Download Physics Practice 10 | FINAL STUDY GUIDE

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Physics Practice 10 | FINAL STUDY GUIDE
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
1) A substance registers a temperature change from 20°C to 40°C. To what incremental temperature change
does this correspond?
c 36°F
a 20°F
b 40°F
d 313°F
Name: ______________________
6) An ideal gas system undergoes an adiabatic process in which it expands and does 20 J of work on its
environment. What is the change in the system’s internal energy?
c 0J
a –20 J
b –5 J
d 20 J
7) An ideal gas system undergoes an adiabatic process in which it expands and does 20 J of work on its
environment. How much energy is transferred to the system as heat?
c 5J
a –20 J
b 0J
d 20 J
2) What temperature has the same numerical value on both the Fahrenheit and the Celsius scales?
c 40.0°
a –40.0°
b 0°
d –72.0°
8) An ideal gas system undergoes an isovolumetric process in which 20 J of energy is added as heat to the
gas. What is the change in the system’s internal energy?
c 5J
a –20 J
b 0J
d 20 J
3) A nail is driven into a board with an initial kinetic energy of 150 J. If the potential energy before and after
the event is the same, what is the change in the internal energy of the board and nail?
c 0J
a 150 J
b 75 J
d !150 J
9) A 2.0 m long stretched rope is fixed at both ends. Which wavelength would not produce standing waves
on this rope?
c 4.0 m
a 2.0 m
b 3.0 m
d 6.0 m
4) A slice of bread contains about 4.19 " 10 5 J of energy. If the specific heat capacity of a person is 4.19 "
10 3 J/kg•°C, by how many degrees Celsius would the temperature of a 70.0 kg person increase if all the
energy in the bread were converted to heat?
c 1.43°C
a 2.25°C
b 1.86°C
d 1.00°C
10) Which of the following decibel levels is nearest to the value that you would expect for a running vacuum
cleaner?
c 70 dB
a 10 dB
b 30 dB
d 120 dB
11) How many beats per second are heard when two vibrating tuning forks having frequencies of 342 Hz and
345 Hz are held side by side?
c 5 Hz
a 687 Hz
b 343.5 Hz
d 3 Hz
12) Four beats per second are heard when two notes are sounded. The frequency of one note is 420 Hz.
Which of the following is a possible frequency of the other note?
c 416 Hz
a 418 Hz
b 105 Hz
d 1680 Hz
13) What is the frequency of infrared light of 1.0 " 10 !4 m wavelength?
a
b
3.0 " 10 !13 Hz
2
3.0 " 10 Hz
c
3.0 " 10 4 Hz
d
3.0 " 10 12 Hz
14) What is the wavelength of an infrared wave with a frequency of 4.2 " 10 14 Hz?
a
b
5) Using the figure above, determine which value equals the latent heat required to change the liquid water
into steam.
3
c 30.6 " 10 3 J
a 8.04 " 10 J
3
3
b 22.6 " 10 J
d 31.1 " 10 J
1
7.1 " 10 5 m
6
1.4 " 10 m
c
7.1 " 10 !7 m
d
1.4 " 10 !6 m
15) If a light ray strikes a flat mirror at an angle of 14° from the normal, the reflected ray will be
c 90° from the mirror’s surface.
a 14° from the mirror’s surface.
b 76° from the normal.
d 14° from the normal.
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Name: ______________________
Name: ______________________
16) What type of mirror is used whenever a magnified image of an object is needed?
c convex mirror
a flat mirror
b concave mirror
d two-way mirror
17) A concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis.
How far from the mirror is the corresponding object?
c 7.5 cm
a 20 cm
b 15 cm
d 5.0 cm
18) A concave mirror forms a real image at 25.0 cm from the mirror surface along the principal axis. If the
corresponding object is at a 10.0 cm distance, what is the mirror’s focal length?
c 12.0 cm
a 1.40 cm
b 7.14 cm
d 17.0 cm
19) An object is 29 cm away from a concave mirror’s surface along the principal axis. If the mirror’s focal
length is 9.50 cm, how far away is the corresponding image?
c 29 cm
a 12 cm
b 14 cm
d 36 cm
20) If a virtual image is formed 10.0 cm along the principal axis from a convex mirror with a focal length of
–15.0 cm, what is the object’s distance from the mirror?
c 6.0 cm
a 30 cm
b 12 cm
d 3.0 cm
24) Carbon tetrachloride (n = 1.46) is poured into a container made of crown glass (n = 1.52). If a light ray in
the glass is incident on the glass-to-liquid boundary and makes an angle of 30.0° with the normal, what is
the angle of the corresponding refracted ray with respect to the normal?
c 31.4°
a 25.6°
b 28.7°
d 64.4°
25) A film projector produces a 1.51 m image of a horse on a screen. If the projector lens is 4.00 m from the
screen and the size of the horse on the film is 1.07 cm, what is the magnitude of the magnification of the
image?
c 0.708
a 141
!3
b 14.1
d 7.08 " 10
26) The distance between two slits in a double-slit interference experiment is 0.0050 mm. What is the angle of
the third-order bright fringe (m = 3) produced with light of 550 nm?
c 12°
a 5.0°
b 9.9°
d 19°
27) Light with a wavelength of 500.0 nm passes through a 3.39 " 10 5 lines/m diffraction grating. What is the
first-order angle of diffraction?
c 36.9°
a 9.73°
b 23.5°
d 53.1°
28) A diffraction grating that contains 650 472 lines/m is illuminated by monochromatic light (# = 632.8 nm)
from a helium-neon laser directed perpendicular to the surface of the grating. At what angles would one
observe the first-order and second-order maxima?
c $ = 29.37°; $ = 55.38°
a $ = 2.358°; $ = 4.719°
1
2
1
2
b $ = 24.30°; $ = 55.38°
d $ = 55.38°; $ = 83.078°
1
2
1
2
29) A helium-neon laser shines monochromatic light (# = 632.8 nm) perpendicular to the surface of a
diffraction grating that contains 146 230 lines/m. Find the angles at which one would observe the
second-order and third-order maxima.
c $ = 10.66°; $ = 18.43°
a $ = 5.307°; $ = 10.66°
1
2
1
2
b $ = 10.66°; $ = 16.11°
d $ = 13.25°; $ = 26.50°
1
2
1
2
21) In the diagram shown above, the image of object B would be
c virtual, reduced, and inverted.
a real, reduced, and upright.
b virtual, enlarged, and upright.
d virtual, reduced, and upright.
22) When a light ray passes from zircon (n = 1.923) into fluorite (n = 1.434) at an angle of 60°, its path is
c parallel to the normal.
a bent toward the normal.
b bent away from the normal.
d not bent.
23) A ray of light in air is incident on an air-to-glass boundary at an angle of exactly 30.0° with the normal. If
the index of refraction of the glass is 1.65, what is the angle of the refracted ray within the glass with
respect to the normal?
c 34.4°
a 58.3°
b 37.3°
d 18.0°
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Name: ______________________
Name: ______________________
36) When compared in a given time interval with other lightbulbs connected to a 120 V circuit, a 60 W
lightbulb
a converts the same electrical energy to heat and light as a 40 W lightbulb.
b converts more electrical energy to heat and light than a 100 W lightbulb.
c converts less electrical energy to heat and light than a 40 W lightbulb.
d converts less electrical energy to heat and light than a 100 W lightbulb.
37)
30) The figure shown above demonstrates charging by
c polarization.
a grounding.
b induction.
d contact.
31) A parallel-plate capacitor has a capacitance of C F. If the area of the plates is doubled while the distance
between the plates is halved, the new capacitance will be
C
c
a 4C.
.
2
C
b 2C.
d
.
4
1
32) A 0.50 µF capacitor is connected to a 12 V battery. Use the expression PE = 2 C(%V) 2 to determine how
much electrical potential energy is stored in the capacitor.
!6
c 1.0 " 10 !5 J
a 3.0 " 10 J
b
6.0 " 10 !6 J
3.6 " 10 !5 J
d
33) The current in an electron beam in a cathode-ray tube is 7.0 " 10
5.0 s?
3
c 3.5 " 10 !4 C
a 2.8 " 10 C
b
5.6 " 10 !2 C
!5
A. How much charge hits the screen in
5.3 " 10 !6 C
d
34) A flashlight bulb with a potential difference of 4.5 V across it has a resistance of 8.0 &. How much
current is in the bulb filament?
c 1.8 A
a 36 A
b 9.4 A
d 0.56 A
35) Which of the following wires would have the greatest resistance?
a an aluminum wire 10 cm in length and 3 cm in diameter
b an aluminum wire 5 cm in length and 3 cm in diameter
c an aluminum wire 10 cm in length and 5 cm in diameter
d an aluminum wire 5 cm in length and 5 cm in diameter
5
What happens when the switch is closed in the circuit shown above?
a The lamp lights because current from the battery flows through the lamp.
b Current from the battery flows through the resistor.
c Current from the battery flows through both the lamp and the resistor.
d The lamp goes out, because the battery terminals connect to each other.
38) If the potential difference across a pair of batteries used to power a flashlight is 6.0 V, what is the potential
difference across the flashlight bulb?
c 9.0 V
a 3.0 V
b 6.0 V
d 12 V
39) Three resistors with values of 4.0 &, 6.0 &, and 8.0 &, respectively, are connected in series. What is their
equivalent resistance?
c 6.0 &
a 18 &
b 8.0 &
d 1.8 &
40) Three resistors connected in series carry currents labeled I 1 , I 2 , and I 3 , respectively. Which of the
following expresses the total current, I t , in the system made up of the three resistors in series?
c It = I1 = I2 = I3
a It = I1 + I2 + I3
ˆ˜
ˆ !1
ÁÊÁÁ 1
˜
ÁÊÁÁ 1
1
1 ˜
1
1 ˜˜˜
b It = Á
ÁÁÁ +
d It = Á
ÁÁÁ +
+ ˜˜˜˜
+ ˜˜˜˜
ÁÁ I
˜
ÁÁ I
˜
Ë 1 I 2 I 3 ˜¯
Ë 1 I 2 I 3 ˜¯
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Name: ______________________
Name: ______________________
41) Three resistors connected in series have potential differences across them labeled %V 1 , %V 2 , and %V 3 .
Which of the following expresses the potential difference taken over the three resistors together?
c %Vt = %V1 = %V2 = %V3
a %Vt = %V1 + %V2 + %V3
ÊÁÁ
ˆ˜˜
ÊÁÁ
ˆ !1
1
1
1
1
1 ˜˜˜˜
ÁÁ
˜˜˜
ÁÁ 1
b %Vt = Á
d %Vt = Á
˜˜
+
+
+
+
ÁÁÁ
˜˜˜
ÁÁÁ
ÁË %V 1 %V 2 %V 3 ˜¯
ÁË %V 1 %V 2 %V 3 ˜˜˜¯
42) Three resistors with values of 4.0 &, 6.0 &, and 10.0 & are connected in parallel. What is their equivalent
resistance?
c 6.0 &
a 20.0 &
b 7.3 &
d 1.9 &
45) Which compass needle orientation in the figure above might correctly describe the magnet’s field at that
point?
c c
a a
b b
d d
46) According to Lenz’s law, if the applied magnetic field changes,
a the induced field attempts to keep the total field strength constant.
b the induced field attempts to increase the total field strength.
c the induced field attempts to decrease the total field strength.
d the induced field attempts to oscillate about an equilibrium value.
47) A coil with 150 turns and a cross-sectional area of 1.00 m 2 experiences a magnetic field whose strength
increases by +0.65T in 1.80 s. The plane of the coil is perpendicular to the plane of the applied magnetic
field. What is the induced emf in the coil?
c !27 V
a 54 V
b 0.36 V
d !54 V
43) What is the equivalent resistance of the resistors in the figure shown above?
c 16 &
a 7.5 &
b 10 &
d 18 &
44) Three resistors connected in parallel have individual values of 4.0 &, 6.0 &, and 10.0 &, as shown above.
If this combination is connected in series with a 12.0 V battery and a 2.0 & resistor, what is the current in
the 10.0 & resistor?
c 11 A
a 0.58 A
b 1.0 A
d 16 A
7
48) A circular coil with an area of 5.0 " 10 !2 m 2 and with 500 turns of wire is placed in a uniform magnetic
field perpendicular to the plane of the coil. If the field changes in value from –0.100 T to +0.150 T in an
interval of 0.500 s, what is the induced emf in the coil?
c 188 V
a –12.5 V
b 125 V
d 252 V
49) Which of the following options can be used to generate electricity?
a Move the circuit loop into and out of a magnetic field.
b Change the magnetic field strength around the circuit loop.
c Change the orientation of the circuit loop with respect to the magnetic field.
d all of the above
50) Two loops of wire are arranged so that a changing current in the primary will induce a current in the
secondary. The secondary loop has twice as many turns as the primary loop. As long as the current in the
primary is steady at 3.0 A, the current in the secondary will be
c 1.5 A.
a 6.0 A.
b 3.0 A.
d zero.
51) A generator’s maximum output is 220 V. What is the rms potential difference?
c 160 V
a 110 V
b 150 V
d 310 V
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Name: ______________________
Name: ______________________
71) What is the wavelength of the fundamental frequency of a tube with open ends and length L?
Short Answer
72) What is the wavelength of the third harmonic in a tube of length L that has a closed end?
52) What does temperature measure?
53) What condition is necessary for two bodies that are in physical contact with each other to be in thermal
equilibrium?
54) A pan of water at a temperature of 80°C is placed on a block of porcelain at a temperature of 15°C. What
can you state about the temperatures of the objects when they are in thermal equilibrium?
73) The fourth harmonic of a certain vibrating string has a frequency of 1600 Hz. What is the frequency of
the string’s first harmonic?
74) How is the brightness of a light source affected by distance?
55) Explain how thermal expansion makes a mercury thermometer a useful device for measuring temperature.
56) In which direction is energy transferred as heat between two objects at different temperatures?
57) What are materials that easily transfer energy as heat called?
58) Do “heat” and “cold” flow between objects? Explain.
59) How does the principle of conservation of energy take internal energy into account?
60) What is latent heat?
75) What type of reflection is illustrated in the figure shown above?
61) What indicates that energy has been added to or removed as heat or work from a system?
62) How are heat and work similar?
76) What does a positive magnification signify?
77) If two light waves are out of phase, what is the smallest amount by which their phases can differ?
63) A physics textbook is balanced on top of an inflated balloon on a cold morning. As the day passes, the
temperature increases, the balloon expands, and the textbook rises. Is there a transfer of energy as heat? If
so, what is it? Has any work been done? If so, on what?
64) A gas is confined in a cylinder with a piston. What happens when work is done on the gas?
65) How does an adiabatic process compare to an isovolumetric process?
66) A mechanic pushes down very quickly on the plunger of an insulated pump. The air hose is plugged so
that no air escapes. What type of thermodynamic process takes place? What type of energy transfer and
change occurs?
78) What is diffraction?
79) Explain what happens when you vigorously rub your wool socks on a carpeted floor, touch a metal
doorknob, and get a shock.
80) What property was discovered in Millikan’s experiments? Explain this property.
81) When an object loses electrons, it acquires a(n) ____________________ electric charge.
82) How does the electric force between two charged objects change when the charge on one of the objects is
doubled? Explain.
67) How is the conservation of a system’s internal energy mathematically expressed for an isovolumetric
process? Explain your answer.
83) How are gravitational and electric force alike?
68) What is entropy?
84) Draw the lines of force representing the electric field surrounding two objects that have equal magnitude
charges of opposite polarity.
69) Suppose that a pendulum has a period of 4.0 s at Earth’s surface. If the pendulum is taken to the moon,
where the acceleration due to gravity is much less than on Earth, will the pendulum’s period increase,
decrease, or stay the same? Explain your answer.
85) How are electric potential and electrical potential energy related?
70) Electromagnetic waves can move through empty space, but mechanical waves require a
____________________ through which to travel.
87) Explain why there is a limit to the amount of charge that can be stored in a capacitor.
9
86) What is capacitance?
10
Name: ______________________
Name: ______________________
88) The spacing between the plates of a particular capacitor is reduced from 1.5 " 10 !2 cm to 1.5 " 10 !5 cm.
How will this spacing change affect the capacitor?
89) List two ways to increase the electrical potential energy that can be stored in a capacitor.
96) Use the right-hand rule to determine the direction of the magnetic field within the loop from the figure
above. Since the magnetic field of a current-carrying loop resembles that of a bar magnet, is the north pole
of the current-carrying loop above or below the loop?
90) The curve above shows a plot of current versus potential difference for a material. What type of material
does the curve illustrate?
91) What are the characteristics of alternating current?
92) What is electric power?
93) If the head of an iron nail touches a magnet, the nail will become magnetized. If the nail touches the north
pole of the magnet, what kind of pole is at the head of the nail? Explain.
97) Find the direction of the force on an electron moving through the magnetic field shown above.
94) The magnetic field of a bar magnet is shown in the figure above. Is the magnet’s north pole at A or B?
95) Define magnetic flux, ' M , and describe how the figure above, magnetic flux, and magnetic field strength
relate to one another.
98) Use the right-hand rule and the figure above to explain why the two wires exert an attractive force on each
other.
99) What is the process of causing a current in a circuit using a changing magnetic field called?
100) List three ways to induce a current in a circuit loop, using only a magnet.
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Name: ______________________
Name: ______________________
101) Lenz’s law states that the magnetic field of the induced ____________________ is in a direction to
produce a field that ____________________ the change causing it.
102) Explain the relationship between maximum emf and each of the four variables in the following equation:
maximum emf = NAB(.
103) What kind of transformer has more turns in the primary coil than in the secondary coil?
114) An ideal gas is maintained at a constant pressure of 7.0 " 10 4 N/m 2 while its volume decreases by 0.20
m 3 . What work is done by the system on its environment?
115) A cylinder has a radius of 0.080 m. How much work is done by a gas in the cylinder if the gas exerts a
constant pressure of 7.8 " 10 5 Pa on the piston, moving it a distance of 0.060 m?
116) Gas within a cylinder expands outward against a piston with a radius of 0.13 m so that 6750 J of work is
to be done by the gas. If the net pressure exerted on the gas is 3.5 " 10 5 Pa, how far is the piston
displaced?
Problem
104) A warm day has a high temperature of 38.0°C. What is this temperature in degrees Fahrenheit?
105) The temperature of an object is measured as 500.0 K. What is this temperature in degrees Celsius?
106) Liquid oxygen has a temperature of –183°C. What is this temperature in kelvins?
107) A hammer drives a nail at a speed of 0.30 m/s into a piece of wood. The wood does not move during this
action. If the mass of the nail is 75.0 g and half of its mechanical energy is transferred to the wood as heat,
how much does the internal energy of the wood change?
108) A 3.00 " 10 !3 kg lead bullet travels at a speed of 2.40 " 10 2 m/s and hits a wooden post. If half the total
internal energy remains with the bullet, and it takes 128 J of energy to raise the temperature of a kilogram
of lead by 1.00°C, what is the increase in temperature of the embedded bullet?
109) A 5.00 " 10 2 kg object is attached by a rope through a pulley to a paddle-wheel shaft that is placed in a
well-insulated tank holding 25.0 kg of water. The object is allowed to fall, causing the paddle wheel to
rotate, churning the water. Assume the object falls a vertical distance of 1.00 " 10 2 m at constant speed,
that all of the change in internal energy is conveyed to the water, and that it takes 4186 J to raise the
temperature of each kilogram of water by 1.00°C. What is the temperature change of the water? (g = 9.81
m/s 2 )
117) If a force of 50 N stretches a spring 0.10 m, what is the spring constant?
118) How much displacement will a coil spring with a spring constant of 120 N/m achieve if it is stretched by a
60 N force?
119) An amusement park ride has a frequency of 0.05 Hz. What is the ride’s period?
120) Imagine that you could transport a simple pendulum from Earth to the moon, where the free-fall
acceleration is one-sixth that on Earth. By what factor would the pendulum’s frequency be changed?
Express the answer with one significant figure.
121) A mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. If the mass of the
object is 0.20 kg and the spring constant is 130 N/m, what is the frequency?
122) What is the period of a 4.12 m long pendulum with a bob of mass 75.0 kg?
123) A student wishes to construct a mass-spring system that will oscillate with the same frequency as a
swinging pendulum with a period of 3.45 s. The student has a spring with a spring constant of 72.0 N/m.
What mass should the student use to construct the mass-spring system?
124) What is the intensity of sound waves produced by a trumpet at a distance of 1.6 m when the power output
of the trumpet is 0.30 W?
2
110) What is the temperature increase of 4.0 kg of water when it is heated by an 8.0 " 10 W immersion heater
for exactly 10.0 min? (cp = 4186 J/kg•°C)
111) A mixture of 30.2 g of sand and 87.7 g of water has a temperature of 12.1°C. What mass of water at
85.8°C must be added to raise the final temperature of the mixture to 29.3°C? (c p, w !=!4.19!J/g•°C!and
c p, s != 2.01 J/g•°C)
112) A container of gas is at a pressure of 3.7 " 10 5 Pa. How much work is done by the gas if its volume
expands by 1.6 m 3 ?
113) A container of gas is at a pressure of 1.3 " 10 5 Pa and a volume of 6.0 m 3 . How much work is done by
the gas if it expands at constant pressure to twice its initial volume?
13
125) If the intensity of a sound is 8.0 " 10 !4 W/m 2 at a distance of 5.0 m, what is the power of the sound?
126) The noise of a certain lawn mower has a power of 0.11 W. At what distance will the intensity of the lawn
mower’s sound be 8.6 " 10 !4 W/m 2 ?
127) A certain radio wave has a frequency of 2.0 " 10 6 Hz. What is its wavelength?
128) A pencil is located 16.0 cm in front of a convex mirror whose focal length is 11.0 cm. In relation to the
mirror’s surface, where and how far away is the corresponding image located?
129) A concave spherical mirror has a radius of curvature of 10.0 cm. A candle that is 5.0 cm tall is placed 15
cm in front of the mirror. Draw a ray diagram to find the image distance and height. Confirm the results of
your diagram with the mirror equation and the equation for magnification.
14
Name: ______________________
Name: ______________________
130) A ray of light passes from air into ice (n = 1.309) at an angle of 46.0° to the normal. The refracted ray of
light then passes from ice into glycerine (n = 1.473). What is the angle of refraction of the ray of light in
glycerine?
138) What is the electric potential at a distance of 0.15 m from a point charge of 6.0 µC? (k C = 8.99 " 10 9
131) An object is placed along the principal axis of a thin converging lens that has a focal length of 28 cm. If
the distance from the image in front of the lens is 24 cm, what is the distance from the object to the lens?
139) A proton moves north at a velocity of 8.4 " 10 4 m/s and has a magnetic force of 3.0 " 10 !18 N east
exerted on it. If the magnetic field points upward, what is the magnitude of the magnetic field?
132) A telescope has an objective lens with a focal length of 36.0 cm. The image formed by the objective lens is
0.500 cm inside the focal point of the eyepiece. Where does the image of the eyepiece appear to be if the
focal length of the eyepiece is 10.00 cm?
140) A wire 48 m long carries a current of 18 A from west to east. If a magnetic field of 8.3 " 10 !4 T directed
toward the south is acting on the wire, find the direction and magnitude of the magnetic force.
133) The critical angle of refraction for calcite is 68.4° when it forms a boundary with water. Use this
information to determine the speed of light in calcite.
134) A helium-neon laser placed perpendicular to a diffraction grating that contains 149 638 lines/m illuminates
the grating’s surface with monochromatic light with a wavelength of 632.8 nm. At what angle will the
second-order maximum appear?
135) What is the electric force between an electron and a proton that are separated by a distance of 1.0 " 10 !10
m? Is the force attractive or repulsive? (e = 1.60 " 10 !19 C, k C = 8.99 " 10 9 N•m 2 /C 2 )
N•m 2 /C 2 )
141) A coil with 275 turns and a cross-sectional area of 0.750 m 2 experiences a magnetic field whose strength
increases by +0.900 T in 1.25 s. The plane of the coil is perpendicular to the plane of the magnetic field.
What is the induced emf in the coil?
142) A coil with 650 turns produces an emf of 120 V as it passes through a magnetic field in 2.7 s. The
magnetic field change is !0.935 T. The plane of the coil moves perpendicularly to the plane of the
magnetic field. What is area of the coil?
143) A coil with 25 turns of wire moves through a uniform magnetic field of 1.5 T. The plane of the magnetic
field is perpendicular to the plane of the coil. The coil has a cross-sectional area of 0.80 m 2 . The coil
enters and exits the field over 1.0 s. If the coil’s resistance is 2.0 &, what is the induced current?
136) Two point charges having charge values of 2.0 µC and –4.0 µC, respectively, are separated by 1.5 cm.
What is the value of the mutual force between them? (k C = 8.99 " 10 9 N•m 2 /C 2 )
137) Two equal positive charges, both q = 2.0 " 10 !6 C, interact with a third charge, Q = 4.0 " 10 !6 C, as
shown in the figure above. What is the magnitude of the electric force on Q? (k C = 8.99 " 10 9 N•m 2 /C 2 )
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ID: A
Physics Practice 10 | FINAL STUDY GUIDE
Answer Section
MULTIPLE CHOICE
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C
A
A
C
B
A
B
D
B
C
D
C
D
C
D
C
B
B
B
A
D
B
D
C
A
D
A
B
B
C
A
D
C
D
A
D
D
B
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9-1.3
9-1.3
9-2.3
9-3.1
9-3.2
10-2.2
10-2.2
10-2.2
11-4.4
12-2.2
12-3.4
12-3.4
13-1.2
13-1.2
13-2.2
13-3.1
13-3.1
13-3.1
13-3.1
13-3.1
13-3.3
14-1.2
14-1.3
14-1.3
14-2.3
15-1.3
15-2.2
15-2.2
15-2.2
16-1.3
17-2.2
17-2.3
17-3.1
17-3.3
17-3.4
17-4.2
18-1.1
18-1.3
ID: A
39)
40)
41)
42)
43)
44)
45)
46)
47)
48)
49)
50)
51)
ANS:
ANS:
ANS:
ANS:
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ANS:
A
C
A
D
B
A
A
A
D
A
D
D
C
OBJ:
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OBJ:
18-2.1
18-2.1
18-2.1
18-2.2
18-3.1
18-3.2
19-1.2
20-1.3
20-1.3
20-1.3
20-2.1
20-2.3
20-3.2
SHORT ANSWER
52) ANS:
Temperature measures the average kinetic energy of the particles in a substance.
OBJ: 9-1.1
53) ANS:
The temperature of the two objects must be the same.
OBJ: 9-1.2
54) ANS:
Both objects will have the same final temperature, which will be somewhere between 15°C and 80°C.
OBJ: 9-1.2
55) ANS:
The volume of many substances, including mercury, increases in proportion to the increase in its
temperature. Therefore, by confining mercury to a tube with a constant cross-sectional area, the increase in
its volume due to thermal expansion results in a uniform increase in the height of the column, which can be
calibrated to given temperatures.
OBJ: 9-1.2
56) ANS:
Energy is transferred as heat from the object at higher temperature to the object at lower temperature.
OBJ: 9-2.1
57) ANS:
thermal conductors
OBJ: 9-2.1
1
2
ID: A
58) ANS:
No, “heat” and “cold” do not flow between objects. Energy transferred between objects changes the
temperature of the objects.
OBJ: 9-2.1
59) ANS:
Mechanical energy is not always conserved. But when the change in internal energy is taken into account
along with changes in kinetic and potential energy, the total energy is conserved.
OBJ: 9-2.3
60) ANS:
Latent heat is the energy per unit mass that is transferred to or from a substance during a phase change of
that substance.
OBJ: 9-3.2
61) ANS:
The internal energy of the system changes, as indicated by a change in the system’s temperature.
OBJ: 10-1.1
62) ANS:
Heat and work are both ways in which energy can be transferred to or from a substance.
OBJ: 10-1.1
63) ANS:
Energy from the air was transferred as heat into the balloon. The balloon did work on the book.
ID: A
68) ANS:
Entropy is a measure of the disorder of a system.
OBJ: 10-3.3
69) ANS:
The period will increase because the restoring force is a component of the gravitational force acting on the
pendulum bob (the bob’s weight). Because the restoring force is less, but the mass remains the same, the
acceleration of the pendulum bob is less.
OBJ: 11-2.3
70) ANS:
medium
OBJ: 11-3.1
71) ANS:
2L
OBJ: 12-3.2
72) ANS:
4/3 L
OBJ: 12-3.2
73) ANS:
400 Hz (Note that the first harmonic is the fundamental frequency of the string.)
OBJ: 12-3.2
74) ANS:
Brightness decreases by the square of the distance from the source.
OBJ: 10-1.1
64) ANS:
The volume of the gas decreases.
OBJ: 10-1.2
65) ANS:
The internal energy of a system changes during both processes. In an adiabatic process, no energy is
transferred as heat, but work is done on or by the system. In an isovolumetric process, no work is done,
but energy is transferred as heat to or from the system.
OBJ: 10-1.3
66) ANS:
The process is adiabatic, because no energy is transferred into or out of the system as heat. Work is done
on the air in the system, which causes the internal energy of the air to increase.
OBJ: 10-1.3
67) ANS:
W = 0 for an isovolumetric process, so %U = Q.
OBJ: 10-2.1
OBJ: 13-1.4
75) ANS:
diffuse
OBJ: 13-2.1
76) ANS:
The image is upright and virtual.
OBJ: 14-2.3
77) ANS:
180°
OBJ: 15-1.1
78) ANS:
Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an
edge.
OBJ: 15-2.1
3
4
ID: A
79) ANS:
Loosely held electrons are transferred from the carpet to the socks when the socks are rubbed against the
carpet. The body and socks have an excess of electrons and are negatively charged. Touching the
doorknob allows the electrons to escape. The shock felt is the sudden movement of charges as the body
and socks return to a neutral state.
OBJ: 16-1.1
80) ANS:
Millikan discovered that charge is quantized. This means that when any object is charged, the net charge is
always a multiple of a fundamental unit of charge. The fundamental unit of charge, which is the charge on
the electron, is –1.60 " 10 !19 C. The charge on a proton is +1.60 " 10 !19 C.
OBJ: 16-1.1
81) ANS:
positive
ID: A
86) ANS:
Capacitance is the measure of the ability of a device to store energy in the form of electrically separated
charges.
OBJ: 17-2.1
87) ANS:
The potential difference between the plates can become so great that a spark discharge or an electrical
breakdown will occur.
OBJ: 17-2.1
88) ANS:
The capacitance of the capacitor will increase to 100 times its original value.
OBJ: 17-2.2
89) ANS:
Increase the area of the plates or place a dielectric material between the plates.
OBJ: 16-1.1
82) ANS:
The force increases to twice its previous value because the force is directly proportional to the charge on
each of the objects.
OBJ: 17-2.3
90) ANS:
ohmic material
OBJ: 16-2.1
83) ANS:
Both gravitational force and electric force are field forces. Both forces obey the inverse square law.
OBJ: 17-3.4
91) ANS:
The motion of charges continually changes in the forward and reverse directions. There is no net motion
of the charge carriers; they simply vibrate back and forth.
OBJ: 16-2.2
84) ANS:
OBJ: 17-4.1
92) ANS:
Electric power is the rate at which charge carriers convert electric potential energy to other forms of
energy.
OBJ: 17-4.2
93) ANS:
The end of the magnetized nail touching the north pole of the magnet must be a south pole. Otherwise, the
magnet would repel the nail.
OBJ: 19-1.1
94) ANS:
A
OBJ: 16-3.2
85) ANS:
Electric potential is the electrical potential energy associated with a charged particle divided by the charge
of the particle.
OBJ: 17-1.1
OBJ: 19-1.2
95) ANS:
Magnetic flux is defined as the number of field lines that cross a certain area. As the figure above
indicates, wherever more magnetic field lines are present, the greater the magnetic flux or magnetic field
strength.
OBJ: 19-1.2
5
6
ID: A
96) ANS:
With the thumb in the direction of the current, the fingers will curl down around the loop. Thus the
magnetic field points downward around the loop. The north pole of the loop is below the loop, since the
magnetic field appears to be exiting the loop area.
OBJ: 19-2.2
97) ANS:
to the right
ID: A
PROBLEM
104) ANS:
100.4°F
Given
T C = 38.0°C
OBJ: 19-3.2
98) ANS:
According to the right-hand rule, as the current in one wire interacts with the magnetic field in the other
wire, opposing (i.e., attractive) magnetic forces are induced. Since the current is flowing in the same
direction in both wires, the resulting magnetic forces are attractive.
OBJ: 19-3.3
99) ANS:
electromagnetic induction
Solution
TF =
9
5
T C + 32.0
Ê9
ˆ
T F = ÁÁÁÁ 5 (38.0) + 32.0 ˜˜˜˜ °F = (68.4 + 32.0)°F = 100.4°F
Ë
¯
OBJ: 9-1.3
105) ANS:
226.8°C
OBJ: 20-1.1
100) ANS:
Move the circuit loop into or out of the magnetic field. Rotate the circuit loop in the magnetic field so that
the angle between the plane of the circuit loop and magnetic field changes. Vary the intensity of the
magnetic field by rotating the magnet.
OBJ: 20-1.2
101) ANS:
current, opposes
Given
T = 500.0 K
Solution
T = T C + 273.15
T C = T ! 273.15 = (500.0 ! 273.15)°C = 226.8°C
OBJ: 20-1.3
102) ANS:
The equation indicates that there is a direct relationship between maximum emf and each of the four
variables. It also indicates that maximum emf is dependent on (or a function of) four things: the number of
loops, N; the area of the loop, A; the magnetic field strength, B; and the angular frequency of the rotation of
the loop, (.
OBJ: 9-1.3
106) ANS:
9.0 " 10 1 K
Given
T C = –183°C
Solution
T = T C + 273.15
OBJ: 20-2.1
103) ANS:
step-down transformer
T = (!183 + 273.15) K = 9.0 " 10 1 K
OBJ: 20-3.3
OBJ: 9-1.3
7
8
ID: A
107) ANS:
!3
1.7 " 10 J
ID: A
108) ANS:
112°C
Given
Given
v = v i = 0.30 m/s
v f = 0 m/s
m bullet = 3.00 " 10 !3 kg
v = v i = 2.40 " 10 2 m/s
m nail = 75.0 g
1
%U bullet = 2 %U
1
%U wood = 2 %U
%U bullet
128 J/kg
=
Solution
%PE + %KE + %U = 0
m bullet %T
The potential energy of the nail and wood is unchanged during the process, and the final kinetic energy of
the nail and wood is zero. The above equation thus takes the following form:
Solution
%PE + %KE + %U = 0
%KE + %U = KE f ! KE i + %U = 0 ! KE i + %U = 0
The potential energy of the bullet and post is unchanged during the process, and the final kinetic energy of
the bullet and post is zero. The above equation thus takes the following form:
The internal energy change of the wood equals half the internal change of the overall system.
%KE + %U = KE f ! KE i + %U = 0 ! KE i + %U = 0
%U wood
%U wood
Ê1
ˆ
= 2 %U = 2 KE i = 2 ÁÁÁÁ 2 m nail v 2 ˜˜˜˜
Ë
¯
ÊÁÁ 1 kg ˆ˜˜
1
˜˜ = 1.7 " 10 !3 J
= 4 (75.0 g)(0.30 m/s) 2 " ÁÁÁÁ
ÁÁ 1000 g ˜˜˜˜
Ë
¯
1
1
1
OBJ: 9-2.3
1.00°C
The internal energy change of the bullet equals half the internal change of the overall system.
%U bullet =
1
2
%U =
1
2
KE i =
1
2
ÊÁÁ 1
2ˆ
ÁÁ 2 m bullet v ˜˜˜˜
Ë
¯
ÊÁÁ 128 J/kg ˆ˜˜
˜˜ m
%U bullet = ÁÁÁÁ
%T
ÁË 1.00°C ˜˜˜¯ bullet
1
%T =
m bullet v 2
(2.40 " 10 2 m/s) 2
=
= 112°C
ÁÊÁÁ 128 J/kg ˜ˆ˜˜
ÁÊÁ 128 J/kg ˜ˆ˜˜
˜˜˜
˜˜˜
m bullet ÁÁÁ
(4) ÁÁÁÁ
ÁË 1.00°C ˜¯
ÁË 1.00°C ˜¯
4
OBJ: 9-2.3
9
10
ID: A
109) ANS:
4.69°C
ID: A
110) ANS:
29°C
Given
Given
m = 4.0 kg
P = 8.0 " 10 2 W
%t = 10.0 min
c p = 4186 J/kg°C
m object = 5.00 " 10 2 kg
m water = 25.0 kg
h = 1.00 " 10 2 m
%U
4186 J/kg
=
m water %T
1.00°C
Solution
Heat equals the power delivered multiplied by the time interval.
g = 9.81 m/s 2
Q = P%t
Solution
%PE + %KE + %U = 0
Q = c p m%T
The object falls at a constant speed, so the change in kinetic energy is zero. Assuming the final potential
energy has a value of zero, the above equation takes the following form:
%T =
0 ! PE i + %U = 0
OBJ: 9-3.1
P%t
cp m
=
ÊÁÁ 60 s ˆ˜˜
˜˜ = 29°C
" ÁÁÁÁ
ÁË 1 min ˜˜˜¯
(4186 J/kg°C)(4.0 kg)
(8.0 " 10 2 W)(10.0 min)
%U = PE i = m object gh
ÊÁÁ 4186 J/kg ˆ˜˜
˜˜ m
%U = ÁÁÁÁ
%T
ÁË 1.00°C ˜˜˜¯ water
%T =
m object
m water
gh
(5.00 " 10 2 kg)(9.81 m/s 2 )(1.00 " 10 2 m)
= 4.69°C
ÊÁÁ 4186 J/kg ˆ˜˜ =
ÊÁÁ 4186 J/kg ˆ˜˜
ÁÁÁ
˜˜˜
ÁÁÁ
˜˜˜
(25.0
kg)
ÁÁ 1.00°C ˜˜
ÁÁ 1.00°C ˜˜
Ë
¯
Ë
¯
OBJ: 9-2.3
11
12
ID: A
111) ANS:
31.1 g
ID: A
112) ANS:
5
5.9 " 10 J
Given
m s = 30.2 g
m w = 87.7 g
T s = T w = 12.1°C
Given
P = 3.7 " 10 5 Pa
%V = 1.6 m 3
T added water = 85.8°C
T f = 29.3°C
Solution
W = P%V = (3.7 " 10 5 Pa)(1.6 m 3 ) = 5.9 " 10 5 J
c p, w = 4.19 J/g•°C
c p, s = 2.01 J/g•°C
Solution
From conservation of energy, the energy absorbed as heat by the water and the sand equals the energy
given up as heat by the added water.
Q w + Q s = !Q added water
c p, w m w %T w + c p, s m s %T s = !c p, w m added water %T added water
c p, w m w (T f ! T w ) + c p, s m s (T f ! T s ) = !c p, w m added water (T f ! T added water ) = c p, w m added water (T added water ! T f )
m added water =
m added water =
m added water =
m added water =
c p, w m w (T f ! T w ) + c p, s m s (T f ! T s )
c p, w (T added water ! T f )
(4.19 J/g°C)(87.7 g)(29.3°C ! 12.1°C) + (2.01 J/g°C)(30.2 g)(29.3°C ! 12.1°C)
(4.19 J/g°C)(85.8°C ! 29.3°C)
(4.19 J/g°C)(87.7 g)(17.2°C) + (2.01 J/g°C)(30.2 g)(17.2°C)
(4.19 J/g°C)(56.5°C)
6.32 " 10 3 J + 1.04 " 10 3 J
(4.19 J/g°C)(56.5°C)
= 31.1 g
OBJ: 10-1.2
113) ANS:
7.8 " 10 5 J
Given
P = 1.3 " 10 5 Pa
V i = 6.0 m 3
V f = 2V i
Solution
W = P%V
%V = V f ! V i = 2V i ! V i = V i
W = (1.3 " 10 5 Pa)(6.0 m 3 ) = 7.8 " 10 5 J
OBJ: 10-1.2
114) ANS:
–1.4 " 10 4 J
Given
P = 7.0 " 10 4 N/m 2
%V = !0.20 m 3
OBJ: 9-3.1
Solution
The volume decreases, so %V, and thus W, are negative.
W = P%V = (7.0 " 10 4 N/m 2 )(!0.20 m 3 ) = !1.4 " 10 4 J
OBJ: 10-1.2
13
14
ID: A
115) ANS:
2
9.4 " 10 J
ID: A
117) ANS:
500 N/M
Given
F elastic = 50 N
x = –0.10 m
Given
r = 0.080 m
P = 7.8 " 10 5 Pa
d = 0.060 m
Solution
F elastic = –kx
Solution
Work is done by the gas, so W is positive.
W = P%V = PAd
!F elastic
!50 N
=
x
!0.10 m
k = 500 N/m
k=
A = )r 2
W = PAd = P) r 2 d = (7.8 " 10 5 Pa)() )(0.080 m) 2 (0.060 m)
P = 9.4 " 10 2 J
OBJ: 10-1.2
116) ANS:
0.36 m = 36 cm
OBJ: 11-1.3
118) ANS:
–0.5 m
Given
k = 120 N/m
F elastic = 60 N
Given
r = 0.13 m
W = 6750 J
Solution
F elastic = –kx
P = 3.5 " 10 5 Pa
x=!
k
x = –0.5 m
Solution
Work is done by the gas, so W is positive.
W = P%V = PAd
W
PA
=
W
P) r 2
=
=!
60 N
120 N/m
OBJ: 11-1.3
119) ANS:
20 s
A = )r 2
d=
F elastic
6750 J
Given
f = 0.05 Hz
(3.5 " 10 5 Pa)() )(0.13 m) 2
P = 0.36 m = 36 cm
Solution
1
T= 1 =
f
0.05 Hz
OBJ: 10-1.2
T = 20s
OBJ: 11-2.3
15
16
ID: A
120) ANS:
0.4
ID: A
121) ANS:
4.0 Hz
Given
1
ag = g
6
Given
x = 8.0 cm (This value is not relevant to the problem.)
m = 0.20 kg
k = 130 N/m
Solution
Solution
L
T = 2)
ag
m
T = 2)
k
Because L and 2) remain constant when the pendulum is
moved to the moon,T moon *
1
ag
,where a g is the gravitational
1
f=
2)
acceleration of the moon.
T moon
T earth
=
1
=
g
ag
1
T
, so
f moon
f earth
=
ag
g
, so
T
1
k
2)
m
=
1
130 N/m
2)
0.20 kg
= 4.1 Hz
k
Given
L = 4.12 m
m = 75.0 kg (This mass is not relevant to the problem.)
g
f=
=
1
OBJ: 11-2.3
122) ANS:
4.07 s
1
ag
m
and f =
=
1
6
= 0.4
Solution
T = 2)
OBJ: 11-2.3
L
ag
= 2)
4.12 m
9.81 m/s 2
= 4.07 s
OBJ: 11-2.3
17
18
ID: A
123) ANS:
21.7 kg
125) ANS:
0.25 W
Given
T pendulum = 3.45 s
Given
Intensity = 8.0 "10 !4 W/m 2
r = 5.0 m
k = 72.0 N/m
Solution
If both systems have the same frequency, they will also have the same period.
Therefore, the given period may be substituted into the equation for a mass-spring system.
T = 2)
m
k
ÊÁ m ˜ˆ˜
Á
T 2 = 4) 2 ÁÁÁÁ ˜˜˜˜
ÁË k ˜¯
m=
T2k
4) 2
ID: A
4) 2
Given
P = 0.11 W
Intensity = 8.6 "10 !4 W/m 2
= 21.7 kg
OBJ: 11-2.3
124) ANS:
Solution
Intensity =
9.3 " 10 !3 W/m 2
r2 =
Given
P = 0.30 W
r = 1.6 m
r=
Solution
Intensity =
Intensity =
P
; P = 4) r 2 (Intensity)
4) r 2
P = 4) (5.0 m) 2 (8.0 "10 !4 W/m 2 ) = 0.25 W
Intensity =
OBJ: 12-2.1
126) ANS:
3.2 m
(3.45 s) 2 (72.0 N/m)
=
Solution
P
4) r 2
P
Intensity " 4)
;r=
P
Intensity " 4)
0.11 W
(8.6 " 10 !4 W/m 2 )(4) )
= 3.2 m
P
4) r 2
0.30 W
4) (1.6 m) 2
OBJ: 12-2.1
!3
= 9.3 " 10 W/m
2
OBJ: 12-2.1
19
20
ID: A
127) ANS:
2
1.5 " 10 m or 150 m
ID: A
129) ANS:
q = 7.50 cm
h+ = !2.5 cm
Given
f = 2.0 " 10 6 Hz = 2.0 " 10 6 s !1
c = 3.00 " 10 8 m/s
Solution
Rearrange the wave speed equation,c = f# , to isolate # , and calculate.
#=
ÊÁÁ
ˆ
8
ÁË 3.00 " 10 m/s ˜˜˜¯
= Ê
= 1.5 " 10 2 m or 150 m
ÁÁÁ 2.0 " 10 6 s !1 ˆ˜˜˜
f
Ë
¯
c
OBJ: 13-1.2
128) ANS:
!6.52 cm
Given
Given
h = 5.0 cm
f = !11.0 cm
R = 10.0 cm
p = 16.0 cm
p = 15 cm
Solution
Rearrange the mirror equation,
1
q
=
1
f
!
1
p
=
1
!11.0 cm
!
1
p
+
1
16.0 cm
1
q
=
=!
1
f
Solution
,and solve for q.
0.0909
1 cm
!
0.0625
1 cm
Since R = 10.0 cm, f = 5.00 cm.
=!
0.1534
Rearrange the mirror equation,
1 cm
1
q = !6.52 cm
Since q is negative,the image is located 6.52 cm behind the mirror.
q
=
1
f
!
1
p
=
1
5.00 cm
!
1
+
p
1
15 cm
=
1
q
=
1
f
3
15 cm
,and solve for q.
!
1
15 cm
=
2
15 cm
q = 7.50 cm
OBJ: 13-3.1
Since q is positive,the image is located 7.50 cm in front of the mirror.
Solve the equation for magnification,M =
h+ = !
qh
p
=!
(7.50 cm) (5.0 cm)
(15 cm)
h+
h
q
= ! ,for h+.
p
= !2.5 cm
Since h+ is negative,the image is located below the principal axis.
21
22
ID: A
OBJ: 13-3.2
130) ANS:
29.2°
132) ANS:
!200 cm
Given
F o = 36.0 cm
Given
$ i = 46.0°
F e = 10.00 cm
n air = 1.000
p e = 10.00 cm ! 0.50 cm = 9.50 cm
n ice = 1.309
Solution
The focal length of the objective lens does not enter into the calculation.
n glycerine = 1.473
Solution
First, determine the angle of refraction in ice.
Rearrange Snell’s law, n i sin$ i = n r sin$ r , and solve for $ r .
ÈÍÍ
˘˙˙
ÈÍÍ
˘˙˙
ÍÈ
˙˘
ÍÍ n i
˙˙
ÍÍ n air ÁÊ
ˆ ˙˙˙˙˙ = sin !1 ÍÍÍÍÍ 1.000 (sin 46.0°) ˙˙˙˙˙ = 33.3°
˜
$ r(ice) = sin !1 ÍÍÍÍ ÁÊË sin$ i ˜ˆ¯ ˙˙˙˙ = sin !1 ÍÍÍÍ
sin
$
i¯˙
ÍÍÍ 1.309
˙˙˙
ÍÍÍ n
˙˙˙
ÍÍÍ n Ë
˙˙˙
Î
˚
Î r
˚
Î ice
˚
Second,use $ r(ice) as the angle of incidence for the ice-glycerine boundary and solve for $ r(glycerine) .
$ r(glycerine)
ID: A
ÈÍÍ
˙˘˙˙
ÍÈÍÍ n
˙˘˙
ÈÍÍ
˘˙˙
Í n ice Ê
1.309
Í i ÊÁ
˙˙˙
ˆ˜ ˙˙˙˙˙ = sin !1 ÍÍÍÍ
ÁÁ sin$ r(ice) ˜ˆ˜ ˙˙˙˙˙ = sin !1 ÍÍÍÍ
= sin ÍÍÍÍ
sin
$
(
sin
33.3°
)
˙˙˙ = 29.2°
i
Í
¯ ˙˙˙
ÍÍÍ 1.453
Ë
¯ ˙˙
ÍÍÍ n
ÍÍÍ n Ë
˙˚
˙˙˚
Î
˙˚
ÍÎ glycerine
Î r
!1
The image of the objective lens is the object of the eyepiece lens.
Rearrange the thin-lens equation,
1
qe
=
1
fe
!
1
pe
=
1
10.00 cm
!
1
p
+
1
9.50 cm
1
q
=
=
1
f
,and solve for q.
0.1000
1 cm
!
OBJ: 14-2.2
Given
q = !24 cm (q is negative,since the image is in front of the lens)
f = 28 cm (f is positive,since this is a converging lens)
Solution
1
1
p
=
1
f
!
1
q
=
1
28 cm
!
1
!24 cm
p
=
+
1
q
=
0.036
1 cm
1
f
+
,and solve for p.
0.042
1 cm
=
0.078
1 cm
p = 13 cm
OBJ: 14-2.2
23
1 cm
=!
0.005
1 cm
q e = !200 cm (since q is negative,the image is virtual and in front of the lens)
OBJ: 14-1.3
131) ANS:
13 cm
Rearrange the thin-lens equation,
0.105
24
ID: A
133) ANS:
8
2.10 " 10 m/s
ID: A
135) ANS:
2.3 " 10
!8
N; attractive
Given
$ c = 68.4°
Given
q e = –e = !1.60 " 10 !19 C
n water = 1.333
q p = +e = +1.60 " 10 !19 C
Solution
r = 1.0 " 10 !10 m
k C = 8.99 " 10 9 N•m 2 /C 2
Rearrange the critical angle equation,sin $ c =
n calcite =
n water
=
sin $ c
1.333
sin 68.4°
=
1.333
0.930
nr
ni
c
n calcite
Solution
= 1.43
Rearrange the index of refraction equation,n =
v light in calcite =
,to find the index of refraction of calcite.
F electric
c
v
,to find velocity of light in calcite.
ÊÁÁ
ˆ
8
ÁË 3.00 " 10 m/s ˜˜˜¯
=
= 2.10 " 10 8 m/s
(1.43)
OBJ: 14-3.1
134) ANS:
$ = 10.91°
ÁÊÁ m# ˜ˆ˜˜
˜
$ = sin !1 ÁÁÁÁ
ÁË d ˜˜˜¯
ÊÁÁ
ˆÊ
ˆ
!19
!19
ÁË !1.60 " 10 C ˜˜˜¯ ÁÁÁË +1.60 " 10 C ˜˜˜¯
ÊÁÁ
ˆ2
!10
ÁË 1.0 " 10 m ˜˜˜¯
OBJ: 16-2.1
136) ANS:
3.2 " 10 2 N
Given
q 1 = 2.0 " 10 !6 C
q 2 = !4.0 " 10 !6 C
r = 1.5 cm = 0.015 m
k C = 8.99 " 10 9 N•m 2 /C 2
Given
# = 632.8 nm = 6.328 " 10 !7 m
1
1
d=
=
m
5 lines
1.496
38
" 10 5
1.496 38 " 10
m
m=2
Solution
dsin $ = m#
F electric
ÁÊÁÁ
ÊÁ
ˆ˜ ÁÁÁÁÁ
9
2
2˜
Á
= k C 2 = Á 8.99 " 10 Nm /C ˜ ÁÁ
Ë
¯ ÁÁÁ
r
ÁÁ
Ë
!8
= !2.3 " 10 N
qeqp
ÁÊÁÁ
ÁÁÁ
ÁÁ
!1 Á
= sin ÁÁÁÁ
ÁÁÁ
ÁÁÁ
ÁÁ
Ë
(2)(6.328 " 10 !7 m)
ÁÊÁÁ
˜ˆ˜
1
ÁÁÁ
m ˜˜
ÁË 1.496 38 " 10 5 ˜˜˜¯
Solution
F electric = k C
˜ˆ˜˜
˜˜˜
˜˜˜
˜˜˜ = 10.91°
˜˜˜
˜˜˜
˜˜˜
¯
q1 q2
r2
ÊÁÁ ÊÁ
Á Á 2.0 " 10 !6 C ˜ˆ˜˜ ÁÊÁÁ !4.0 " 10 !6 C ˜ˆ˜˜
Ê
ˆ ÁÁÁ ÁË
¯Ë
¯
= ÁÁÁ 8.99 " 10 9 Nm 2 /C 2 ˜˜˜ ÁÁÁÁ
2
Ë
¯ ÁÁÁ
(
0.015
m
)
ÁÁ
Ë
F electric = !3.2 " 10 2 N
OBJ: 16-2.1
OBJ: 15-2.2
25
26
˜ˆ˜˜
˜˜˜
˜˜˜
˜˜˜
˜˜˜
¯
˜ˆ˜˜
˜˜˜
˜˜˜
˜˜˜
˜˜˜
¯
ID: A
137) ANS:
0.46 N
ID: A
138) ANS:
5
3.6 " 10 V
Given
q 1 = 2.0 " 10 !6 C
Given
d = 0.15 m
q = 6.0 µC = 6.0 " 10 !6 C
k C = 8.99 " 10 9 N•m 2 /C 2
q 2 = 2.0 " 10 !6 C
Q = 4.0 " 10 !6 C
r q 1 ,Q = r q 2 ,Q = 0.50 m
Solution
r = d = 0.15 m
$ = tan !1 (0.30 m/0.40 m) = 37°
k C = 8.99 "10 9 Nm 2 /C 2
V = kC
Solution
ÊÁÁ ÁÊ
Á Á 2.0 " 10 !6 C ˜ˆ˜˜ ÁÊÁÁ 4.0 " 10 !6 C ˜ˆ˜˜
ÊÁ
ˆ˜ ÁÁÁÁÁ ÁË
¯Ë
¯
9
2
2˜
Á
F q 1 ,Q = F q 2 ,Q = k C
=
8.99
"
10
Nm
/C
Á
˜
2
¯ ÁÁÁÁÁ
ÊÁÁ r ˆ˜˜ 2 Ë
(
0.50
m
)
ÁÁ
Ë q,Q ¯
Ë
For F q 1 , Q : F x,q 1 ,Q = (F q 1 ,Q )(cos 37°) = (0.29 N)(cos 37°) = 0.23 N
qQ
ˆ˜˜
˜˜˜
˜˜˜
˜˜˜ = 0.29 N
˜˜˜
˜¯
ÊÁÁ
ˆ
!6
Á 6.0 " 10 C ˜˜˜˜˜
= (8.99 " 10 9 Nm/C 2 ) ÁÁÁÁ
= 3.6 " 10 5 V
ÁÁÁ 0.15 m ˜˜˜˜˜
r
Ë
¯
q
OBJ: 17-1.2
139) ANS:
2.2 " 10 !4 T
F y,q 1 ,Q = (F q 1 ,Q )(sin 37°) = (0.29 N)(sin 37°) = 0.17 N
Given
For F q 2 , Q : F x,q 1 ,Q = (F q 2 ,Q )(cos 37°) = (0.29 N)(cos 37°) = 0.23 N
v = 8.4 " 10 4 m/s,north
F y,q 2 ,Q = –(F q 2 ,Q )(sin 37°) = –(0.29 N)(sin 37°) = –0.17 N
q proton = 1.60 " 10 !19 C
F x,tot = F x,q 1 ,Q + F x,q 2 ,Q = 0.23 N + 0.23 N = 0.46 N
F y,tot = F y,q 1 ,Q + F y,q 2 ,Q = 0.17 N +(–0.17 N) = 0 N
ÊÁÁ F ˆ˜˜ 2 + ÊÁÁ F ˆ˜˜ 2 =
Ë x,tot ¯
Ë y,tot ¯
= 0.46 N
F Q,tot =
F Q,tot
(0.46 N) 2 + 0 = 0.46 N
F magnetic = 3.0 " 10 !18 N,east
Solution
ÊÁÁ
ˆ
!18
ÁË 3.0 " 10 N ˜˜˜¯
B=
= Ê
= 2.2 " 10 !4 T
ÁÁÁ 1.60 " 10 !19 C ˜ˆ˜˜ ÁÊÁÁ 8.4 " 10 4 m/s ˜ˆ˜˜
q proton v
Ë
¯Ë
¯
F magnetic
OBJ: 16-2.3
OBJ: 19-3.1
27
28
ID: A
140) ANS:
!1
7.2 " 10 N,downward
142) ANS:
2
0.53 m
Given
Given
B = 8.3 " 10
ID: A
!4
N = 650 turns
T
I = 18 A
%B = !0.935 T
l = 48 m
%t = 2.7 s
$ = 0.00°
Solution
Ê
ˆ
F magnetic = BIl = ÁÁÁ 8.3 " 10 !4 T ˜˜˜ (18 A) (48 m) = 7.2 " 10 !1 N,downward
Ë
¯
OBJ: 19-3.3
141) ANS:
!149 V
emf = 120 V
Solution
Rearrange Faraday’s law of magnetic induction and solve for A.
%' M
%ABcos$
Acos$%B
emf = !N
= !N
= !N
%t
%t
%t
A=
Given
N = 275 turns
emf%t
!Ncos$%B
=
(120 V) (2.7 s)
! (650 turns) (cos 0.00°) (!0.935 T)
OBJ: 20-1.3
A = 0.750 m 2
%B = +0.900 T
%t = 1.25 s
$ = 0.00°
Solution
Substitute values into Faraday’s law of magnetic induction.
%' M
%ABcos$
%B
emf = !N
= !N
= !NAcos$
%t
%t
%t
(0.900 T)
Ê
ˆ
= ! (275 turns) ÁÁÁ 0.750 m 2 ˜˜˜ (cos 0.00°)
= !149 V
Ë
¯
(1.25 s)
OBJ: 20-1.3
29
30
= 0.53 m 2
ID: A
143) ANS:
15 A
Given
N = 25 turns
A = 0.80 m 2
B = +1.5 T
%t = 1.0 s
$ = 0.00°
R = 2.0 &
Solution
Use Faraday’s law of magnetic induction to calculate emf.
%' M
%ABcos$
%B
emf = !N
= !N
= !NAcos$
%t
%t
%t
(+1.5 T)
Ê
ˆ
= ! (25 turns) ÁÁÁ 0.80 m 2 ˜˜˜ (cos 0.00°)
= 30 V
Ë
¯
(1.0 s)
Substitute the induced emf into the definition of resistance to determine the induced current in the coil.
I=
emf
R
=
(30 V)
(2.0 &)
= 15 A
OBJ: 20-1.3
31