Instructor’s Manual
For
The Physical Universe
Eighteenth Edition
Konrad Krauskopf
Arthur Beiser
Elizabeth Shay Carter
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Notes on Math in
The Physical Universe, Eighteenth Edition
Physics and chemistry are quantitative sciences and some math is essential to understand and appreciate
them. How much math is appropriate for a given classroom is for the instructor to decide. The math level
of The Physical Universe is quite low, but it may still be too high in places for some students. Here is a
list of material that can be omitted without loss of continuity. In most cases the instructor may well
choose to outline the material in class without requiring the students to work out relevant problems.
2.2. Vectors. Pythagorean theorem. Can omit.
2.4. Distance, Time, and Acceleration. Although can be omitted, it is useful in providing the origin of
Eq. 2-14 and especially in deriving the KE formula in Sec. 3.3. But calculations need not be
required.
2.12. Circular Motion. Discuss, but need not require calculations.
5.2. Heat. Discuss Eq. 5-3, but need not require calculations.
5.7. The Gas Laws. Discuss, but need not require calculations.
6.3. Coulomb’s Law. Discuss, but need not require calculations.
6.19. Transformers. Discuss, but need not require calculations.
7.12. Refraction. Can omit calculations that involve index of refraction.
7.13. Lenses. Can omit ray tracing, but really quite easy.
7.17. Diffraction. Discuss resolving power, but need not require calculations.
8.7. Binding Energy. Discuss, but need not require calculations.
9.2. Photons. Keep quantum energy calculations, but can omit those that involve the photoelectric
effect.
10.17. Chemical Equations. Discuss, but need not require equation balancing.
12.3. The Mole. 12.4 Formula Units. Could omit entirely, but better to give at least brief discussion since
this is how chemical ideas meet the real world.
Chapter 13. Organic Chemistry. Discuss the various aspects of structural formulas, but need not require
solving problems that involve them.
14.2. Atmospheric Moisture. Discuss relative humidity, but need not require calculations using the graph.
Answers to Even-Numbered Exercises 1
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Answers to Even-Numbered
Exercises
14. The Copernican model, because in it the
CHAPTER ONE distances from the earth, and hence the
apparent brightnesses of the other planets,
Exercises: vary with time.
2. The reliance of the scientific method on 16. Only elliptical orbits agree with observa-
experiment and observation. tional data.
18. a. No explanation is possible in the
4. Because a model isolates the most important
ptolemaic system, in which the stars are
features of a complex phenomenon, it
fixed at the same distance from the earth in
may permit scientists to determine the
a crystal ball that revolves around a
fundamental origin of the phenomenon
stationary earth.
with out being confused by secondary
b. In the copernican system the explanation
details.
follows from the orbital motion of the earth
relative to stars at different distances away.
6. A year is the time the sun takes to complete
a circuit across the sky relative to the stars. 20. The earth would then be more flattened at
the poles and bulge to a greater extent at
the equator.
8. If the moon is seen near a particular star on
one evening, by the next evening it will be 22. Yesterday, because the length of the day
some distance east of that star. has been increasing steadily since the
earth’s formation.
10. a. A year does not correspond to a whole
number of days. In order that the seasons 24. The moon.
do not shift around the calendar, an extra
day must be added to every fourth year 26. (291 km)(0.621 mi/km) = 181 mi
with further adjustments at longer intervals.
b. A year does not correspond to a whole 28. 1 mm = 10−3 m so d = (104 )(10−3 m) = 10 m
number of days. In order that the seasons and (10 m)(3.28 ft/m) = 32.8 ft
do not shift around the calendar, an extra
day must be added to every fourth year
30. (20.0 m)(7.00 m)(2.00 m)(3.28 ft/m ) =
3
with further adjustments at longer intervals.
9.88 103 ft 3
12. These observations suggest that the mem-
bers of the solar system all lie in or near a 32. 2 m 35 s = (155 s)/(3600 s/h ) = 0.0431 h;
plane not far from the earth’s equator and 1 mi = 1.61 km, speed = (1.61 km)/
(0.0431 h ) = 37.4 km/h
that all move in the same direction about
the sun or, in the case of the moon, about
the earth. 34. 42; 7.5 105 ; 3.0 105
2 Answers to Even-Numbered Exercises
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, 28. v = 2 gh = 15.3 m/s.
CHAPTER TWO
30. h = 12 gt 2 = 78.4 m.
Exercises:
32. Time of rise = time of fall = t = v /g = 1.0 s.
2. t = d /v = 0.029 s.
Hence the total time of flight = 2t = 2.0 s.
4. The snake covers 105 m in t = d /v = 70 s.
Therefore your speed must be greater than 34. a. t = v /g = 2.04 s.
v = (100 m )/( 70 s ) = 1.43 m/s . c. h = 12 gt 2 = 20.4 m.
6. 30 lb; 0. 36. t = d /v, h = 12 gt 2 = 12 gd 2 /v 2 = 0.10 m =
10 cm.
8. a. Directly across the river.
b. t = d /v(boat ) = 0.1875 h = 11 min 15 s . 38. v ( vert ) = 2 gh = 19.8 m/s; v ( horiz ) =
c. d = v ( river ) t = 0.94 km .
30 m/s; v = v(vert) 2 + v(horiz) 2 =
35.9 m/s.
10. F = F12 + F22 = 7.1 tons .
40. The time of fall is t = 2h /g = 10.1 s . In
12. No. An example is the curved path of a ball
this time the pump will have moved
thrown at an angle with the ground.
horizontally d = vt = 606 m.
14. v2 – at = 15 m/s.
42. a. The tensions are the same.
b. The front coupling is under greater
16. a. a = (v f − v0 ) / t = −3.5 m/s2 . tension because it is accelerating a greater
b. t = (v f − v0 ) / a = 5.71 s. mass.
c. t = (v f − v0 ) / a = 2.86 s.
44. a. v = 55.6 m/s, a = v /t = 18.5 m/s 2 =
1.89 g.
18. a. d = v1t + 12 at , a = 2d /t – 2v1 /t =
2 2
b. F = ma = 222 kN.
–0.178 m/s 2 .
b. v2 = v1 + at = 6.65 m/s. 46. m = F /a = 4 kg.
a. F = ma = 4 N.
20. Yes. b. F = ma = 40 N.
22. T = total time of flight = 2 2h /g . Since 48. F = ma = mv f /t , t = mv f /F = 0.015 s.
g is smaller on Venus than on the earth, T
will be greater and the ball will return to 50. v2 − v1 = −4.17 m/s, F = ma = m ( v2 − v1 ) /t =
the ground later.
−667 N.
24. a. The crate appears to move vertically
downward because both the crate and the 52. F = w = mg , a = F /m = g.
observer have the same horizontal speed.
b. The crate appears to move in a curved 54. a = F /m = (m1 – m2 ) g / (m1 + m2 ) = 1.96
path downward, as in Fig. 2-12. m/s 2 .
26. a. The distance remains the same. 56. A force of ma is needed in addition to the
b. The distance increases. force w = mg needed just to lift the box
Answers to Even-Numbered Exercises 3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
For
The Physical Universe
Eighteenth Edition
Konrad Krauskopf
Arthur Beiser
Elizabeth Shay Carter
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Notes on Math in
The Physical Universe, Eighteenth Edition
Physics and chemistry are quantitative sciences and some math is essential to understand and appreciate
them. How much math is appropriate for a given classroom is for the instructor to decide. The math level
of The Physical Universe is quite low, but it may still be too high in places for some students. Here is a
list of material that can be omitted without loss of continuity. In most cases the instructor may well
choose to outline the material in class without requiring the students to work out relevant problems.
2.2. Vectors. Pythagorean theorem. Can omit.
2.4. Distance, Time, and Acceleration. Although can be omitted, it is useful in providing the origin of
Eq. 2-14 and especially in deriving the KE formula in Sec. 3.3. But calculations need not be
required.
2.12. Circular Motion. Discuss, but need not require calculations.
5.2. Heat. Discuss Eq. 5-3, but need not require calculations.
5.7. The Gas Laws. Discuss, but need not require calculations.
6.3. Coulomb’s Law. Discuss, but need not require calculations.
6.19. Transformers. Discuss, but need not require calculations.
7.12. Refraction. Can omit calculations that involve index of refraction.
7.13. Lenses. Can omit ray tracing, but really quite easy.
7.17. Diffraction. Discuss resolving power, but need not require calculations.
8.7. Binding Energy. Discuss, but need not require calculations.
9.2. Photons. Keep quantum energy calculations, but can omit those that involve the photoelectric
effect.
10.17. Chemical Equations. Discuss, but need not require equation balancing.
12.3. The Mole. 12.4 Formula Units. Could omit entirely, but better to give at least brief discussion since
this is how chemical ideas meet the real world.
Chapter 13. Organic Chemistry. Discuss the various aspects of structural formulas, but need not require
solving problems that involve them.
14.2. Atmospheric Moisture. Discuss relative humidity, but need not require calculations using the graph.
Answers to Even-Numbered Exercises 1
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Answers to Even-Numbered
Exercises
14. The Copernican model, because in it the
CHAPTER ONE distances from the earth, and hence the
apparent brightnesses of the other planets,
Exercises: vary with time.
2. The reliance of the scientific method on 16. Only elliptical orbits agree with observa-
experiment and observation. tional data.
18. a. No explanation is possible in the
4. Because a model isolates the most important
ptolemaic system, in which the stars are
features of a complex phenomenon, it
fixed at the same distance from the earth in
may permit scientists to determine the
a crystal ball that revolves around a
fundamental origin of the phenomenon
stationary earth.
with out being confused by secondary
b. In the copernican system the explanation
details.
follows from the orbital motion of the earth
relative to stars at different distances away.
6. A year is the time the sun takes to complete
a circuit across the sky relative to the stars. 20. The earth would then be more flattened at
the poles and bulge to a greater extent at
the equator.
8. If the moon is seen near a particular star on
one evening, by the next evening it will be 22. Yesterday, because the length of the day
some distance east of that star. has been increasing steadily since the
earth’s formation.
10. a. A year does not correspond to a whole
number of days. In order that the seasons 24. The moon.
do not shift around the calendar, an extra
day must be added to every fourth year 26. (291 km)(0.621 mi/km) = 181 mi
with further adjustments at longer intervals.
b. A year does not correspond to a whole 28. 1 mm = 10−3 m so d = (104 )(10−3 m) = 10 m
number of days. In order that the seasons and (10 m)(3.28 ft/m) = 32.8 ft
do not shift around the calendar, an extra
day must be added to every fourth year
30. (20.0 m)(7.00 m)(2.00 m)(3.28 ft/m ) =
3
with further adjustments at longer intervals.
9.88 103 ft 3
12. These observations suggest that the mem-
bers of the solar system all lie in or near a 32. 2 m 35 s = (155 s)/(3600 s/h ) = 0.0431 h;
plane not far from the earth’s equator and 1 mi = 1.61 km, speed = (1.61 km)/
(0.0431 h ) = 37.4 km/h
that all move in the same direction about
the sun or, in the case of the moon, about
the earth. 34. 42; 7.5 105 ; 3.0 105
2 Answers to Even-Numbered Exercises
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, 28. v = 2 gh = 15.3 m/s.
CHAPTER TWO
30. h = 12 gt 2 = 78.4 m.
Exercises:
32. Time of rise = time of fall = t = v /g = 1.0 s.
2. t = d /v = 0.029 s.
Hence the total time of flight = 2t = 2.0 s.
4. The snake covers 105 m in t = d /v = 70 s.
Therefore your speed must be greater than 34. a. t = v /g = 2.04 s.
v = (100 m )/( 70 s ) = 1.43 m/s . c. h = 12 gt 2 = 20.4 m.
6. 30 lb; 0. 36. t = d /v, h = 12 gt 2 = 12 gd 2 /v 2 = 0.10 m =
10 cm.
8. a. Directly across the river.
b. t = d /v(boat ) = 0.1875 h = 11 min 15 s . 38. v ( vert ) = 2 gh = 19.8 m/s; v ( horiz ) =
c. d = v ( river ) t = 0.94 km .
30 m/s; v = v(vert) 2 + v(horiz) 2 =
35.9 m/s.
10. F = F12 + F22 = 7.1 tons .
40. The time of fall is t = 2h /g = 10.1 s . In
12. No. An example is the curved path of a ball
this time the pump will have moved
thrown at an angle with the ground.
horizontally d = vt = 606 m.
14. v2 – at = 15 m/s.
42. a. The tensions are the same.
b. The front coupling is under greater
16. a. a = (v f − v0 ) / t = −3.5 m/s2 . tension because it is accelerating a greater
b. t = (v f − v0 ) / a = 5.71 s. mass.
c. t = (v f − v0 ) / a = 2.86 s.
44. a. v = 55.6 m/s, a = v /t = 18.5 m/s 2 =
1.89 g.
18. a. d = v1t + 12 at , a = 2d /t – 2v1 /t =
2 2
b. F = ma = 222 kN.
–0.178 m/s 2 .
b. v2 = v1 + at = 6.65 m/s. 46. m = F /a = 4 kg.
a. F = ma = 4 N.
20. Yes. b. F = ma = 40 N.
22. T = total time of flight = 2 2h /g . Since 48. F = ma = mv f /t , t = mv f /F = 0.015 s.
g is smaller on Venus than on the earth, T
will be greater and the ball will return to 50. v2 − v1 = −4.17 m/s, F = ma = m ( v2 − v1 ) /t =
the ground later.
−667 N.
24. a. The crate appears to move vertically
downward because both the crate and the 52. F = w = mg , a = F /m = g.
observer have the same horizontal speed.
b. The crate appears to move in a curved 54. a = F /m = (m1 – m2 ) g / (m1 + m2 ) = 1.96
path downward, as in Fig. 2-12. m/s 2 .
26. a. The distance remains the same. 56. A force of ma is needed in addition to the
b. The distance increases. force w = mg needed just to lift the box
Answers to Even-Numbered Exercises 3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
