Solutions Manual for Physics for
Scientists & Engineers with
Modern Physics 5th Global
Edition By Douglas C. Giancoli
(All Chapters 1-44, 100%
Original Verified, A+ Grade)
Chapter 42 Solutions Manual Not Available.
This is The Only Original and
Complete Solutions Manual for
5th Global Edition, All Other
Files in the Market are
Fake/Old/Wrong Edition.
,CHAPTER 1: Introduction, Measurement, Estimating
Responses to Questions
1. (a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general
public; not invariable (could change size with age, time of day, etc.); not indestructible.
(b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have
different size feet); not invariable (could change size with age, time of day, etc.); not
indestructible.
Neither of these options would make a good standard.
2. The distance in miles is given to one significant figure, and the distance in kilometers is given to five
significant figures! The value in kilometers indicates more precision than really exists or than is
meaningful. The last digit represents a distance on the same order of magnitude as a car’s length!
The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of
significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty.
3. The number of digits you present in your answer should represent the precision with which you
know a measurement; it says very little about the accuracy of the measurement. For example, if you
measure the length of a table to great precision, but with a measuring instrument that is not
calibrated correctly, you will not measure accurately. Accuracy is a measure of how close a
measurement is to the true value.
4. If you measure the length of an object and you report that it is “4,” you haven’t given enough
information for your answer to be useful. There is a large difference between an object that is 4
meters long and one that is 4 km long. Units are necessary to give meaning to a numerical answer.
5. You should report a result of 8.32 cm. Your measurement had three significant figures. When you
multiply by 2, you are really multiplying by the integer 2, which is an exact value. The number of
significant figures is determined by the measurement.
6. The correct number of significant figures is three: sin 30.0º = 0.500.
7. Useful assumptions include the population of the city, the fraction of people who own cars, the
average number of visits to a mechanic that each car makes in a year, the average number of weeks a
mechanic works in a year, and the average number of cars each mechanic can see in a week.
(a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of
these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a
year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be
seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a
week. The final estimate, then, is 800 car mechanics in San Francisco.
(b) Answers will vary. But following the same reasoning, the estimate is 1/1000 of the population.
,Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition Instructor Solutions Manual
Responses to MisConceptual Questions
1. (c) As stated in the text, scientific laws are descriptive – they are meant to describe how nature
behaves. Since our understanding of nature evolves, so do the laws of physics, when evidence
can convince the community of physicists. The laws of physics are not permanent, and are not
subject to political treaties. In fact, there have been major changes in the laws of physics since
1900 – particularly due to relativity and quantum mechanics. The laws of physics apply in
chemistry and other scientific fields, since those areas of study are based on physics. Finally,
the laws of physics are man-made, not a part of nature. They are our “best description” of
nature as we currently understand it. As stated in the text, “Laws are not lying there in nature,
waiting to be discovered.”
2. (e) The first product is 142.08 m, which is only accurate to the 10’s place, since 37 m/s has only
two significant figures. The second product is 74.73 m, which is only accurate to the 1’s place,
since 5.3 s has only two significant figures. Thus the sum of the two terms can only be accurate
to the 10’s place. 142.08 + 74.73 = 216.81, which to the 10’s place is 220 m.
3. (a) The total number of digits present does not determine the precision, as the leading zeros in (c)
and (d) are only place holders. Rewriting all the measurements in units of meters shows that (a)
implies a precision of 0.0001m, (b) and (c) both imply a precision of 0.001 m, and (d) implies a
precision of 0.01 m. Note that since the period is shown, the trailing zeros are significant. If all
the measurements are expressed in meters, (a) has 4 significant figures, (b) and (c) each have 3
significant figures, and (d) has 2 significant figures.
4. (b) The leading zeros are not significant. Rewriting this number in scientific notation, 7.8 10−3 ,
shows that it only has two significant digits.
5. (b) When you add or subtract numbers, the final answer should contain no more decimal places
than the number with the fewest decimal places. Since 25.2 has one decimal place, the answer
must be rounded to one decimal place, or to 26.6. Thus the answer has 3 significant figures.
6. (b) The word “accuracy” is often misused. If a student repeats a measurement multiple times and
obtains the same answer each time, it is often assumed to be accurate. In fact, students are
frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However,
systematic errors may cause each measurement to be inaccurate. A poorly working instrument
may also limit the accuracy of your measurement.
7. (a) Quoting the textbook, “precision” refers to the repeatability of the measurement using a given
instrument. Precision and accuracy are often confused. “Accuracy” is defined by answer (b).
8. (d) This addresses misconceptions about squared units and about which factor should be in the
numerator of the conversion. This error can be avoided by treating the units as algebraic
symbols that must be cancelled out.
9. (e) When making estimates, the estimator may frequently believe that their answers are more
significant than they actually are. This question helps the estimator realize what an order-of-
magnitude estimation is NOT supposed to accomplish.
10. (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system
of units.
,Chapter 1 Introduction, Measurement, Estimating
Solutions to Problems
1. (a) 777 3 significant figures
(b) 81.60 4 significant figures
(c) 7.03 3 significant figures
(d) 0.03 1 significant figure
(e) 0.0086 2 significant figures
(f) 6465 4 significant figures
(g) 8700 2 significant figures
2. (a) 5.859 = 5.859 100
(b) 21.8 = 2.18 101
(c) 0.0068 = 6.8 10−3
(d) 328.65 = 3.2865 102
(e) 0.219 = 2.19 10−1
(f) 444 = 4.44 102
3. (a) 8.69 105 = 869, 000
(b) 9.1103 = 9100
(c) 2.5 10−1 = 0.25
(d) 4.76 102 = 476
(e) 3.62 10−5 = 0.0000362
0.35 m
4. % uncertainty = 100% = 11%
3.25 m
0.2 s
5. (a) % uncertainty = 100% = 4.444% 4%
4.5s
0.2 s
(b) % uncertainty = 100% = 0.4444% 0.4%
45s
(c) The time of 4.5 minutes is 270 seconds.
0.2 s
% uncertainty = 100% = 0.0741% 0.07%
270 s
,Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition Instructor Solutions Manual
6. To add values with significant figures, adjust all values to be added so that their exponents are all
the same.
( 9.2 103 s ) + ( 6.3 104 s ) + ( 0.008 106 s ) = ( 9.2 103 s ) + ( 63 103 s ) + (8 103 s )
= ( 9.2 + 63 + 8 ) 103 s = 80.2 103 s = 8.0 104 s
When adding, keep the least accurate value, and so keep to the “ones” place in the last set of
parentheses.
7. When you multiply, the result should have as many digits as the number with the least number of
significant digits used in the calculation.
( 4.079 10 m )(0.057 10
2 −1
m ) = 2.325m2 2.3 m2
8. The “best value” uncertainty is taken to be 0.01 m.
0.01m 2
% uncertainty = 100% = 0.787% 0.8%
1.27 m 2
9. (radians) sin( ) tan( )
0 0.00 0.00 Keeping 2 significant figures in the angle, and
0.10 0.10 0.10 expressing the angle in radians, the largest angle that has
0.12 0.12 0.12 the same sine and tangent is 0.24 radians . In degrees,
0.20 0.20 0.20 the largest angle (keeping 2 significant figure) is 12 .
0.24 0.24 0.24
0.25 0.25 0.26
10. (a) To find the number of students, multiply the sample size times the percentage.
( 215 students )( 0.372 ) = 79.98 students
But the number of students must be an integer, so the number is 80 students .
(b) The original statement gave the percentage with too many significant figures. The calculation is
= 0.372093023… , but that result should only have 2 significant figures, since the value
of 80 (as an integer, and not an estimate) has only 2 significant figures. The percentage should
have been quoted as 37%. That would still give the correct number of students.
( 215 students )( 0.37 ) = 79.55 students, which rounds to 80 students.
11. We calculate what fraction 9 seconds is of an entire year.
9s 1yr −5 −5
1yr 3.156 107s 100% = 2.85 10 % 3 10 %
12. In order to find the approximate uncertainty in the area, calculate the area for the specified radius, the
minimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the area is
then half this variation in area. The uncertainty in the radius is assumed to be 0.1 104 cm .
( )
2
Aspecified = rspecified
2
= 5.1 104 cm = 8.17 109 cm 2
( ) = 7.85 10 cm
2
Amin = rmin
2
= 5.0 104 cm 9 2
= ( 5.2 10 cm ) = 8.49 10 cm
2
Amax = rmax
2 4 9 2
,Chapter 1 Introduction, Measurement, Estimating
A = 1
2 ( Amax − Amin ) = 12 (8.49 109 cm2 − 7.85 109 cm2 ) = 0.32 109 cm2
Note that the last value above has only 1 significant figure. Thus the area should be quoted as
A = (8.2 0.3) 109 cm2 .
13. In order to find the approximate uncertainty in the volume, calculate the volume for the minimum
radius and the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the
volume is then half of this variation in volume.
= 43 ( 0.64 m ) = 1.098 m3
3
Vspecified = 43 rspecified
3
= 43 ( 0.60 m ) = 0.905 m 3
3
Vmin = 43 rmin
3
= 43 ( 0.68 m ) = 1.317 m 3
3
Vmax = 43 rmax
3
V = 1
2 (Vmax − Vmin ) = 12 (1.317 m3 − 0.905 m3 ) = 0.206 m3
Note that the last value above has only 1 significant figure. Thus the percent uncertainty is
V 0.206 m 3
= 100 = 18.6 20 % .
Vspecified 1.098 m 3
14. (a) 286.6 mm 286.6 10−3 m 0.286 6 m
(b) 74 V 74 10−6 V 0.000 074 V
(c) 430 mg 430 10 −3 g 0.43g (if last zero is not significant)
(d) 47.2 ps 47.2 10−12 s 0.000 000 000 047 2 s
(e) 22.5 nm 22.5 10−9 m 0.000 000 022 5 m
(f) 2.50 gigavolts 2.5 109 volts 2, 500, 000, 000 volts
Note that in part (f) in particular, the correct number of significant digits cannot be determined when
you write the number in this format.
15. (a) 3 106 volts 3megavolt = 3 Mvolt
(b) 2 10−6 meters 2 micrometers = 2 m
(c) 5 103 days 5 kilodays = 5 kdays
(d) 18 102 bucks 18 hectobucks = 18 hbucks or 1.8 kilobucks
(e) −7
9 10 seconds 900 nanoseconds = 900 ns or 0.9 s
16. Assuming a height of 5 feet 10 inches, then 5'10" = ( 70 in )(1 m 39.37 in ) = 1.8 m . Assuming a
weight of 165 lbs, then (165 lbs )( 0.456 kg 1 lb ) = 75.2 kg . Technically, pounds and mass
measure two separate properties. To make this conversion, we have to assume that we are at a
location where the acceleration due to gravity is 9.80 m/s2.
Scientists & Engineers with
Modern Physics 5th Global
Edition By Douglas C. Giancoli
(All Chapters 1-44, 100%
Original Verified, A+ Grade)
Chapter 42 Solutions Manual Not Available.
This is The Only Original and
Complete Solutions Manual for
5th Global Edition, All Other
Files in the Market are
Fake/Old/Wrong Edition.
,CHAPTER 1: Introduction, Measurement, Estimating
Responses to Questions
1. (a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general
public; not invariable (could change size with age, time of day, etc.); not indestructible.
(b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have
different size feet); not invariable (could change size with age, time of day, etc.); not
indestructible.
Neither of these options would make a good standard.
2. The distance in miles is given to one significant figure, and the distance in kilometers is given to five
significant figures! The value in kilometers indicates more precision than really exists or than is
meaningful. The last digit represents a distance on the same order of magnitude as a car’s length!
The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of
significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty.
3. The number of digits you present in your answer should represent the precision with which you
know a measurement; it says very little about the accuracy of the measurement. For example, if you
measure the length of a table to great precision, but with a measuring instrument that is not
calibrated correctly, you will not measure accurately. Accuracy is a measure of how close a
measurement is to the true value.
4. If you measure the length of an object and you report that it is “4,” you haven’t given enough
information for your answer to be useful. There is a large difference between an object that is 4
meters long and one that is 4 km long. Units are necessary to give meaning to a numerical answer.
5. You should report a result of 8.32 cm. Your measurement had three significant figures. When you
multiply by 2, you are really multiplying by the integer 2, which is an exact value. The number of
significant figures is determined by the measurement.
6. The correct number of significant figures is three: sin 30.0º = 0.500.
7. Useful assumptions include the population of the city, the fraction of people who own cars, the
average number of visits to a mechanic that each car makes in a year, the average number of weeks a
mechanic works in a year, and the average number of cars each mechanic can see in a week.
(a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of
these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a
year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be
seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a
week. The final estimate, then, is 800 car mechanics in San Francisco.
(b) Answers will vary. But following the same reasoning, the estimate is 1/1000 of the population.
,Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition Instructor Solutions Manual
Responses to MisConceptual Questions
1. (c) As stated in the text, scientific laws are descriptive – they are meant to describe how nature
behaves. Since our understanding of nature evolves, so do the laws of physics, when evidence
can convince the community of physicists. The laws of physics are not permanent, and are not
subject to political treaties. In fact, there have been major changes in the laws of physics since
1900 – particularly due to relativity and quantum mechanics. The laws of physics apply in
chemistry and other scientific fields, since those areas of study are based on physics. Finally,
the laws of physics are man-made, not a part of nature. They are our “best description” of
nature as we currently understand it. As stated in the text, “Laws are not lying there in nature,
waiting to be discovered.”
2. (e) The first product is 142.08 m, which is only accurate to the 10’s place, since 37 m/s has only
two significant figures. The second product is 74.73 m, which is only accurate to the 1’s place,
since 5.3 s has only two significant figures. Thus the sum of the two terms can only be accurate
to the 10’s place. 142.08 + 74.73 = 216.81, which to the 10’s place is 220 m.
3. (a) The total number of digits present does not determine the precision, as the leading zeros in (c)
and (d) are only place holders. Rewriting all the measurements in units of meters shows that (a)
implies a precision of 0.0001m, (b) and (c) both imply a precision of 0.001 m, and (d) implies a
precision of 0.01 m. Note that since the period is shown, the trailing zeros are significant. If all
the measurements are expressed in meters, (a) has 4 significant figures, (b) and (c) each have 3
significant figures, and (d) has 2 significant figures.
4. (b) The leading zeros are not significant. Rewriting this number in scientific notation, 7.8 10−3 ,
shows that it only has two significant digits.
5. (b) When you add or subtract numbers, the final answer should contain no more decimal places
than the number with the fewest decimal places. Since 25.2 has one decimal place, the answer
must be rounded to one decimal place, or to 26.6. Thus the answer has 3 significant figures.
6. (b) The word “accuracy” is often misused. If a student repeats a measurement multiple times and
obtains the same answer each time, it is often assumed to be accurate. In fact, students are
frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However,
systematic errors may cause each measurement to be inaccurate. A poorly working instrument
may also limit the accuracy of your measurement.
7. (a) Quoting the textbook, “precision” refers to the repeatability of the measurement using a given
instrument. Precision and accuracy are often confused. “Accuracy” is defined by answer (b).
8. (d) This addresses misconceptions about squared units and about which factor should be in the
numerator of the conversion. This error can be avoided by treating the units as algebraic
symbols that must be cancelled out.
9. (e) When making estimates, the estimator may frequently believe that their answers are more
significant than they actually are. This question helps the estimator realize what an order-of-
magnitude estimation is NOT supposed to accomplish.
10. (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system
of units.
,Chapter 1 Introduction, Measurement, Estimating
Solutions to Problems
1. (a) 777 3 significant figures
(b) 81.60 4 significant figures
(c) 7.03 3 significant figures
(d) 0.03 1 significant figure
(e) 0.0086 2 significant figures
(f) 6465 4 significant figures
(g) 8700 2 significant figures
2. (a) 5.859 = 5.859 100
(b) 21.8 = 2.18 101
(c) 0.0068 = 6.8 10−3
(d) 328.65 = 3.2865 102
(e) 0.219 = 2.19 10−1
(f) 444 = 4.44 102
3. (a) 8.69 105 = 869, 000
(b) 9.1103 = 9100
(c) 2.5 10−1 = 0.25
(d) 4.76 102 = 476
(e) 3.62 10−5 = 0.0000362
0.35 m
4. % uncertainty = 100% = 11%
3.25 m
0.2 s
5. (a) % uncertainty = 100% = 4.444% 4%
4.5s
0.2 s
(b) % uncertainty = 100% = 0.4444% 0.4%
45s
(c) The time of 4.5 minutes is 270 seconds.
0.2 s
% uncertainty = 100% = 0.0741% 0.07%
270 s
,Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition Instructor Solutions Manual
6. To add values with significant figures, adjust all values to be added so that their exponents are all
the same.
( 9.2 103 s ) + ( 6.3 104 s ) + ( 0.008 106 s ) = ( 9.2 103 s ) + ( 63 103 s ) + (8 103 s )
= ( 9.2 + 63 + 8 ) 103 s = 80.2 103 s = 8.0 104 s
When adding, keep the least accurate value, and so keep to the “ones” place in the last set of
parentheses.
7. When you multiply, the result should have as many digits as the number with the least number of
significant digits used in the calculation.
( 4.079 10 m )(0.057 10
2 −1
m ) = 2.325m2 2.3 m2
8. The “best value” uncertainty is taken to be 0.01 m.
0.01m 2
% uncertainty = 100% = 0.787% 0.8%
1.27 m 2
9. (radians) sin( ) tan( )
0 0.00 0.00 Keeping 2 significant figures in the angle, and
0.10 0.10 0.10 expressing the angle in radians, the largest angle that has
0.12 0.12 0.12 the same sine and tangent is 0.24 radians . In degrees,
0.20 0.20 0.20 the largest angle (keeping 2 significant figure) is 12 .
0.24 0.24 0.24
0.25 0.25 0.26
10. (a) To find the number of students, multiply the sample size times the percentage.
( 215 students )( 0.372 ) = 79.98 students
But the number of students must be an integer, so the number is 80 students .
(b) The original statement gave the percentage with too many significant figures. The calculation is
= 0.372093023… , but that result should only have 2 significant figures, since the value
of 80 (as an integer, and not an estimate) has only 2 significant figures. The percentage should
have been quoted as 37%. That would still give the correct number of students.
( 215 students )( 0.37 ) = 79.55 students, which rounds to 80 students.
11. We calculate what fraction 9 seconds is of an entire year.
9s 1yr −5 −5
1yr 3.156 107s 100% = 2.85 10 % 3 10 %
12. In order to find the approximate uncertainty in the area, calculate the area for the specified radius, the
minimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the area is
then half this variation in area. The uncertainty in the radius is assumed to be 0.1 104 cm .
( )
2
Aspecified = rspecified
2
= 5.1 104 cm = 8.17 109 cm 2
( ) = 7.85 10 cm
2
Amin = rmin
2
= 5.0 104 cm 9 2
= ( 5.2 10 cm ) = 8.49 10 cm
2
Amax = rmax
2 4 9 2
,Chapter 1 Introduction, Measurement, Estimating
A = 1
2 ( Amax − Amin ) = 12 (8.49 109 cm2 − 7.85 109 cm2 ) = 0.32 109 cm2
Note that the last value above has only 1 significant figure. Thus the area should be quoted as
A = (8.2 0.3) 109 cm2 .
13. In order to find the approximate uncertainty in the volume, calculate the volume for the minimum
radius and the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the
volume is then half of this variation in volume.
= 43 ( 0.64 m ) = 1.098 m3
3
Vspecified = 43 rspecified
3
= 43 ( 0.60 m ) = 0.905 m 3
3
Vmin = 43 rmin
3
= 43 ( 0.68 m ) = 1.317 m 3
3
Vmax = 43 rmax
3
V = 1
2 (Vmax − Vmin ) = 12 (1.317 m3 − 0.905 m3 ) = 0.206 m3
Note that the last value above has only 1 significant figure. Thus the percent uncertainty is
V 0.206 m 3
= 100 = 18.6 20 % .
Vspecified 1.098 m 3
14. (a) 286.6 mm 286.6 10−3 m 0.286 6 m
(b) 74 V 74 10−6 V 0.000 074 V
(c) 430 mg 430 10 −3 g 0.43g (if last zero is not significant)
(d) 47.2 ps 47.2 10−12 s 0.000 000 000 047 2 s
(e) 22.5 nm 22.5 10−9 m 0.000 000 022 5 m
(f) 2.50 gigavolts 2.5 109 volts 2, 500, 000, 000 volts
Note that in part (f) in particular, the correct number of significant digits cannot be determined when
you write the number in this format.
15. (a) 3 106 volts 3megavolt = 3 Mvolt
(b) 2 10−6 meters 2 micrometers = 2 m
(c) 5 103 days 5 kilodays = 5 kdays
(d) 18 102 bucks 18 hectobucks = 18 hbucks or 1.8 kilobucks
(e) −7
9 10 seconds 900 nanoseconds = 900 ns or 0.9 s
16. Assuming a height of 5 feet 10 inches, then 5'10" = ( 70 in )(1 m 39.37 in ) = 1.8 m . Assuming a
weight of 165 lbs, then (165 lbs )( 0.456 kg 1 lb ) = 75.2 kg . Technically, pounds and mass
measure two separate properties. To make this conversion, we have to assume that we are at a
location where the acceleration due to gravity is 9.80 m/s2.
