Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Table of contents
Textbook solutions

All Solutions

Page 127: Practice Problems

Exercise 20
Step 1
1 of 4
$newcommand{tx}$[1]${text{#1}}$

#### Known

Taking into account that the position of a vector is not relevant, we can add two or more vectors moving them parallel to the point where we want to add it to another vector. That is, moving the vectors while maintaining the direction they have. For example:

In Figure 1 we have a vector $vec{A}$ and a vector $vec{B}$. If we want to add $vec{A}+vec{B}$, we move vector $vec{B}$ keeping its direction until we put the tail of $vec{B}$ on the head of $vec{A}$. Then the resulting vector is the vector that goes from the tail of $vec{A}$ to the head of $vec{B}$. If we want to calculate the sum $vec{B}+vec{A}$ we do the same as in the previous case, but now we translate vector $vec{A}$, we see that the result will be the same even though the resulting vectors are in different positions. Figure 2.

Step 2
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Exercise scan
Step 3
3 of 4
$star$ Note: This method only works in flat spaces, in curved spaces it is a bit more complicated.

#### Conclusion

We can add two or more vectors moving them parallel to the point where we want to add it to another vector. That is, moving the vectors while maintaining the direction they have.

Result
4 of 4
We can add two or more vectors moving them parallel to the point where we want to add it to another vector. That is, moving the vectors while maintaining the direction they have.
Exercise 21
Step 1
1 of 2
$newcommand{tx}$[1]${text{#1}}$

#### Known

The vector resulting from adding two or more vectors has as the $x$ component, the sum of all the $x$ components of the vectors in the sum, as the $y$ component, the sum of all the $y$ components of the vectors in the sum. If we have more coordinates besides $x$ and $y$, the method is the same.

For example, if we have:

$$
begin{align*}
vec{A}=left(A_x, A_y, A_zright), vec{B}=left(B_x, B_y, B_zright) tx{and} vec{C}=left(C_x, C_y, C_zright)
end{align*}
$$

The sum $vec{D}=vec{A}+vec{B}+vec{C}$, has a components:

$$
begin{align*}
&D_x=A_x+B_x+C_x\
&D_y=A_x+B_y+C_y\
&D_z=A_z+B_z+C_z
end{align*}
$$

With:

$$
begin{align*}
boxed{vec{D}=left(D_x, D_y, D_zright)}
end{align*}
$$

#### Conclusion

The vector resulting from adding two or more vectors has as the $x$ component, the sum of all the $x$ components of the vectors in the sum, as the $y$ component, the sum of all the $y$ components of the vectors in the sum. If we have more coordinates besides $x$ and $y$, the method is the same.

Result
2 of 2
The vector resulting from adding two or more vectors has as the $x$ component, the sum of all the $x$ components of the vectors in the sum, as the $y$ component, the sum of all the $y$ components of the vectors in the sum. If we have more coordinates besides $x$ and $y$, the method is the same.
Exercise 22
Step 1
1 of 4
$newcommand{tx}$[1]${text{#1}}$

#### Known

The location of a vector does not matter.

Vectors $vec{A}$, $vec{D}$ and $vec{E}$ are shown in Figure 1. The sums $vec{K}=vec{A}+vec{D}$ and $vec{L}=vec{A}+vec{E}$ are shown in Figure 2.

From Figure 2 we can see that the length (magnitude) of $vec{K}$ is greater than the length of $vec{L}$, taking into account that $vec{D}$ and $vec{E}$ have the same length.

We can also see this analytically using the law of cosines. Where:

$$
begin{align*}
K^2=A^2+D^2-2AD tx{cos}(theta)
end{align*}
$$

And

$$
begin{align*}
L^2=A^2+E^2-2AE tx{cos}(phi)
end{align*}
$$

With $0<phi<90^circ$ and $90^circ<theta<180^circ$.

Step 2
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Exercise scan
Step 3
3 of 4
Therefore:

$$
begin{align*}
boxed{L<K}
end{align*}
$$

#### Conclusion

The magnitude of $vec{K}=vec{A}+vec{D}$ is greater than the magnitude of $vec{L}=vec{A}+vec{E}$.

Result
4 of 4
The magnitude of $vec{K}=vec{A}+vec{D}$ is greater than the magnitude of $vec{L}=vec{A}+vec{E}$.
Exercise 23
Solution 1
Solution 2
Step 1
1 of 4
$newcommand{tx}$[1]${text{#1}}$

#### Known

The location of a vector does not matter.

Vectors $vec{B}$, $vec{C}$, $vec{F}$, $-vec{C}$ and $-vec{F}$ are shown in Figure 1. The subtractions $vec{P}=vec{B}-vec{C}$ and $vec{Q}=vec{B}-vec{F}$ are shown in Figure 2.

From Figure 2 we can see that the length (magnitude) of $vec{Q}$ is greater than the length of $vec{P}$, taking into account that $vec{C}$ and $vec{F}$ have the same length.

We can also see this analytically using the law of cosines. Where:

$$
begin{align*}
P^2=B^2+C^2-2BC tx{cos}(phi)
end{align*}
$$

And

$$
begin{align*}
Q^2=B^2+F^2-2BF tx{cos}(theta)
end{align*}
$$

With $0<phi<90^circ$ and $90^circ<theta<180^circ$.

Step 2
2 of 4
Exercise scan
Step 3
3 of 4
Therefore:

$$
begin{align*}
boxed{P<Q}
end{align*}
$$

#### Conclusion

The magnitude of $vec{P}=vec{B}-vec{C}$ is less than the magnitude $vec{Q}=vec{B}-vec{F}$.

Result
4 of 4
The magnitude of $vec{P}=vec{B}-vec{C}$ is less than the magnitude $vec{Q}=vec{B}-vec{F}$.
Step 1
1 of 2
The magnitude of the vector $vec{B}-vec{C}$ is less than the vector $vec{B}-vec{F}$.
Result
2 of 2
less.
Exercise 24
Step 1
1 of 1
Exercise scan
Exercise 25
Step 1
1 of 4
$newcommand{tx}$[1]${text{#1}}$

#### Known

A vector length (magnitude) and direction. The position of the vector is not relevant.
#### Calculation

Figure 1 shown vectors $vec{A}$ and $vec{B}$ of the problem.

Step 2
2 of 4
Exercise scan
Step 3
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$newcommand{tx}$[1]${text{#1}}$

From the figure, we can see that the angle that $vec{C}$ makes below the positive $x$-axis is approximately equal to a little less than half of $55^circ$. Also from the figure we can infer that the magnitude of $vec{C}$ is a little less than double the value of the magnitude of $vec{A}$, $27 tx{m}$. That is, $vec{C}$ points approximately in a direction $22^circ$ below the positive $x$-axis and has a magnitude of approximately $48 tx{m}$.

An exact result can be reached either using the component method or geometry (law of cosines and law of sines, etc).

#### Conclusion

$vec{C}$ points approximately in a direction $22^circ$ below the positive $x$-axis and has a magnitude of approximately $48 tx{m}$.

Result
4 of 4
$vec{C}$ points approximately in a direction $22^circ$ below the positive $x$-axis and has a magnitude of approximately $48 text{m}$.
Exercise 26
Step 1
1 of 2
$newcommand{tx}$[1]${text{#1}}$

#### Known

The sum of two vector is equal to a vector whose components is the sum of the respective components of the two vectors. This is:

$$
begin{align}
vec{C}=vec{A}+vec{B}implies C_x=A_x+B_x tx{and} C_y=A_y+B_y
end{align}
$$

#### Calculation

Givens: $A=27 tx{m}$ with $theta=32^circ$ above the positive $x$ axis. $B=35 tx{m}$ with $phi=55^circ$ below the positive $x$ axis.

Calculating the components of $vec{A}$ and $vec{B}$.

$$
begin{align*}
&A_x=A tx{cos}(theta)=(27 tx{m}) tx{cos}(32^circ)=22.9 tx{m}approx 23 tx{m}\
&A_y=A tx{sin}(theta)=(27 tx{m}) tx{sin}(32^circ)=14.3 tx{m}approx 14 tx{m}\
&B_x=B tx{cos}(phi)=(35 tx{m}) tx{cos}(55^circ)=20.1 tx{m}approx 20 tx{m}\
&B_y=-B tx{sin}(phi)=(-35 tx{m}) tx{sin}(55^circ)=-28.7 tx{m}approx -29 tx{m}\
end{align*}
$$

The negative sign of the $y$ component of $vec{B}$ is due to the fact that $vec{B}$ has a direction of $55^circ$ below the $x$-axis, therefore its $y$ component is on the negative axis of the $y$-axis.

From (1) we have:

$$
begin{align*}
&C_x=22.9 tx{m}+20.1 tx{m}=43 tx{m}\
&C_y=14.3 tx{m}-28.7 tx{m}=-14.4 tx{m}approx -14 tx{m}\
&implies vec{C}=left(C_x, C_yright)=(43 tx{m}, -14 tx{m})
end{align*}
$$

The magnitude and direction of $vec{C}$ can be calculated as:

$$
begin{align*}
C=sqrt{C^2_x+C^2_y}=sqrt{(43 tx{m})^2+(-14.4 tx{m})^2}=45.3 tx{m}approx 45 tx{m}
end{align*}
$$

$$
begin{align*}
theta&=tx{tan}^{-1}left(frac{C_y}{C_x}right)=tx{tan}^{-1}left(frac{-14.4 tx{m}}{43 tx{m}}right)\
&=-18.5^circapprox-19^circ tx{or} 19^circ tx{below the} x-tx{axis}.
end{align*}
$$

#### Conclusion

Vector $vec{C}$ has a magnitude of $45 tx{m}$ and point in a direction $19^circ$ below the positive $x$ axis.

Result
2 of 2
Vector $vec{C}$ has a magnitude of $45 text{m}$ and point in a direction $19^circ$ below the positive $x$ axis.
Exercise 27
Step 1
1 of 5
$newcommand{tx}$[1]${text{#1}}$

.
#### Known

The magnitude and direction of a vector $vec{A}=left(A_x, A_yright)$ are determined by:

$$
begin{align*}
A=sqrt{A^2_x+A^2_y} tx{and} theta=tx{tan}^{-1}left(frac{A_y}{A_x}right)
end{align*}
$$

#### Calculation

Givens: $A=12 tx{m}$ and points in the positive direction of the $y$-axis. $B=33 tx{m}$ and points in the negative direction of the $x$-axis.

Since the position of a vector is not relevant, only the magnitude and its direction matter. We can place vectors $vec{A}$ and $vec{B}$ as shown in the figure, this for convenience.

From the figure, we can see that:

$$
begin{align*}
&vec{A}=left(A_x, A_yright)=(0, 12 tx{m})\
&vec{B}=left(B_x, B_yright)=(-33 tx{m}, 0)
end{align*}
$$

Therefore:

a)

$$
begin{align*}
vec{C}=vec{A}+vec{B}
end{align*}
$$

With:

$$
begin{align*}
&C_x=A_x+B_x=0 tx{m}-33 tx{m}=-33 tx{m}\
&C_y=A_y+B_y=12 tx{m}+0 tx{m}=12 tx{m}\
&implies vec{C}=left(C_x, C_yright)=(-33 tx{m}, 12 tx{m})\
&implies C=sqrt{(-33 tx{m})^2+(12 tx{m})^2}=35 tx{m}
end{align*}
$$

And

$$
begin{align*}
theta=tx{tan}^{-1}left(frac{12 tx{m}}{-33 tx{m}}right)=&160^circ tx{above the positive} x-tx{axis or} 20^circ tx{above the negative} xtx{axis}.
end{align*}
$$

Other vector directions can also be specified.

Step 2
2 of 5
$newcommand{tx}$[1]${text{#1}}$
We will use:

$theta=20^circ$ above the negative $x$ axis.

b)

$$
begin{align*}
vec{D}=vec{A}-vec{B}
end{align*}
$$

With:

$$
begin{align*}
&D_x=A_x-B_x=0 tx{m}-(-33 tx{m})=33 tx{m}\
&D_y=A_y-B_y=12 tx{m}-0 tx{m}=12 tx{m}\
&implies vec{D}=left(D_x, D_yright)=(33 tx{m}, 12 tx{m})\
&implies D=sqrt{(33 tx{m})^2+(12 tx{m})^2}=35 tx{m}
end{align*}
$$

And

$$
begin{align*}
theta=tx{tan}^{-1}left(frac{12 tx{m}}{33 tx{m}}right)=&20^circ tx{above the positive} x tx{axis}
end{align*}
$$

c)

$$
begin{align*}
vec{F}=vec{B}-vec{A}
end{align*}
$$

With:

$$
begin{align*}
&F_x=B_x-A_x=-33 tx{m}-0 tx{m}=-33 tx{m}\
&F_y=B_y-A_y=0 tx{m}-12 tx{m}=-12 tx{m}\
&implies vec{F}=left(F_x, F_yright)=(-33 tx{m}, -12 tx{m})\
&implies F=sqrt{(-33 tx{m})^2+(-12 tx{m})^2}=35 tx{m}
end{align*}
$$

And

$$
begin{align*}
theta=tx{tan}^{-1}left(frac{-12 tx{m}}{-33 tx{m}}right)=&20^circ tx{below the negative} x-tx{axis}
end{align*}
$$

Step 3
3 of 5
$newcommand{tx}$[1]${text{#1}}$
Note that:

$$
begin{align*}
tx{tan}^{-1}left(frac{12 tx{m}}{33 tx{m}}right)neq tx{tan}^{-1}left(frac{-12 tx{m}}{-33 tx{m}}right)
end{align*}
$$

The first is in the first quadrant and the second is in the third quadrant.

#### Conclusion

a) $vec{C}=vec{A}+vec{B}$ has magnitude of $35 tx{m}$ and points in a direction $20^circ$ above the negative $x$ axis.

b) $vec{D}=vec{A}-vec{B}$ has magnitude of $35 tx{m}$ and points in a direction $20^circ$ above the positive $x$ axis.

c) $vec{F}=vec{B}-vec{A}$ has magnitude of $35 tx{m}$ and points in a direction $20^circ$ below the negative $x$ axis.

The following figures show each of the cases.

Step 4
4 of 5
Exercise scan
Result
5 of 5
a) $vec{C}=vec{A}+vec{B}$ has magnitude of $35 text{m}$ and points in a direction $20^circ$ above the negative $x$ axis.

b) $vec{D}=vec{A}-vec{B}$ has magnitude of $35 text{m}$ and points in a direction $20^circ$ above the positive $x$ axis.

c) $vec{F}=vec{B}-vec{A}$ has magnitude of $35 text{m}$ and points in a direction $20^circ$ below the negative $x$ axis.

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Chapter 1: Introduction to Physics
Section 1.1: Physics and the Scientific Method
Section 1.2: Physics and Society
Section 1.3: Units and Dimensions
Section 1.4: Basic Math for Physics
Page 38: Assessment
Page 41: Standardized Test Prep
Chapter 2: Introduction to Motion
Section 2.1: Describing Motion
Section 2.2: Speed and Velocity
Section 2.3: Position-Time Graphs
Section 2.4: Equation of Motion
Page 66: Assessment
Page 71: Standardized Test Prep
Page 45: Practice Problems
Page 47: Practice Problems
Page 47: Lesson Check
Page 49: Practice Problems
Page 52: Practice Problems
Page 53: Lesson Check
Page 56: Practice Problems
Page 57: Lesson Check
Page 59: Practice Problems
Page 60: Practice Problems
Page 62: Practice Problems
Page 62: Lesson Check
Chapter 3: Acceleration and Acceleration Motion
Section 3.1: Acceleration
Section 3.2: Motion with Constant Acceleration
Section 3.3: Position-Time Graphs for Constant Acceleration
Section 3.4: Free Fall
Page 105: Assessment
Page 111: Standardized Test Prep
Chapter 4: Motion in Two Dimensions
Section 4.1: Vectors in Physics
Section 4.2: Adding and Subtracting Vectors
Section 4.3: Relative Motion
Section 4.4: Projectile Motion
Page 144: Assessment
Page 149: Standardized Test Prep
Chapter 5: Newton’s Laws of Motion
Section 5.1: Newton’s Laws of Motion
Section 5.2: Applying Newton’s Laws
Section 5.3: Friction
Page 180: Assessment
Page 187: Standardized Test Prep
Chapter 6: Work and Energy
Section 6.1: Work
Section 6.2: Work and Energy
Section 6.3: Conservation of Energy
Section 6.4: Power
Page 220: Assessment
Page 227: Standardized Test Prep
Page 191: Practice Problems
Page 193: Practice Problems
Page 196: Lesson Check
Page 196: Practice Problems
Page 199: Practice Problems
Page 201: Practice Problems
Page 203: Practice Problems
Page 204: Practice Problems
Page 205: Practice Problems
Page 206: Lesson Check
Page 209: Practice Problems
Page 211: Lesson Check
Page 213: Practice Problems
Page 214: Practice Problems
Page 215: Practice Problems
Page 216: Lesson Check
Chapter 7: Linear Momentum and Collisions
Section 7.1: Momentum
Section 7.2: Impulse
Section 7.3: Conservation of Momentum
Section 7.4: Collisions
Page 260: Assessment
Page 265: Standardized Test Prep
Chapter 8: Rotational Motion and Equilibrium
Section 8.1: Describing Angular Motion
Section 8.2: Rolling Motion and the Moment of Inertia
Section 8.3: Torque
Section 8.4: Static Equilibrium
Page 300: Assessment
Page 305: Standardized Test Prep
Page 269: Practice Problems
Page 271: Practice Problems
Page 272: Practice Problems
Page 275: Practice Problems
Page 275: Lesson Check
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Page 280: Lesson Check
Page 284: Practice Problems
Page 286: Practice Problems
Page 287: Practice Problems
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Page 294: Practice Problems
Page 295: Practice Problems
Page 296: Lesson Check
Chapter 9: Gravity and Circular Motion
Section 9.1: Newton’s Law of Universal Gravity
Section 9.2: Applications of Gravity
Section 9.3: Circular Motion
Section 9.4: Planetary Motion and Orbits
Page 336: Assessment
Page 341: Standardized Test Prep
Chapter 10: Temperature and Heat
Section 10.1: Temperature, Energy, and Heat
Section 10.2: Thermal Expansion and Energy Transfer
Section 10.3: Heat Capacity
Section 10.4: Phase Changes and Latent Heat
Page 378: Assessment
Page 383: Standardized Test Prep
Chapter 11: Thermodynamics
Section 11.1: The First Law of Thermodynamics
Section 11.2: Thermal Processes
Section 11.3: The Second and Third Laws of Thermodynamics
Page 410: Assessment
Page 413: Standardized Test Prep
Chapter 12: Gases, Liquids, and Solids
Section 12.1: Gases
Section 12.2: Fluids at Rest
Section 12.3: Fluids in Motion
Section 12.4: Solids
Page 446: Assessment
Page 451: Standardized Test Prep
Chapter 13: Oscillations and Waves
Section 13.1: Oscillations and Periodic Motion
Section 13.2: The Pendulum
Section 13.3: Waves and Wave Properties
Section 13.4: Interacting Waves
Page 486: Assessment
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Chapter 14: Sound
Section 14.1: Sound Waves and Beats
Section 14.2: Standing Sound Waves
Section 14.3: The Doppler Effect
Section 14.4: Human Perception of Sound
Page 523: Assessment
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Page 495: Practice Problems
Page 496: Practice Problems
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Page 501: Lesson Check
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Page 506: Lesson Check
Page 510: Practice Problems
Page 511: Practice Problems
Page 512: Lesson Check
Page 514: Practice Problems
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Page 517: Practice Problems
Page 519: Lesson Check
Chapter 15: The Properties of Lights
Section 15.1: The Nature of Light
Section 15.2: Color and the Electromagnetic Spectrum
Section 15.3: Polarization and Scattering of Light
Page 557: Assessment
Page 563: Standardized Test Prep
Chapter 16: Reflection and Mirrors
Section 16.1: The Reflection of Light
Section 16.2: Plane Mirrors
Section 16.3: Curved Mirrors
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Chapter 17: Refraction and Lenses
Section 17.1: Refraction
Section 17.2: Applications of Refraction
Section 17.3: Lenses
Section 17.4: Applications of Lenses
Page 629: Assessment
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Chapter 18: Interference and Diffraction
Section 18.1: Interference
Section 18.2: Interference in Thin Films
Section 18.3: Diffraction
Section 18.4: Diffraction Gratings
Page 668: Assessment
Page 673: Standardized Test Prep
Chapter 19: Electric Charges and Forces
Section 19.1: Electric Charge
Section 19.2: Electric Force
Section 19.3: Combining Electric Forces
Page 698: Assessment
Page 703: Standardized Test Prep
Chapter 20: Electric Fields and Electric Energy
Section 20.1: The Electric Field
Section 20.2: Electric Potential Energy and Electric Potential
Section 20.3: Capacitance and Energy Storage
Page 738: Assessment
Page 743: Standardized Test Prep
Chapter 21: Electric Current and Electric Circuits
Section 21.1: Electric Current, Resistance, and Semiconductors
Section 21.2: Electric Circuits
Section 21.3: Power and Energy in Electric Circuits
Page 775: Assessment
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Chapter 22: Magnetism and Magnetic Fields
Section 22.1: Magnets and Magnetic Fields
Section 22.2: Magnetism and Electric Currents
Section 22.3: The Magnetic Force
Page 810: Assessment
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Chapter 23: Electromagnetic Induction
Section 23.1: Electricity from Magnetism
Section 23.2: Electric Generators and Motors
Section 23.3: AC Circuits and Transformers
Page 844: Assessment
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Chapter 24: Quantum Physics
Section 24.1: Quantized Energy and Photons
Section 24.2: Wave-Particle Duality
Section 24.3: The Heisenberg Uncertainty Principle
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Page 881: Standardized Test Prep
Chapter 26: Nuclear Physics
Section 26.1: The Nucleus
Section 26.2: Radioactivity
Section 26.3: Applications of Nuclear Physics
Section 26.4: Fundamental Forces and Elementary Particles
Page 944: Assessment
Page 947: Standardized Test Prep