Test Bank
for Fundamentals of
PHYSICS
6th Edition
By David Halliday, Robert Resnick, Jearl Walker
,Table of Contents
Chapter 1: Measurement
Chapter 2: Motion Along a Straight Line
Chapter 3: Vectors
Chapter 4: Motion in Two and Three Dimensions
Chapter 5: Force and Motion—I
Chapter 6: Force and Motion—II
Chapter 7: Kinetic Energy and Work
Chapter 8: Potential Energy and Conservation of Energy
Chapter 9: Systems of Particles
Chapter 10: Collisions
Chapter 11: Rotation
Chapter 12: Rolling, Torque, and Angular Momentum
Chapter 13: Equilibrium and Elasticity
Chapter 14: Gravitation
Chapter 15: Fluids
Chapter 16: Oscillations
Chapter 17: Waves—I
Chapter 18: Waves—II
Chapter 19: Temperature, Heat, and the First Law of Thermodynamics
Chapter 20: The Kinetic Theory of Gases
Chapter 21: Entropy and the Second Law of Thermodynamics
Chapter 22: Electric Charge
Chapter 23: Electric Fields
Chapter 24: Gauss' Law
Chapter 25: Electric Potential
Chapter 26: Capacitance
Chapter 27: Current and Resistance
Chapter 28: Circuits
Chapter 29: Magnetic Fields
Chapter 30: Magnetic Fields Due to Currents
Chapter 31: Induction and Inductance
Chapter 32: Magnetism of Matter: Maxwell's Equation
Chapter 33: Electromagnetic Oscillations and Alternating Current
Chapter 34: Electromagnetic Waves
Chapter 35: Images
Chapter 36: Interference
Chapter 37: Diffraction
Chapter 38: Special Theory of Relativity
Chapter 39: Photons and Matter Waves
Chapter 40: More About Matter Waves
Chapter 41: All About Atoms
Chapter 42: Conduction of Electricity in Solids
Chapter 43: Nuclear Physics
Chapter 44: Energy from the Nucleus
Chapter 45: Quarks, Leptons, and the Big Bang
,Chapter 1
1. The metric prefixes (micro, pico, nano, . . .) are given for ready reference on the inside front cover of the
textbook (see also Table 1-2).
(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 µm,
1 km = 103 m = (103 m)(106 µm/m) = 109 µm .
The given measurement is 1.0 km (two significant figures), which implies our result should be
written as 1.0 × 109 µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,
1 cm = 10−2 m = (10−2 m)(106 µm/m) = 104 µm .
We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0 × 10−4 .
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = (0.91 m)(106 µm/m) = 9.1 × 105 µm .
2. The customer expects 20 × 7056 in3 and receives 20 × 5826 in3 , the difference being 24600 cubic inches,
or
2.54 cm 3
3 1L
24600 in = 403 L
1 inch 1000 cm3
where Appendix D has been used (see also Sample Problem 1-2).
3. Using the given conversion factors, we find
(a) the distance d in rods to be
(4.0 furlongs)(201.168 m/furlong)
d = 4.0 furlongs = = 160 rods ,
5.0292 m/rod
(b) and that distance in chains to be
(4.0 furlongs)(201.168 m/furlong)
d= = 40 chains .
20.117 m/chain
4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain
1 inch 6 picas 12 points
(0.80 cm) ≈ 23 points ,
2.54 cm 1 inch 1 pica
(b) and
1 inch 6 picas
(0.80 cm) ≈ 1.9 picas .
2.54 cm 1 inch
1
, 2 CHAPTER 1.
5. Various geometric formulas are given in Appendix E.
(a) Substituting
R = 6.37 × 106 m 10−3 km/m = 6.37 × 103 km
into circumference = 2πR, we obtain 4.00 × 104 km.
(b) The surface area of Earth is
2
4πR2 = 4π 6.37 × 103 km = 5.10 × 108 km2 .
(c) The volume of Earth is
4π 3 4π 3
R = 6.37 × 103 km = 1.08 × 1012 km3 .
3 3
6. (a) Using the fact that the area A of a rectangle is width×length, we find
Atotal = (3.00 acre) + (25.0 perch)(4.00 perch)
(40 perch)(4 perch)
= (3.00 acre) + 100 perch2
1 acre
= 580 perch2 .
We multiply this by the perch2 → rood conversion factor (1 rood/40 perch2 ) to obtain the answer:
Atotal = 14.5 roods.
(b) We convert our intermediate result in part (a):
2
16.5 ft
Atotal = (580 perch2 ) = 1.58 × 105 ft2 .
1 perch
Now, we use the feet → meters conversion given in Appendix D to obtain
2
5 2 1m
= 1.47 × 104 m2 .
Atotal = 1.58 × 10 ft
3.281 ft
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The
semicircle area is A = πr2 /2, where r is the radius. Therefore, the volume is
π 2
V = r z
2
where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
3 2
10 m 10 cm
r = (2000 km) = 2000 × 105 cm .
1 km 1m
In these units, the thickness becomes
102 cm
z = (3000 m) = 3000 × 102 cm .
1m
Therefore,
π 2
2000 × 105 cm 3000 × 102 cm = 1.9 × 1022 cm3 .
V =
2
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area
A = 20 × 12 = 240 m2 ) in addition to a rectangular box (height h′ = 6.0 m and same base). Therefore,
1 ′ h
V = hA + h A = + h A = 1800 m3 .
′
2 2
for Fundamentals of
PHYSICS
6th Edition
By David Halliday, Robert Resnick, Jearl Walker
,Table of Contents
Chapter 1: Measurement
Chapter 2: Motion Along a Straight Line
Chapter 3: Vectors
Chapter 4: Motion in Two and Three Dimensions
Chapter 5: Force and Motion—I
Chapter 6: Force and Motion—II
Chapter 7: Kinetic Energy and Work
Chapter 8: Potential Energy and Conservation of Energy
Chapter 9: Systems of Particles
Chapter 10: Collisions
Chapter 11: Rotation
Chapter 12: Rolling, Torque, and Angular Momentum
Chapter 13: Equilibrium and Elasticity
Chapter 14: Gravitation
Chapter 15: Fluids
Chapter 16: Oscillations
Chapter 17: Waves—I
Chapter 18: Waves—II
Chapter 19: Temperature, Heat, and the First Law of Thermodynamics
Chapter 20: The Kinetic Theory of Gases
Chapter 21: Entropy and the Second Law of Thermodynamics
Chapter 22: Electric Charge
Chapter 23: Electric Fields
Chapter 24: Gauss' Law
Chapter 25: Electric Potential
Chapter 26: Capacitance
Chapter 27: Current and Resistance
Chapter 28: Circuits
Chapter 29: Magnetic Fields
Chapter 30: Magnetic Fields Due to Currents
Chapter 31: Induction and Inductance
Chapter 32: Magnetism of Matter: Maxwell's Equation
Chapter 33: Electromagnetic Oscillations and Alternating Current
Chapter 34: Electromagnetic Waves
Chapter 35: Images
Chapter 36: Interference
Chapter 37: Diffraction
Chapter 38: Special Theory of Relativity
Chapter 39: Photons and Matter Waves
Chapter 40: More About Matter Waves
Chapter 41: All About Atoms
Chapter 42: Conduction of Electricity in Solids
Chapter 43: Nuclear Physics
Chapter 44: Energy from the Nucleus
Chapter 45: Quarks, Leptons, and the Big Bang
,Chapter 1
1. The metric prefixes (micro, pico, nano, . . .) are given for ready reference on the inside front cover of the
textbook (see also Table 1-2).
(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 µm,
1 km = 103 m = (103 m)(106 µm/m) = 109 µm .
The given measurement is 1.0 km (two significant figures), which implies our result should be
written as 1.0 × 109 µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,
1 cm = 10−2 m = (10−2 m)(106 µm/m) = 104 µm .
We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0 × 10−4 .
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = (0.91 m)(106 µm/m) = 9.1 × 105 µm .
2. The customer expects 20 × 7056 in3 and receives 20 × 5826 in3 , the difference being 24600 cubic inches,
or
2.54 cm 3
3 1L
24600 in = 403 L
1 inch 1000 cm3
where Appendix D has been used (see also Sample Problem 1-2).
3. Using the given conversion factors, we find
(a) the distance d in rods to be
(4.0 furlongs)(201.168 m/furlong)
d = 4.0 furlongs = = 160 rods ,
5.0292 m/rod
(b) and that distance in chains to be
(4.0 furlongs)(201.168 m/furlong)
d= = 40 chains .
20.117 m/chain
4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain
1 inch 6 picas 12 points
(0.80 cm) ≈ 23 points ,
2.54 cm 1 inch 1 pica
(b) and
1 inch 6 picas
(0.80 cm) ≈ 1.9 picas .
2.54 cm 1 inch
1
, 2 CHAPTER 1.
5. Various geometric formulas are given in Appendix E.
(a) Substituting
R = 6.37 × 106 m 10−3 km/m = 6.37 × 103 km
into circumference = 2πR, we obtain 4.00 × 104 km.
(b) The surface area of Earth is
2
4πR2 = 4π 6.37 × 103 km = 5.10 × 108 km2 .
(c) The volume of Earth is
4π 3 4π 3
R = 6.37 × 103 km = 1.08 × 1012 km3 .
3 3
6. (a) Using the fact that the area A of a rectangle is width×length, we find
Atotal = (3.00 acre) + (25.0 perch)(4.00 perch)
(40 perch)(4 perch)
= (3.00 acre) + 100 perch2
1 acre
= 580 perch2 .
We multiply this by the perch2 → rood conversion factor (1 rood/40 perch2 ) to obtain the answer:
Atotal = 14.5 roods.
(b) We convert our intermediate result in part (a):
2
16.5 ft
Atotal = (580 perch2 ) = 1.58 × 105 ft2 .
1 perch
Now, we use the feet → meters conversion given in Appendix D to obtain
2
5 2 1m
= 1.47 × 104 m2 .
Atotal = 1.58 × 10 ft
3.281 ft
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The
semicircle area is A = πr2 /2, where r is the radius. Therefore, the volume is
π 2
V = r z
2
where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
3 2
10 m 10 cm
r = (2000 km) = 2000 × 105 cm .
1 km 1m
In these units, the thickness becomes
102 cm
z = (3000 m) = 3000 × 102 cm .
1m
Therefore,
π 2
2000 × 105 cm 3000 × 102 cm = 1.9 × 1022 cm3 .
V =
2
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area
A = 20 × 12 = 240 m2 ) in addition to a rectangular box (height h′ = 6.0 m and same base). Therefore,
1 ′ h
V = hA + h A = + h A = 1800 m3 .
′
2 2
