name: class: date: id: a - d2ct263enury6r.cloudfront.net · electric field lines can start on a...

30
Name: ________________________ Class: ___________________ Date: __________ ID: A 1 Physics & H-Physics Practice test Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. (2 points) The distance between wave fronts of plane waves corresponds to ____ of a sound wave. a. two amplitudes b. one wavelength c. two rarefactions d. one compression ____ 2. (2 points) A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the a. acceleration reach a maximum. b. velocity reach zero. c. acceleration reach zero. d. velocity reach a maximum. ____ 3. (2 points) Which is the most correct statement regarding the drawing of electric field lines? a. Electric field lines always form closed loops. b. Electric field lines can start on a charge of either polarity. c. Electric field lines always connect from one charge to another. d. Electric field lines never cross each other. ____ 4. (2 points) The quality of a musical tone of a certain pitch results from a combination of a. harmonics. b. fundamental frequencies. c. velocities. d. transverse waves. ____ 5. (2 points) Under which of the following conditions is the net magnetic force on a charged particle equal to zero? a. when the particle is moving parallel to the magnetic field b. when the particle is stationary c. when the particle is not charged d. all of the above ____ 6. (2 points) Two mechanical waves can occupy the same space at the same time because waves a. do not cause interference patterns. b. are matter. c. cannot pass through one another. d. are displacements of matter. ____ 7. (2 points) A generator’s maximum output is 220 V. What is the rms potential difference? a. 110 V b. 160 V c. 310 V d. 150 V ____ 8. (2 points) 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? a. I t = I 1 = I 2 = I 3 b. I t = 1 I 1 + 1 I 2 + 1 I 3 Ê Ë Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ c. I t = I 1 + I 2 + I 3 d. I t = 1 I 1 + 1 I 2 + 1 I 3 Ê Ë Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ -1 ____ 9. (2 points) Which of the following is not a characteristic of electrical potential energy? a. It results from a single charge. b. It is a form of mechanical energy. c. It is associated with a charge in an electric field. d. It results from the interaction between charges. ____ 10. (2 points) A potential difference of 115 V across the primary of a step-down transformer provides a potential difference of 2.3 V across the secondary. What is the ratio of the number of turns of wire on the primary to the number of turns on the secondary? a. 50:1 b. 1:7 c. 7:1 d. 1:50 ____ 11. (2 points) An electric current that changes directions at regular intervals is called a. reversible current. b. fluctuating current. c. directional current. d. alternating current. ____ 12. (2 points) Refraction is the bending of a wave disturbance as it passes at an angle from one ____ into another. a. medium b. boundary c. glass d. area

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Page 1: Name: Class: Date: ID: A - d2ct263enury6r.cloudfront.net · Electric field lines can start on a charge ... Consider two identical wave pulses on a rope ... and then meets the second

Name: ________________________ Class: ___________________ Date: __________ ID: A

1

Physics & H-Physics Practice test

Multiple Choice

Identify the choice that best completes the statement or answers the question.

____ 1. (2 points) The distance between wave fronts of plane waves corresponds to ____ of a sound wave.

a. two amplitudes b. one wavelength c. two rarefactions d. one compression

____ 2. (2 points) A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and

the

a. acceleration reach a maximum. b. velocity reach zero. c. acceleration reach zero. d. velocity reach a

maximum.

____ 3. (2 points) Which is the most correct statement regarding the drawing of electric field lines?

a. Electric field lines always form closed loops. b. Electric field lines can start on a charge of either polarity.

c. Electric field lines always connect from one charge to another. d. Electric field lines never cross each

other.

____ 4. (2 points) The quality of a musical tone of a certain pitch results from a combination of

a. harmonics. b. fundamental frequencies. c. velocities. d. transverse waves.

____ 5. (2 points) Under which of the following conditions is the net magnetic force on a charged particle equal to zero?

a. when the particle is moving parallel to the magnetic field b. when the particle is stationary c. when the

particle is not charged d. all of the above

____ 6. (2 points) Two mechanical waves can occupy the same space at the same time because waves

a. do not cause interference patterns. b. are matter. c. cannot pass through one another. d. are

displacements of matter.

____ 7. (2 points) A generator’s maximum output is 220 V. What is the rms potential difference?

a. 110 V b. 160 V c. 310 V d. 150 V

____ 8. (2 points) Three resistors connected in series carry currents labeled I1 , I2 , and I3 , respectively. Which of the

following expresses the total current, I t , in the system made up of the three resistors in series?

a. It = I1 = I2 = I3 b. It = 1

I1

+1

I2

+1

I3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜ c. It = I1 + I2 + I3 d. It =

1

I1

+1

I2

+1

I3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜

−1

____ 9. (2 points) Which of the following is not a characteristic of electrical potential energy?

a. It results from a single charge. b. It is a form of mechanical energy. c. It is associated with a charge in

an electric field. d. It results from the interaction between charges.

____ 10. (2 points) A potential difference of 115 V across the primary of a step-down transformer provides a potential

difference of 2.3 V across the secondary. What is the ratio of the number of turns of wire on the primary to the

number of turns on the secondary?

a. 50:1 b. 1:7 c. 7:1 d. 1:50

____ 11. (2 points) An electric current that changes directions at regular intervals is called

a. reversible current. b. fluctuating current. c. directional current. d. alternating current.

____ 12. (2 points) Refraction is the bending of a wave disturbance as it passes at an angle from one ____ into another.

a. medium b. boundary c. glass d. area

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Name: ________________________ ID: A

2

The figure above shows the pattern of a double-slit interference experiment. The center of the pattern is located

at E.

____ 13. (2 points) In the double-slit interference experiment that produced the figure above, the two slits are moved closer.

Which of the following would occur to the pattern shown in the figure?

a. F would shift to the right. b. E would shift to the right. c. F would shift to the left. d. E would shift to

the left.

____ 14. (2 points) In the figure above, which fringe represents a second-order minimum?

a. G b. F c. E d. H

____ 15. (2 points) Light with a wavelength of 546.1 nm passes through a 6.62 × 103 lines/cm diffraction grating. What is

the first-order angle of diffraction?

a. 39.2° b. 34.6° c. 41.6° d. 21.2°

____ 16. (2 points) The equivalent resistance of a complex circuit is usually determined by

a. adding and subtracting individual resistances. b. simplifying the circuit into groups of series and parallel

circuits. c. inspection. d. dividing the sum of the individual resistances by the number of resistances.

____ 17. (2 points) Light with a wavelength of 400.0 nm passes through a 1.00 × 104 lines/cm diffraction grating. What is

the second-order angle of diffraction?

a. 72.1° b. 56.5° c. 21.3° d. 53.1°

____ 18. (2 points) A capacitor consists of two metal plates; ____ is stored on one plate, and ____ is stored on the other.

a. potential energy, kinetic energy b. residual charge, induced charge c. potential difference, internal

resistance d. negative charge, positive charge

____ 19. (2 points) The direction of the force on a current-carrying wire in an external magnetic field is

a. perpendicular to the magnetic field only. b. perpendicular to the current only. c. perpendicular to both

the current and the magnetic field. d. parallel to the current and to the magnetic field.

____ 20. (2 points) One end of a taut rope is fixed to a post. What type of wave is produced if the free end is quickly raised

and lowered one time?

a. sine wave b. longitudinal wave c. periodic wave d. pulse wave

____ 21. (2 points) Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end

of the rope, is reflected back, and then meets the second pulse. When the two pulses overlap exactly, what will

be the amplitude of the resultant pulse?

a. zero b. double the amplitude of the original pulses c. same as the original pulses d. half the amplitude

of the original pulses

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3

____ 22. (2 points) Tripling the current in a circuit with constant resistance has the effect of changing the power by what

factor?

a. 1

9 b. 9 c.

1

3 d. 3

____ 23. (2 points) For a conductor that is in electrostatic equilibrium, any excess charge

a. resides entirely on the conductor’s outer surface. b. flows to the ground. c. resides entirely on the

conductor’s interior. d. resides entirely in the center of the conductor.

____ 24. (2 points) In the diagram above, use the superposition principle to find the resultant wave of waves X and Y.

a. d b. c c. b d. a

____ 25. (2 points) Which of the following wavelengths would produce standing waves on a string approximately 3.5 m

long?

a. 2.33 m b. 3.75 m c. 4.55 m d. 2.85 m

____ 26. (2 points) Waves arriving at a fixed boundary are

a. neither reflected nor inverted. b. inverted but not reflected. c. reflected but not inverted. d. reflected

and inverted.

____ 27. (2 points) A solenoid is in an upright position on a table. A clockwise current of electrons—when viewed from

above—causes the solenoid to have a ____ magnetic pole at its bottom end. If a compass is placed at the top of

the solenoid, the north pole of the compass would be ____.

a. north; attracted b. south; repelled c. north; repelled d. south; attracted

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4

____ 28. (2 points) Each compression in the waveform of the longitudinal wave shown above corresponds to what feature

of the transverse wave below it?

a. wavelength b. amplitude c. troughs d. crests

____ 29. (2 points) An electron that moves with a speed of 3.0 × 104

m/s perpendicular to a uniform magnetic field of 0.40

T experiences a force of what magnitude? (q e = 1.60 × 10−19

C)

a. 0 N b. 1.9 × 1015

N c. 4.8 × 1014

N d. 2.2 × 1024

N

____ 30. (2 points) Which of the following situations is not true for magnets?

a. A north pole and a south pole will attract each other. b. North poles repel each other. c. Like poles repel

each other. d. Unlike poles repel each other.

____ 31. (2 points) Vibration of an object about an equilibrium point is called simple harmonic motion when the restoring

force is proportional to

a. displacement. b. mass. c. a spring constant. d. time.

____ 32. (2 points) Which of the following is not true for both gravitational and electric forces?

a. Forces are proportional to physical properties. b. The inverse square distance law applies. c. Forces are

either attractive or repulsive. d. Potential energy is a function of distance of separation.

____ 33. (2 points) Which of the following types of interference will occur when the pulses in the figure above meet?

a. complete destructive interference b. no interference c. complete constructive interference d. partial

interference

____ 34. (2 points) Light with a wavelength of 500.0 nm passes through a 3.39 × 105 lines/m diffraction grating. What is

the first-order angle of diffraction?

a. 23.5° b. 9.73° c. 53.1° d. 36.9°

____ 35. (2 points) Which is not correct when describing the formation of rainbows?

a. A rainbow is really spherical in nature. b. All wavelengths refract at the same angle. c. Sunlight is

internally reflected on the back side of a raindrop. d. Sunlight is spread into a spectrum when it enters a

spherical raindrop.

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5

____ 36. (2 points) Waves arriving at a free boundary are

a. neither reflected nor inverted. b. reflected but not inverted. c. inverted but not reflected. d. reflected

and inverted.

____ 37. (2 points) A circuit is composed of resistors wired in series. What is the relationship between the equivalent

resistance of the circuit and the resistance of the individual resistors?

a. The equivalent resistance is equal to the largest resistance in the circuit. b. The equivalent resistance is

greater than the sum of all the resistances in the circuit. c. The equivalent resistance is less than the smallest

resistance in the circuit. d. The equivalent resistance is equal to the sum of the individual resistances.

____ 38. (2 points) When light passes at an angle to the normal from one material into another material in which its speed is

higher,

a. it is unaffected. b. it is bent away from the normal to the surface. c. it always lies along the normal to

the surface. d. it is bent toward the normal to the surface.

____ 39. (2 points) A stationary positive charge, Q, is located in a magnetic field, B, which is directed toward the right.

What is the direction of the magnetic force on Q?

a. toward the right b. down c. There is no magnetic force. d. up

____ 40. (2 points) In the figure shown above, a beat occurs at

a. t1 . b. t2 . c. t1 and t3 . d. t3 .

____ 41. (2 points) The perceived loudness of a sound is measured in

a. watts per square meter. b. decibels. c. hertz. d. watts.

____ 42. (2 points) If the charge is tripled for two identical charges maintained at a constant separation, the electric force

between them will be changed by what factor?

a. 1

9 b. 18 c. 9 d.

2

3

____ 43. (2 points) For a standing wave in an air column in a pipe that is open at both ends, there must be at least

a. two antinodes and one node. b. one node and one antinode. c. two nodes and one antinode. d. two

nodes and two antinodes.

____ 44. (2 points) A solenoid is in an upright position on a table. A counterclockwise current of electrons—when viewed

from above—causes the solenoid to have a ____ magnetic pole at its bottom end. If a compass is placed at the

top of the solenoid, the north pole of the compass would be ____.

a. south; repelled b. south; attracted c. north; attracted d. north; repelled

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6

____ 45. (2 points) Three resistors with values of R1 , R2 , and R3 are connected in series. Which of the following expresses

the total resistance, Req , of the three resistors?

a. Req = R1 + R2 + R3 b. Req = 1

R1

+1

R2

+1

R3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜ c. Req = R1 = R2 = R3 d. Req =

1

R1

+1

R2

+1

R3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜

−1

____ 46. (2 points) Electricity may be generated by rotating a loop of wire between the poles of a magnet. The induced

current is greatest when

a. the plane of the loop makes an angle of 45° with the magnetic field. b. the magnetic flux through the loop

is a minimum. c. the plane of the loop is parallel to the magnetic field. d. the plane of the loop is

perpendicular to the magnetic field.

____ 47. (2 points) In a double-slit interference experiment, a wave from one slit arrives at a point on a screen one

wavelength behind the wave from the other slit. What is observed at that point?

a. gray fringe, neither dark nor bright b. bright fringe c. multicolored fringe d. dark fringe

____ 48. (2 points) If the batteries in a portable CD player provide a terminal voltage of 12 V, what is the potential

difference across the entire CD player?

a. 4.0 V b. 6.0 V c. 3.0 V d. 12 V

____ 49. (2 points) According to the right-hand rule, if a current-carrying wire is grasped in the right hand with the thumb

in the direction of the current, the four fingers will curl in the direction of

a. the current’s velocity, v. b. the magnetic force, F magnetic. c. the current’s path, P. d. the magnetic

field, B.

____ 50. (2 points) A flashlight bulb with a potential difference of 4.5 V across it has a resistance of 8.0 Ω. How much

current is in the bulb filament?

a. 0.56 A b. 36 A c. 1.8 A d. 9.4 A

____ 51. (2 points) One useful way to model magnetic field strength is to define a quantity called magnetic flux Φ M . Which

of the following definitions for magnetic flux, Φ M , is correct?

a. AB cos θ b. (surface area) × (magnetic field component normal to the plane of surface) c. the number of

field lines that cross a certain area d. all of the above

____ 52. (2 points) If the potential difference across the bulb in a camping lantern is 9.0 V, what is the potential difference

across the battery used to power it?

a. 3.0 V b. 18 V c. 1.0 V d. 9.0 V

____ 53. (2 points) When a positive charge moves in the direction of the electric field, what happens to the electrical

potential energy associated with the charge?

a. It decreases. b. It sharply increases, and then decreases. c. It remains the same. d. It increases.

____ 54. (2 points) The acronym laser stands for light amplification by ____ emission of radiation.

a. spontaneous b. similar c. simultaneous d. stimulated

____ 55. (2 points) When two mechanical waves coincide, the amplitude of the resultant wave is always ____ the

amplitudes of each wave alone.

a. greater than b. less than c. the same as d. the sum of

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7

____ 56. (2 points) 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?

a. 37.3° b. 34.4° c. 18.0° d. 58.3°

____ 57. (2 points) An ac generator has a maximum output emf of 120 V. The generator is connected to a 125 Ω resistor.

What is the rms current through the resistor?

a. 1.73 A b. 0.68 A c. 2.43 A d. 1.47 A

____ 58. (2 points) What is the path of an electron moving perpendicular to a uniform magnetic field?

a. circle b. parabola c. ellipse d. straight line

____ 59. (2 points) A sound twice the intensity of the faintest audible sound is not perceived as twice as loud because the

sensation of loudness in human hearing

a. is proportional to frequency. b. depends on the speed of sound. c. is approximately logarithmic. d. is

approximately exponential.

____ 60. (2 points) A steam turbine at an electric power plant delivers 4500 kW of power to an electrical generator that

converts 95 percent of this mechanical energy into electrical energy. What is the current delivered by the

generator if it delivers energy at 3600 V?

a. 1.0 × 103 A b. 5.9 × 10

3 A c. 0.66 × 10

3 A d. 1.2 × 10

3 A

____ 61. (2 points) If a lamp has a resistance of 136 Ω when it operates at a power of 1.00 × 102 W, what is the potential

difference across the lamp?

a. 220 V b. 136 V c. 125 V d. 117 V

____ 62. (2 points) In a two-coil system, the mutual inductance depends on

a. neither the geometrical properties of the coils nor their orientation to each other. b. both the geometrical

properties of the coils and their orientation to each other. c. only the geometrical properties of the coils.

d. only the orientation of the coils to each other.

____ 63. (2 points) Two positive point charges are initially separated by a distance of 2 cm. If their separation is increased

to 6 cm, the resultant electrical potential energy is equal to what factor multiplied by the initial electrical

potential energy?

a. 9 b. 1

9 c. 3 d.

1

3

____ 64. (2 points) A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50

T experiences a force of what magnitude?

a. 0.30 N b. 0.0 N c. 0.60 N d. 0.15 N

____ 65. (2 points) Which statement describes Earth’s magnetic declination?

a. the angle between Earth’s magnetic field and Earth’s surface b. the angle between true north and north

indicated by a compass c. the tendency for Earth’s field to reverse itself d. Earth’s magnetic field strength

at the equator

____ 66. (2 points) The distance between two slits in a double-slit interference experiment is 2.9 × 10−6

m. The first-order

bright fringe is measured on a screen at an angle of 12° from the central maximum. What is the wavelength of

the light?

a. 6.0 × 102 nm b. 3.0 × 10

2 nm c. 1.2 × 10

2 nm d. 4.6 × 10

2 nm

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____ 67. (2 points) How many focal points and focal lengths do converging and diverging lenses have?

a. two, one b. one, two c. one, one d. two, two

____ 68. (2 points) What is rms (root-mean-square) current?

a. the value of alternating current that gives the same heating effect that the corresponding value of direct

current does b. an important measure of the current in an ac circuit c. the amount of direct current that

would dissipate the same energy in a resistor as is dissipated by the instantaneous alternating current over a

complete cycle d. all of the above

____ 69. (2 points) Which of the following statements about electromagnetic radiation is true?

a. It transfers energy to objects in the path of the electromagnetic waves. b. It can be converted to other

energy forms. c. It transports the energy of electromagnetic waves. d. all of the above

____ 70. (2 points) In what direction does a focal ray from an object proceed after passing through a converging lens?

a. The ray passes through the center of the lens. b. The ray passes through the focal point, F. c. The ray

intersects with the center of curvature, C. d. The ray exits the lens parallel to the principal axis.

____ 71. (2 points) 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

a. zero. b. 6.0 A. c. 1.5 A. d. 3.0 A.

____ 72. (2 points) According to Lenz’s law, the magnetic field of an induced current in a conductor will

a. heat the conductor. b. increase the potential difference. c. enhance the applied field. d. oppose a

change in the applied magnetic field.

____ 73. (2 points) An electron moves across Earth’s equator at a speed of 2.5 × 106 m/s and in a direction 35° north of

east. At this point, Earth’s magnetic field has a direction due north, is parallel to the surface, and has a value of

0.10 × 10−4

T. What is the magnitude of the force acting on the electron due to its interaction with Earth’s

magnetic field? (q e = 1.60 × 10−19 C)

a. 2.3 × 10−18

N b. 4.0 × 10−18

N c. 5.1 × 10−18

N d. 3.3 × 10−18

N

____ 74. (2 points) Both insulators and conductors can be charged by

a. contact. b. grounding. c. induction. d. polarization.

____ 75. (2 points) A microscopic magnetic region composed of a group of atoms whose magnetic fields are aligned in a

common direction is called a(n) ____. In most materials, when these groups are randomly distributed, the

substance will show ____ no magnetism.

a. cell; unusual b. pole; some c. domain; no d. ion; strong

____ 76. (2 points) A step-down transformer has 2500 turns on its primary and 5.0 × 101 turns on its secondary. If the

potential difference across the primary is 4850 V, what is the potential difference across the secondary?

a. 1.0 V b. 97 V c. 240 V d. 110 V

____ 77. (2 points) Two resistors with values of 6.0 Ω and 12 Ω are connected in parallel. This combination is connected in

series with a 4.0 Ω resistor. What is the equivalent resistance of this combination?

a. 0.50 Ω b. 8.0 Ω c. 2.0 Ω d. 22 Ω

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____ 78. (2 points) Two resistors having the same resistance value are wired in parallel. How does the equivalent resistance

compare to the resistance value of a single resistor?

a. The equivalent resistance is the same as a single resistor. b. The equivalent resistance is half the value of a

single resistor. c. The equivalent resistance is twice the value of a single resistor. d. The equivalent

resistance is greater than that of a single resistor.

____ 79. (2 points) Pitch depends on the ____ of a sound wave.

a. frequency b. power c. speed d. amplitude

____ 80. (2 points) A coil with a wire that is wound around a 2.0 m2 hollow tube 35 times. A uniform magnetic field is

applied perpendicular to the plane of the coil. If the field changes uniformly from 0.00 T to 0.55 T in 0.85 s,

what is the induced emf in the coil?

a. 33 V b. –33 V c. 45 V d. –45 V

____ 81. (2 points) Three resistors connected in series have potential differences across them labeled ∆V1 , ∆V2 , and ∆V3 .

Which of the following expresses the potential difference taken over the three resistors together?

a. ∆Vt = 1

∆V1

+1

∆V2

+1

∆V3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜ b. ∆Vt =

1

∆V1

+1

∆V2

+1

∆V3

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜

−1

c. ∆Vt = ∆V1 + ∆V2 + ∆V3 d. ∆Vt = ∆V1 =

∆V2 = ∆V3

____ 82. (2 points) A surface charge can be produced on insulators by

a. grounding. b. polarization. c. induction. d. contact.

____ 83. (2 points) In general, sound travels faster through

a. empty space than through matter. b. gases than through solids. c. solids than through gases. d. gases

than through liquids.

____ 84. (2 points) Three resistors with values of 3.0 Ω, 6.0 Ω, and 12 Ω are connected in parallel. What is the equivalent

resistance of this combination?

a. 9.0 Ω b. 0.26 Ω c. 21 Ω d. 1.7 Ω

____ 85. (2 points) An ac generator has a maximum emf output of 150 V. What is the rms current in the circuit when the

generator is connected to a 35 Ω resistor?

a. 2.6 A b. 1.5 A c. 3.1 A d. 1.2 A

____ 86. (2 points) Increasing the potential difference between the plates of a capacitor will produce what effect on the

capacitor?

a. It will increase the charge on each plate. b. It will increase the capacitance. c. It will decrease the

capacitance. d. It will decrease the charge on each plate.

____ 87. (2 points) Which of the following statements about a battery as a source of electric current is not true?

a. A battery is a source of emf. b. The terminal voltage of a battery is the voltage it delivers to the load.

c. A battery provides the energy that moves charge. d. The terminal voltage of a battery is equal to its emf.

____ 88. (2 points) Which of the following is the region of a sound wave in which the density and pressure are greater than

normal?

a. wavelength b. rarefaction c. amplitude d. compression

____ 89. (2 points) Which set of information will allow you to calculate the kilowatt•hr usage?

a. the current and the time the circuit operates b. the resistance, the current, and the time the circuit operates

c. the voltage and current in the circuit d. the voltage and the resistance of the circuit

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____ 90. (2 points) 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

a. C

2. b. 4C. c. 2C. d.

C

4.

____ 91. (2 points) To produce a sustained interference pattern by light waves from multiple sources, which condition or

conditions must be met?

a. Sources must be monochromatic. b. Sources must be coherent and monochromatic. c. Sources must be

coherent. d. Sources must be neither coherent nor monochromatic.

____ 92. (2 points) All of the following statements about magnetic field lines around a permanent magnet are true except

which one?

a. In a permanent magnet, the field lines actually continue within the magnet itself. b. Magnetic field lines

have no beginning or end. c. Magnetic field lines appear to end at the north pole of a magnet. d. Magnetic

field lines always form a closed loop.

____ 93. (2 points) Two beams of coherent light are shining on the same sheet of white paper. When referring to the crests

and troughs of such waves, where will darkness appear on the paper?

a. where the crest from one wave overlaps the trough from the other b. where the troughs from both waves

overlap c. Darkness cannot occur because the two waves are coherent. d. where the crest from one wave

overlaps the crest from the other

____ 94. (2 points) Of the following materials, sound waves travel fastest through

a. copper at 0°C. b. air at 100°C. c. air at 0°C. d. helium at 0°C.

____ 95. (2 points) When electrons move through a metal conductor,

a. they move in zigzag patterns because of repeated collisions with the vibrating metal atoms. b. they move

in a straight line through the conductor. c. they move at the speed of light in a vacuum. d. the temperature

of the conductor decreases.

____ 96. (2 points) When a light ray passes from water (n = 1.333) into diamond (n = 2.419) at an angle of 45°, its path is

a. bent away from the normal. b. parallel to the normal. c. not bent. d. bent toward the normal.

____ 97. (2 points) Which of the following wavelengths would not produce standing waves on a rope whose length is 1 m?

a. 2 1/4 m b. 2/3 m c. 1 m d. 2 m

____ 98. (2 points) A generator supplies an rms current of 1.66 A. If the resistance of the circuit is 66.0 Ω, what is the

maximum emf?

a. 38.7 V b. 156 V c. 77.5 V d. 125 V

____ 99. (2 points) When comparing the net charge of a charged capacitor with the net charge of the same capacitor when it

is uncharged, the net charge is

a. greater or less in the charged capacitor, but never equal. b. equal in both capacitors. c. greater in the

charged capacitor. d. less in the charged capacitor.

____ 100. (2 points) If atmospheric refraction did not occur, how would the apparent time of sunrise and sunset be changed?

a. Both would be earlier. b. Sunrise would be earlier, and sunset would be later. c. Sunrise would be later,

and sunset would be earlier. d. Both would be later.

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Problem

101. (5 points) A uniform electric field with a magnitude of 5.0 × 102 N/C is directed parallel to the positive x-axis.

What is the change in electrical energy of a proton (q = 1.60 × 10−19

C) as it moves from x = 5.0 m to x = 2.2

m?

102. (5 points) What is the current in the 2 Ω resistor in the circuit shown in the figure above?

103. (5 points) Two equal charges are separated by 3.7 × 10−10

m. The force between the charges has a magnitude of

2.37 × 10−3

N. What is the magnitude of q on the charges? (kC = 8.99 × 109 N•m

2/C

2 )

104. (5 points) A generator supplies an rms current of 9.50 A. If the resistance of the circuit is 24.0 Ω, what is the

maximum emf?

105. (5 points) Two charges are located on the positive x-axis of a coordinate system. Charge q1 = 2.00 × 10−9

C, and it

is 0.020 m from the origin. Charge q2 = –3.00 × 10−9

C, and it is 0.040 m from the origin. What is the electric

force exerted by these two charges on a third charge, q3 = 5.00 × 10−9

, located at the origin? (kC = 8.99 × 109

N•m2/C

2)

106. (5 points) A coil with 275 turns and a cross-sectional area of 0.750 m2 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?

107. (5 points) A uniform electric field with a magnitude of 5.0 × 102 N/C is directed parallel to the positive x-axis. If

the potential at x = 5.0 m is 2.5 × 103 V, what is the potential at x = 2.0 m?

108. (5 points) A 3.2 µF capacitor has a potential difference of 21.0 V between its plates. How much additional charge

flows into the capacitor if the potential difference is increased to 47.0 V?

109. (5 points) An inverted image 1.8 cm in height is formed in back of a lens. The object used to form the image is

0.45 cm in height. What is the magnification of the image? Is the image real or virtual?

110. (5 points) When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4 A. What

is the resistance of the resistor?

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111. (5 points) The distance between two slits in a double-slit experiment is 4.2 × 10−6

m. The first-order bright fringe

is measured on a screen at an angle of 8.0° from the central maximum. What is the wavelength of the light?

112. (5 points) Three resistors with values of 9.0 Ω, 6.0 Ω, 12 Ω, respectively, are connected in parallel. What is their

equivalent resistance?

113. (5 points) An object is placed along the principal axis of a thin converging lens that has a focal length of 16 cm. If

the distance from the object to the lens is 24 cm, what is the distance from the image to the lens?

114. (5 points) A 4.7 × 108 Ω resistor has a current of 1.27 × 10

−6 A flowing in it. What is the potential difference

across the resistor?

115. (5 points) 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?

116. (5 points) 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?

117. (5 points) A 2.2 kΩ resistor has 0.042 A of current in it. What is the potential difference across the resistor?

118. (5 points) When an electron (q = –1.6 × 10−19

C) moves 0.10 m along the direction of an electric field with a

strength of 3.0 N/C, what is the magnitude of the potential difference between the electron’s initial and final

points?

119. (5 points) The potential difference across a resistor is 4.4 × 102 V. If the resistance of the resistor is 1.8 kΩ, how

much current flows through the resistor?

120. (5 points) An amusement park ride has a frequency of 0.05 Hz. What is the ride’s period?

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Physics & H-Physics Practice test

Answer Section

MULTIPLE CHOICE

1. ANS: B PTS: 2 DIF: II OBJ: 12-1.4

2. ANS: A PTS: 2 DIF: I OBJ: 11-1.2

3. ANS: D PTS: 2 DIF: II OBJ: 16-3.2

4. ANS: A PTS: 2 DIF: I OBJ: 12-3.3

5. ANS: D PTS: 2 DIF: II OBJ: 19-3.1

6. ANS: D PTS: 2 DIF: I OBJ: 11-4.1

7. ANS: B

Given

∆Vmax = 220 V

Solution

∆Vrms = 0.707∆Vmax = 0.707( ) 220 V( ) = 160 V

PTS: 2 DIF: IIIA OBJ: 20-3.2

8. ANS: A PTS: 2 DIF: II OBJ: 18-2.1

9. ANS: A PTS: 2 DIF: I OBJ: 17-1.1

10. ANS: A

Given

∆V1 = 115 V

∆V2

= 2.3 V

Solution

N1 :N2 =∆V1

∆V2

=115 V( )

2.3 V( )= 50:1

PTS: 2 DIF: IIIA OBJ: 20-3.3

11. ANS: D PTS: 2 DIF: I OBJ: 20-2.1

12. ANS: A PTS: 2 DIF: I OBJ: 14-1.1

13. ANS: A PTS: 2 DIF: II OBJ: 15-1.3

14. ANS: D PTS: 2 DIF: I OBJ: 15-1.2

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15. ANS: D

Given

d =1

6.62 × 103 lines

cm

=1

6.62 × 103

cm1 m

100 cm

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ =

1

6.62 × 105

m

m = 1

λ = 546.1 nm = 5.461 × 10−7

m

Solution

d sinθ = mλ

θ = sin−1 mλd

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ = sin−1

(5.461 × 10−7

m)

1

6.62 × 105

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜

Ê

Ë

ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜˜˜˜˜˜

= 21.2°

PTS: 2 DIF: IIIB OBJ: 15-2.2

16. ANS: B PTS: 2 DIF: I OBJ: 18-3.1

17. ANS: D

Given

d =1

1.0 × 104 lines

cm

=1

1.0 × 104

cm =1

1.0 × 104

cm1 m

100 cm

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ =

1

1.0 × 106

m

m = 2

λ = 400.0 nm = 4.000 × 10−7

m

Solution

d sinθ = mλ

θ = sin−1 mλd

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ = sin−1

(2)(4.000 × 10−7

m)

1

1.00 × 10−6

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜

Ê

Ë

ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜˜˜˜˜˜

= 53.1°

PTS: 2 DIF: IIIB OBJ: 15-2.2

18. ANS: D PTS: 2 DIF: I OBJ: 17-2.1

19. ANS: C PTS: 2 DIF: I OBJ: 19-3.2

20. ANS: D PTS: 2 DIF: I OBJ: 11-3.2

21. ANS: A PTS: 2 DIF: IIIA OBJ: 11-4.3

22. ANS: B PTS: 2 DIF: II OBJ: 17-4.3

23. ANS: A PTS: 2 DIF: I OBJ: 16-3.3

24. ANS: D PTS: 2 DIF: II OBJ: 11-4.1

25. ANS: A PTS: 2 DIF: IIIC OBJ: 11-4.4

26. ANS: D PTS: 2 DIF: I OBJ: 11-4.3

27. ANS: A PTS: 2 DIF: II OBJ: 19-2.1

28. ANS: D PTS: 2 DIF: I OBJ: 11-3.3

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29. ANS: B

Given

v = 3.0 × 104

m/s

qelectron

= 1.60 × 10−19

C

B = 0.40 T

Solution

Rearrange equation, B =F

magnetic

qv, and solve for Fmagnetic .

Fmagnetic = q electron vB = 1.60 × 10−19

CÊËÁÁÁ

ˆ¯˜ 3.0 × 10

4m/s

ÊËÁÁÁ

ˆ¯˜ 0.40 T( ) = 1.9 × 10

−15N

PTS: 2 DIF: IIIB OBJ: 19-3.1

30. ANS: D PTS: 2 DIF: I OBJ: 19-1.1

31. ANS: A PTS: 2 DIF: II OBJ: 11-1.1

32. ANS: C PTS: 2 DIF: I OBJ: 16-2.2

33. ANS: A PTS: 2 DIF: I OBJ: 11-4.2

34. ANS: B

Given

d =1

3.39 × 105 lines

m

=1

3.39 × 105

m

m = 1

λ = 500.0 nm = 5.000 × 10−7

m

Solution

d sinθ = mλ

θ = sin−1 mλd

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ = sin−1

(5.000 × 10−7

m)

1

3.39 × 105

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜

Ê

Ë

ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜˜˜˜˜˜

= 9.73°

PTS: 2 DIF: IIIB OBJ: 15-2.2

35. ANS: B PTS: 2 DIF: I OBJ: 14-3.3

36. ANS: B PTS: 2 DIF: I OBJ: 11-4.3

37. ANS: D PTS: 2 DIF: II OBJ: 18-2.1

38. ANS: B PTS: 2 DIF: I OBJ: 14-1.2

39. ANS: C PTS: 2 DIF: I OBJ: 19-3.3

40. ANS: A PTS: 2 DIF: II OBJ: 12.3.-4

41. ANS: B PTS: 2 DIF: I OBJ: 12-2.2

42. ANS: C PTS: 2 DIF: II OBJ: 16-2.2

43. ANS: A PTS: 2 DIF: II OBJ: 12-3.1

44. ANS: A PTS: 2 DIF: II OBJ: 19-2.1

45. ANS: A PTS: 2 DIF: II OBJ: 18-2.1

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46. ANS: C PTS: 2 DIF: I OBJ: 20-1.2

47. ANS: B PTS: 2 DIF: I OBJ: 15-1.1

48. ANS: D PTS: 2 DIF: II OBJ: 18-1.3

49. ANS: D PTS: 2 DIF: I OBJ: 19-2.2

50. ANS: A

Given

∆V = 4.5 V

R = 8.0 Ω

Solution

∆V = IR

Rearrange to solve for I.

I =∆V

R=

4.5 V

8.0 Ω= 0.56 A

PTS: 2 DIF: IIIA OBJ: 17-3.3

51. ANS: D PTS: 2 DIF: I OBJ: 19-1.2

52. ANS: D PTS: 2 DIF: I OBJ: 18-1.3

53. ANS: A PTS: 2 DIF: I OBJ: 17-1.1

54. ANS: D PTS: 2 DIF: I OBJ: 15-3.1

55. ANS: D PTS: 2 DIF: I OBJ: 11-4.2

56. ANS: C

Given

θ i = 30.0°

ni

= 1.00

nr

= 1.65

Solution

Rearrange Snell’s law, n i sinθ i = n r sinθ r , and solve for θ r .

θ r = sin−1

ni

n r

sinθ iÊËÁÁ ˆ

¯˜

È

Î

ÍÍÍÍÍÍÍÍÍÍ

˘

˚

˙˙˙˙˙

= sin−1 1.00

1.65sin 3.0 × 10

1°ÊËÁÁÁ

ˆ¯˜

È

Î

ÍÍÍÍÍÍÍÍ

˘

˚

˙˙˙˙

= 18.0°

PTS: 2 DIF: IIIA OBJ: 14-1.3

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57. ANS: B

Given

∆Vmax = 120 V

R = 125 Ω

Solution

∆Vrms

= 0.707∆Vmax

= 0.707( ) 120 V( ) = 85 V

Rearrange the equation, ∆Vrms

= Irms

R, to solve for Irms

.

I rms =∆V

rms

R=

85 V( )

125 Ω( )= 0.68 A

PTS: 2 DIF: IIIB OBJ: 20-3.2

58. ANS: A PTS: 2 DIF: I OBJ: 19-3.2

59. ANS: C PTS: 2 DIF: I OBJ: 12-2.2

60. ANS: D

Given

Pmechanical = 4500 kW

efficiencymechanical to electrical = 95%

∆V = 3600 V

Solution

Pelectrical = 0.95(Pmechanical ) = 0.95(4500 kW) = 4300 kW

Pelectrical = (4300 kW)1.0 × 10

3W

1 kW

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜

= 4.3 × 106

W

P = I∆V

Rearange to solve for I.

I =P

∆V=

4.3 × 106

W

3.6 × 103

V= 1.2 × 103 A

PTS: 2 DIF: IIIB OBJ: 17-4.3

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61. ANS: D

Given

R = 136 Ω

P = 1.00 × 102 W

Solution

P =(∆V)

2

R

Rearrange to solve for ∆V.

∆V = PR = (1.00 × 102

W)(136 Ω) = 117 V

PTS: 2 DIF: IIIB OBJ: 17-4.3

62. ANS: B PTS: 2 DIF: I OBJ: 20-2.3

63. ANS: D PTS: 2 DIF: II OBJ: 17-1.2

64. ANS: B PTS: 2 DIF: IIIB OBJ: 19-3.3

65. ANS: B PTS: 2 DIF: I OBJ: 19-1.3

66. ANS: B

Given

d = 2.9 × 10−6

m

m = 1

θ = 12°

Solution

d sinθ = mλ

λ =d sinθ

m=

(2.9 × 10−6

m)(sin12°)

2= 3.0 × 10

−7m = 3.0 × 10

2nm

PTS: 2 DIF: IIIA OBJ: 15-1.3

67. ANS: A PTS: 2 DIF: I OBJ: 14-2.1

68. ANS: D PTS: 2 DIF: I OBJ: 20-3.1

69. ANS: D PTS: 2 DIF: I OBJ: 20-4.4

70. ANS: D PTS: 2 DIF: I OBJ: 14-2.1

71. ANS: A PTS: 2 DIF: II OBJ: 20-2.3

72. ANS: D PTS: 2 DIF: I OBJ: 20-1.3

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73. ANS: D

Given

v = 2.5 × 106

m/s

qelectron

= 1.60 × 10−19 C

B = 0.10 × 10−4

T

θ = 35° north of east

Solution

Since only the magnetic field component that is perpendicular to the electron’s motion contributes to the

magnetic field strength, Bnet = Bcosθ.

Therefore, Bnet

= 0.10 × 10−4

TÊËÁÁÁ

ˆ¯˜ cos 35°( ) = 8.2 × 10

−6T.

Rearrange equation, B =Fmagnetic

qv, and solve for Fmagnetic.

Fmagnetic

= qelectron

vB = 1.60 × 10−19 CÊËÁÁÁ

ˆ¯˜ 2.5 × 106 m/sÊËÁÁÁ

ˆ¯˜ 8.2 × 10−6 TÊËÁÁÁ

ˆ¯˜ = 3.3 × 10−18 N

PTS: 2 DIF: IIIC OBJ: 19-3.1

74. ANS: A PTS: 2 DIF: I OBJ: 16-1.3

75. ANS: C PTS: 2 DIF: I OBJ: 19-1.2

76. ANS: B

Given

∆V1 = 4850 V

N1

= 2500 turns

N2

= 5.0 × 101

turns

Solution

∆V2

= ∆V1

N2

N1

= 4850 V( )5.0 × 10

1turns

2500 turns

Ê

Ë

ÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜˜

= 97 V

PTS: 2 DIF: IIIA OBJ: 20-3.3

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77. ANS: B

Given

R1 = 6.0 Ω

R2 = 12 Ω

R3 = 4.0 Ω

Solution

1

R1,2

=1

R1

+1

R2

=1

6.0 Ω+

1

12 Ω

1

R1,2

=2.0

12 Ω+

1.0

12 Ω=

3.0

12 Ω

R1,2 =12 Ω

3.0= 4.0 Ω

R1,2,3 = R1,2 + R3 = 4.0 Ω + 4.0 Ω = 8.0 Ω

PTS: 2 DIF: IIIA OBJ: 18-3.1

78. ANS: B PTS: 2 DIF: II OBJ: 18-2.2

79. ANS: A PTS: 2 DIF: I OBJ: 12-1.2

80. ANS: D

Given

N = 35 turns

A = 2.0 m2

B i = 0.00 T

Bf

= 0.55 T

∆t = 0.85 s

θ = 0.00°

Solution

Substitute values into Faraday’s law of magnetic induction.

emf = −N∆Φ

M

∆t= −N

∆ABcosθ∆t

= −NAcosθ∆B

∆t= −NAcosθ

Bf

− Bi

ÊËÁÁ ˆ

¯˜

∆t

= − 35 turns( ) 2.0 m2Ê

ËÁÁÁ

ˆ¯˜ cos 0.00°( )

0.55 T − 0.00 T( )

0.85 s( )

= −45 V

PTS: 2 DIF: IIIB OBJ: 20-1.3

81. ANS: C PTS: 2 DIF: II OBJ: 18-2.1

82. ANS: B PTS: 2 DIF: I OBJ: 16-1.3

83. ANS: C PTS: 2 DIF: I OBJ: 12-1.3

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84. ANS: D

Given

R1 = 3.0 Ω

R2 = 6.0 Ω

R3 = 12 Ω

Solution

1

Req

=1

R1

+1

R2

+1

R3

=1

3.0 Ω+

1

6.0 Ω+

1

12 Ω

1

Req

=4.0

12 Ω+

2.0

12 Ω+

1.0

12 Ω=

7.0

12 Ω

Req =12 Ω

7.0= 1.7 Ω

PTS: 2 DIF: IIIA OBJ: 18-2.2

85. ANS: C

Given

∆Vmax = 150 V

R = 35 Ω

Solution

∆Vrms = 0.707∆Vmax = 0.707( ) 150 V( ) = 110 V

Rearrange the equation, ∆Vrms = I rms R, to solve for I rms .

I rms =∆V

rms

R=

110 V( )

35 Ω( )= 3.1 A

PTS: 2 DIF: IIIB OBJ: 20-3.2

86. ANS: A PTS: 2 DIF: I OBJ: 17-2.1

87. ANS: D PTS: 2 DIF: II OBJ: 18-1.3

88. ANS: D PTS: 2 DIF: I OBJ: 12-1.1

89. ANS: B PTS: 2 DIF: II OBJ: 17-4.2

90. ANS: B PTS: 2 DIF: I OBJ: 17-2.2

91. ANS: B PTS: 2 DIF: I OBJ: 15-1.2

92. ANS: C PTS: 2 DIF: I OBJ: 19-1.2

93. ANS: A PTS: 2 DIF: I OBJ: 15-1.2

94. ANS: A PTS: 2 DIF: II OBJ: 12-1.3

95. ANS: A PTS: 2 DIF: I OBJ: 17-3.2

96. ANS: D PTS: 2 DIF: II OBJ: 14-1.2

97. ANS: A PTS: 2 DIF: I OBJ: 11-4.4

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98. ANS: B

Given

I rms = 1.66 A

R = 66.0 Ω

Solution

∆Vrms

= Irms

R = 1.66 A( ) 66.0 Ω( ) = 110 V

Rearrange the equation, ∆Vrms

= 0.707∆Vmax

, to solve for ∆Vmax

.

∆Vmax =∆V

rms

0.707=

110 V( )

0.707( )= 156 V

PTS: 2 DIF: IIIB OBJ: 20-3.2

99. ANS: B PTS: 2 DIF: II OBJ: 17-2.1

100. ANS: C PTS: 2 DIF: II OBJ: 14-3.2

PROBLEM

101. ANS:

2.0 × 10−16

J

Given

E = 5.0 × 102 N/C

q = 1.60 × 10−19

C

x1 = 5.0 m

x2 = 2.2 m

Solution

d = x 2 − x 1 = (2.2 m) − (5.0 m) = −2.8 m

∆PEelectric = −qEd

∆PEelectric

= −(1.60 × 10−19 C)(5.0 × 102 N/C)(−2.8 m) = 2.2 × 10−16 J

PTS: 5 DIF: IIIB OBJ: 17-1.2

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102. ANS:

4 A

Given

R1 = 2 Ω

R2 = 8 Ω

V1 = 8 V

V2 = 16 V

Solution

The voltage across R1 must be equal to 8 V, because R1 is connected directly across the 8 V battery.

IR1

=V1

R1

=8 V

2 Ω= 4 A

PTS: 5 DIF: IIIB OBJ: 18-3.2

103. ANS:

1.9 × 10−16

C

Given

q1 = q2

Felectric = 2.37×10−3

N

r = 3.7 × 10−10

m

k C = 8.99 × 109 N•m

2/C

2

Solution

Felectric = k C

q1q

2

r2

=k

Cq

2

r2

q = Felectricr

2

kC

= (2.37 × 10

−3N)(3.7 × 10

−10m)

2

8.99 × 109

Nm2C

2

q = (2.37 × 10

−3N)(1.4 × 10

−19m

2)

8.99 × 109

Nm2/C

2

q = 1.9×10−16

C

PTS: 5 DIF: IIIB OBJ: 16-2.1

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104. ANS:

322 V

Given

I rms = 9.50 A

R = 24.0 Ω

Solution

∆Vrms = I rms R = 9.50 A( ) 24.0 Ω( ) = 228 V

Rearrange the equation, ∆Vrms = 0.707∆V max, to solve for ∆V

max.

∆Vmax=∆Vrms

0.707=

228 V( )

0.707( )= 322 V

PTS: 5 DIF: IIIB OBJ: 20-3.2

105. ANS:

1.4 × 10−4 N

Given

q1 = 2.00 ×10−9

C

q2 = –3.00 ×10−9

C

q3 = 5.00 ×10−9

C

r3,1 = 0.020 m = 2.0 ×10−2

m

r3,2 = 0.040 m = 4.0 ×10−2

m

kC = 8.99 ×109

Nm2/C

2

Solution

F3,1 = k C

q 3 q 1

r3,1

ÊËÁÁ ˆ

¯˜

2 = 8.99 × 10

9Nm

2/C

2ÊËÁÁÁ

ˆ¯˜

5.00 × 10−9 CÊËÁÁÁ

ˆ¯˜ 2.00 × 10−9 CÊËÁÁÁ

ˆ¯˜

2.0 × 10−2

mÊËÁÁÁ

ˆ¯˜

2= 2.2 × 10

−4N

F3,2 = k C

q 3 q 2

r3,2

ÊËÁÁ ˆ

¯˜

2 = 8.99 × 10

9Nm

2/C

2ÊËÁÁÁ

ˆ¯˜

5.00 × 10−9 CÊËÁÁÁ

ˆ¯˜ −3.00 × 10−9 CÊËÁÁÁ

ˆ¯˜

4.0 × 10−2

mÊËÁÁÁ

ˆ¯˜

2= −8.4 × 10

−5N

F3 = 1.4 × 10−4

N

PTS: 5 DIF: IIIC OBJ: 16-2.3

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106. ANS:

−149 V

Given

N = 275 turns

A = 0.750 m2

∆B = +0.900 T

∆t = 1.25 s

θ = 0.00°

SolutionSubstitute values into Faraday’s law of magnetic induction.

emf = −N∆Φ M

∆t= −N

∆ABcosθ∆t

= −NAcosθ∆B

∆t

= − 275 turns( ) 0.750 m2ÊËÁÁÁ

ˆ¯˜ cos 0.00°( )

0.900 T( )

1.25 s( )= −149 V

PTS: 5 DIF: IIIB OBJ: 20-1.3

107. ANS:

4.0 × 103 V

Given

E = 5.0 × 102 N/C

V5.0 m = 2.5 × 103 V

x initial = 5.0 m

x final = 2.0 m

Solution

d = (x initial ) − (x final ) = (5.0 m) − (2.0 m) = −3.0 m

∆V = −Ed = −E(x5.0m

− x2.0m

) = −(5.0 × 102 N/C)(5.0 m − 2.0 m)

∆V = −1.5 × 103

V

∆V = V5.0m

− V2.0m

V2.0 m = V5.0 m − ∆V = (2.5 × 103

V) − ( − 1.5 × 103

V) = 4.0 × 103

V

PTS: 5 DIF: IIIC OBJ: 17-1.2

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108. ANS:

8 × 10−5

C

Given

C = 3.2 × 10−6

F

∆V1 = 21.0 V

∆V2 = 47.0 V

Solution

Q1

= C∆V1

= (3.2 × 10−6

F)(21.0 V) = 6.7 × 10−5

C

Q2 = C∆V2 = (3.2 × 10−6

F)(47.0 V) = 1.5 × 10−4

C

Q = Q 2 − Q 1 = (1.5 × 10−4

C) − (6.7 × 10−5

C)

Q = 8 × 10−5

C

PTS: 5 DIF: IIIC OBJ: 17-2.2

109. ANS:

−4.0Since M is negative, the image is real.

Given

h ′ = −1.8 cm

h = 0.45 cm

Solution

Use the magnification of a lens equation, M =h ′

h, to find M.

M =h ′

h=

−1.8 cm( )

0.45 cm( )= −4.0

Since M is negative, the image is real.

PTS: 5 DIF: IIIA OBJ: 14-2.3

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110. ANS:

9.3 Ω

Given

∆V = 13 V

I = 1.4 A

Solution

∆V = I × R

Rearrange to solve for R.

R =∆V

I=

13 V

1.4 A= 9.3 Ω

PTS: 5 DIF: IIIA OBJ: 17-3.3

111. ANS:

580 nm

Given

d = 4.2 × 10−6

m

m = 1

θ = 8.0°

Solution

d sinθ = mλ

λ =d sinθ

m=

(4.2 × 10−6

m)(sin8.0°)

1= 5.8 × 10

−7m = 5.8 × 10

2nm

PTS: 5 DIF: IIIA OBJ: 15-1.3

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112. ANS:

2.8 Ω

Given

R1 = 9.0 Ω

R2 = 6.0 Ω

R3 = 12 Ω

Solution

1

Req

=1

R1

+1

R2

+1

R3

=1

9.0 Ω+

1

6.0 Ω+

1

12 Ω

1

Req

=0.11

1 Ω+

0.17

1 Ω+

0.083

1 Ω=

0.36

1 Ω

Req =1 Ω

0.36= 2.8 Ω

PTS: 5 DIF: IIIA OBJ: 18-2.2

113. ANS:

48 cm

Given

p = 24 cm

f = 16 cm (f is positive, since this is a converging lens)

Solution

Rearrange the thin-lens equation,1

p+

1

q=

1

f, and solve for q.

1

q=

1

f−

1

p=

1

16 cm−

1

24 cm=

3

48 cm−

2

48 cm=

1

48 cm

q = 48 cm (since q is positive, the image is real and in back of the lens)

PTS: 5 DIF: IIIB OBJ: 14-2.2

114. ANS:

6.0 × 102 V

Given

R = 4.7 × 108 Ω

I = 1.27 × 10−6

A

Solution

∆V = I × R = (1.27 × 10−6

A)(4.7 × 108 Ω) = 6.0 × 10

2V

PTS: 5 DIF: IIIA OBJ: 17-3.3

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115. ANS:

θ = 10.91°

Given

λ = 632.8 nm = 6.328 × 10−7

m

d =1

1.496 38 × 105 lines

m

=1

1.496 38 × 105

m

m = 2

Solution

d sinθ = mλ

θ = sin−1 mλd

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜ = sin−1

(2)(6.328 × 10−7

m)

1

1.496 38 × 105

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜˜˜

Ê

Ë

ÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜˜˜˜˜˜

= 10.91°

PTS: 5 DIF: IIIB OBJ: 15-2.2

116. ANS:

9.3 × 10−3

W/m2

Given

P = 0.30 W

r = 1.6 m

Solution

Intensity = P

4πr2

Intensity =0.30 W

4π(1.6 m)2

= 9.3 × 10−3

W/m2

PTS: 5 DIF: IIIB OBJ: 12-2.1

117. ANS:

92 V

Given

R = 2.2 kΩ = 2.2 × 103 Ω

I = 0.042 A

Solution

∆V = I × R = (0.042 A)(2.2 × 103 Ω) = 92 V

PTS: 5 DIF: IIIA OBJ: 17-3.3

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118. ANS:

0.30 V

Given

q = –1.6 × 10−19

C

d = 0.10 m

E = 3.0 N/C

Solution

∆V = −Ed = −(3.0 N/C)(0.10 m) = −0.30 V

Because the magnitude is requested, the sign can be ignored, so

∆V = 0.30 V.

PTS: 5 DIF: IIIC OBJ: 17-1.2

119. ANS:

0.24 A

Given

R = 1.8 kΩ

∆V = 4.4 × 102 V

Solution

R = 1.8 kΩ( )1 × 103 Ω

1 kΩ

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜˜˜˜

= 1.8 × 103 Ω

∆V = I × R

Rearrange to solve for I.

I =∆V

R=

4.4 × 102

V

1.8 × 103 Ω

= 0.24 A

PTS: 5 DIF: IIIA OBJ: 17-3.3

120. ANS:

20 s

Given

f = 0.05 Hz

Solution

T =1

f=

1

0.05 Hz

T = 20s

PTS: 5 DIF: IIIA OBJ: 11-2.3