thus we can substitute the $sqrt{frac{2y}{g}}$ to the $t$ in the equation for $v_y$.

$$
begin{align*}
v_y &= v_1 sin{theta} + gt\
&= v_1 sin{theta} + g left(sqrt{-frac{2y}{g}} right)\
&= v_1 sin{theta} + left(sqrt{-2yg} right)
end{align*}
$$

Step 5
5 of 8
We then plug in the equations for $v_x$ and $v_y$ to $v = sqrt{v_{x}^{2} + v_{y}^{2}}$.
begin{align*}
v &= sqrt{left(v_i cos{theta}right)^{2} + left(v_1 sin{theta} + left(sqrt{-2yg} right)right)^{2}}
intertext{Now, to solve for the magnitude $v$, we can plug in the given values}
v &= sqrt{left((7.0 ,mathrm{frac{m}{s}}) cos{(53.0^{circ})}right)^{2} + left((7.0 ,mathrm{frac{m}{s}}) sin{53.0^{circ}} + left(sqrt{-2(-50.0 ,mathrm{m})(9.8 ,mathrm{frac{m}{s^2}}} right)right)^{2}}\
&= boxed{37.1 ,mathrm{frac{m}{s}}}
end{align*}
Step 6
6 of 8
Now we need to solve for its angle. Since

$$
begin{align}
tan{theta} = frac{v_{iy}}{v_{ix}}
end{align}
$$

Step 7
7 of 8
We can isolate $theta$ and plug in the values to obtain the angle of the velocity vector $v$.

$$
begin{align*}
tan{theta} &= frac{v_{iy}}{v_{ix}}\
theta &= tan^{-1}{left(frac{v_{iy}}{v_{ix}}right)}\
&= tan^{-1}{left(frac{v_1 sin{theta} + left(sqrt{2yg} right)}{v_i cos{theta}}right)}\
&= tan^{-1}{left(frac{(7.0 ,mathrm{frac{m}{s}}) sin{53.0^{circ}} + left(sqrt{2(50.0 ,mathrm{m})(9.8 ,mathrm{frac{m}{s^2}}} right)}{(7.0 ,mathrm{frac{m}{s}}) cos{(53.0^{circ})}}right)}\
&= boxed{83.5^{circ}}text{ from the $+x$-axis}
end{align*}
$$

Result
8 of 8
$$
v = 37.1 ,mathrm{frac{m}{s}} text{ at } 83.5^{circ}text{ from the $+x$-axis}
$$
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Chapter 1: A Physics Toolkit
Section 1.1: Mathematics and Physics
Section 1.2: Measurement
Section 1.3: Graphing Data
Page 24: Assessment
Page 29: Standardized Test Practice
Chapter 3: Accelerated Motion
Section 3.1: Acceleration
Section 3.2: Motion with Constant Acceleration
Section 3.3: Free Fall
Page 80: Assessment
Page 85: Standardized Test Practice
Chapter 4: Forces in One Dimension
Section 4.1: Force and Motion
Section 4.2: Using Newton’s Laws
Section 4.3: Interaction Forces
Page 112: Assessment
Page 117: Standardized Test Practice
Chapter 5: Forces in Two Dimensions
Section 5.1: Vectors
Section 5.2: Friction
Section 5.3: Force and Motion in Two Dimensions
Page 140: Assessment
Page 145: Standardized Test Practice
Chapter 6: Motion in Two Dimensions
Section 6.1: Projectile Motion
Section 6.2: Circular Motion
Section 6.3: Relative Velocity
Page 164: Assessment
Page 169: Standardized Test Practice
Chapter 7: Gravitation
Section 7.1: Planetary Motion and Gravitation
Section 7.2: Using the Law of Universal Gravitation
Page 190: Assessment
Page 195: Standardized Test Practice
Chapter 8: Rotational Motion
Section 8.1: Describing Rotational Motion
Section 8.2: Rotational Dynamics
Section 8.3: Equilibrium
Page 222: Assessment
Page 227: Standardized Test Practice
Chapter 9: Momentum and Its Conservation
Chapter 10: Energy, Work, and Simple Machines
Section 10.1: Energy and Work
Section 10.2: Machines
Page 278: Assessment
Page 283: Standardized Test Practice
Chapter 11: Energy and Its Conservation
Section 11.1: The Many Forms of Energy
Section 11.2: Conservation of Energy
Page 306: Assessment
Page 311: Standardized Test Practice
Chapter 13: State of Matter
Section 13.1: Properties of Fluids
Section 13.2: Forces Within Liquids
Section 13.3: Fluids at Rest and in Motion
Section 13.4: Solids
Page 368: Assessment
Page 373: Standardized Test Practice
Chapter 14: Vibrations and Waves
Section 14.1: Periodic Motion
Section 14.2: Wave Properties
Section 14.3: Wave Behavior
Page 396: Assessment
Page 401: Section Review
Chapter 15: Sound
Section 15.1: Properties of Detection of Sound
Section 15.2: The Physics of Music
Page 424: Assessment
Page 429: Standardized Test Practice
Chapter 17: Reflections and Mirrors
Section 17.1: Reflection from Plane Mirrors
Section 17.2: Curved Mirrors
Page 478: Assessment
Page 483: Standardized Test Practice
Chapter 18: Refraction and lenses
Section 18.1: Refraction of Light
Section 18.2: Convex and Concave Lenses
Section 18.3: Applications of Lenses
Page 508: Assessment
Page 513: Standardized Test Practice
Chapter 21: Electric Fields
Section 21.1: Creating and Measuring Electric Fields
Section 21.2: Applications of Electric Fields
Page 584: Assessment
Page 589: Standardized Test Practice
Chapter 22: Current Electricity
Section 22.1: Current and Circuits
Section 22.2: Using Electric Energy
Page 610: Assessment
Page 615: Standardized Test Practice
Chapter 23: Series and Parallel Circuits
Section 23.1: Simple Circuits
Section 23.2: Applications of Circuits
Page 636: Assessment
Page 641: Standardized Test Practice
Chapter 24: Magnetic Fields
Section 24.1: Magnets: Permanent and Temporary
Section 24.2: Forces Caused by Magnetic Fields
Page 664: Assessment
Page 669: Standardized Test Practice
Chapter 25: Electromagnetic Induction
Section 25.1: Electric Current from Changing Magnetic Fields
Section 25.2: Changing Magnetic Fields Induce EMF
Page 690: Assessment
Page 695: Standardized Test Practice
Chapter 30: Nuclear Physics
Section 30.1: The Nucleus
Section 30.2: Nuclear Decay and Reactions
Section 30.3: The Building Blocks of Matter
Page 828: Assessment
Page 831: Standardized Test Practice