CHAPTER 1: Introduction, Measurement, Estimating
Answers to Questions
1. (a) Fundamental standards should be accessible, invariable, indestructible, and reproducible. A
particular person’s foot would not be very accessible, since the person could not be at more than
one place at a time. The standard would be somewhat invariable if the person were an adult, but
even then, due to swelling or injury, the length of the standard foot could change. The standard
would not be indestructible – the foot would not last forever. The standard could be
reproducible – tracings or plaster casts could be made as secondary standards.
(b) If any person’s foot were to be used as a standard, “standard” would vary significantly
depending on the person whose foot happened to be used most recently for a measurement. The
standard would be very accessible, because wherever a measurement was needed, it would be
very easy to find someone with feet. The standard would be extremely variable – perhaps by a
factor of 2. That also renders the standard as not reproducible, because there could be many
reproductions that were quite different from each other. The standard would be almost
indestructible in that there is essentially a limitless supply of feet to be used.
2. There are various ways to alter the signs. The number of meters could be expressed in one
significant figure, as “900 m (3000 ft)”. Or, the number of feet could be expressed with the same
precision as the number of meters, as “914 m (2999 ft)”. The signs could also be moved to different
locations, where the number of meters was more exact. For example, if a sign was placed where the
elevation was really 1000 m to the nearest meter, then the sign could read “1000 m (3280 ft)”.
3. Including more digits in an answer does not necessarily increase its accuracy. The accuracy of an
answer is determined by the accuracy of the physical measurement on which the answer is based. If
you draw a circle, measure its diameter to be 168 mm and its circumference to be 527 mm, their
quotient, representing , is 3.136904762. The last seven digits are meaningless – they imply a
greater accuracy than is possible with the measurements.
4. The problem is that the precision of the two measurements are quite different. It would be more
appropriate to give the metric distance as 11 km, so that the numbers are given to about the same
precision (nearest mile or nearest km).
5. A measurement must be measured against a scale, and the units provide that scale. Units must be
specified or the answer is meaningless – the answer could mean a variety of quantities, and could be
interpreted in a variety of ways. Some units are understood, such as when you ask someone how old
they are. You assume their answer is in years. But if you ask someone how long it will be until they
are done with their task, and they answer “five”, does that mean five minutes or five hours or five
days? If you are in an international airport, and you ask the price of some object, what does the
answer “ten” mean? Ten dollars, or ten pounds, or ten marks, or ten euros?
6. If the jar is rectangular, for example, you could count the number of marbles along each dimension,
and then multiply those three numbers together for an estimate of the total number of marbles. If the
jar is cylindrical, you could count the marbles in one cross section, and then multiply by the number
of layers of marbles. Another approach would be to estimate the volume of one marble. If we
assume that the marbles are stacked such that their centers are all on vertical and horizontal lines,
then each marble would require a cube of edge 2R, or a volume of 8R3, where R is the radius of a
marble. The number of marbles would then be the volume of the container divided by 8R3.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
1
,Chapter 1 Introduction, Measurement, Estimating
7. The result should be written as 8.32 cm. The factor of 2 used to convert radius to diameter is exact –
it has no uncertainty, and so does not change the number of significant figures.
8. sin 30.0o 0.500
9. Since the size of large eggs can vary by 10%, the random large egg used in a recipe has a size with
an uncertainty of about 5% . Thus the amount of the other ingredients can also vary by about 5%
and not adversely affect the recipe.
10. In estimating the number of car mechanics, the assumptions and estimates needed are:
the population of the city
the number of cars per person in the city
the number of cars that a mechanic can repair in a day
the number of days that a mechanic works in a year
the number of times that a car is taken to a mechanic, per year
We estimate that there is 1 car for every 2 people, that a mechanic can repair 3 cars per day, that a
mechanic works 250 days a year, and that a car needs to be repaired twice per year.
(a) For San Francisco, we estimate the population at one million people. The number of mechanics
is found by the following calculation.
repairs
2
1 car year 1 yr 1 mechanic
1 106 people 1300 mechanics
2 people 1 car 250 workdays repairs
3
workday
(b) For Upland, Indiana, the population is about 4000. The number of mechanics is found by a
similar calculation, and would be 5 mechanics . There are actually two repair shops in Upland,
employing a total of 6 mechanics.
Solutions to Problems
1. (a) 14 billion years 1.4 1010 years
(b) 1.4 1010 y 3.156 107 s 1 y 4.4 1017 s
2. (a) 214 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) 3236 4 significant figures
(g) 8700 2 significant figures
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
2
,Giancoli Physics: Principles with Applications, 6th Edition
3. (a) 1.156 1.156 100
(b) 21.8 2.18 101
3
(c) 0.0068 6.8 10
(d) 27.635 2.7635 101
1
(e) 0.219 2.19 10
(f) 444 4.44 102
4. (a) 8.69 10 4 86, 900
(b) 9.1 103 9,100
1
(c) 8.8 10 0.88
(d) 4.76 10 2 476
5
(e) 3.62 10 0.0000362
5. The uncertainty is taken to be 0.01 m.
0.01 m
% uncertainty 100% 1%
1.57 m
0.25 m
6. % uncertainty 100% 6.6%
3.76 m
0.2 s
7. (a) % uncertainty 100% 4%
5s
0.2 s
(b) % uncertainty 100% 0.4%
50 s
0.2 s
(c) % uncertainty 100% 0.07%
300 s
8. To add values with significant figures, adjust all values to be added so that their exponents are all the
same.
9.2 103 s 8.3 10 4 s 0.008 106 s 9.2 103 s 83 103 s 8 103 s 9.2 83 8 103 s
100 103 s 1.00 105 s
When adding, keep the least accurate value, and so keep to the “ones” place in the parentheses.
9. 2.079 10 2 m 0.082 10 1
1.7 m . When multiplying, the result should have as many digits as
the number with the least number of significant digits used in the calculation.
10. 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 .
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
3
,Chapter 1 Introduction, Measurement, Estimating
2 2
Aspecified rspecified 3.8 10 4 cm 4.5 109 cm 2
2 2
Amin rmin 3.7 104 cm 4.30 109 cm 2
2 2
Amax rmax 3.9 10 4 cm 4.78 109 cm 2
A 1
2
Amax Amin 1
2
4.78 109 cm 2 4.30 109 cm 2 0.24 109 cm 2
Thus the area should be quoted as A 4.5 0.2 109 cm 2
11. To find the approximate uncertainty in the volume, calculate the volume for the specified radius, the
minimum radius, and the maximum radius. Subtract the extreme volumes. The uncertainty in the
volume is then half this variation in volume.
3 3
Vspecified 4
3
rspecified 4
3
2.86 m 9.80 101 m 3
3 3
Vmin 4
3
rmin 4
3
2.77 m 8.903 101 m 3
3 3
Vmax 4
3
rmax 4
3
2.95 m 10.754 101 m 3
V 1
2
Vmax Vmin 1
2
10.754 101 m 3 8.903 101 m 3 0.926 101 m 3
V 0.923 101 m 3
The percent uncertainty is 100 0.09444 9%
Vspecified 9.80 101 m 3
12. (a) 286.6 mm 286.6 10 3 m 0.286 6 m
(b) 85 V 85 10 6 V 0.000 085 V
(c) 760 mg 760 10 6 kg 0.000 760 kg (if last zero is significant)
(d) 60.0 ps 60.0 10 12 s 0.000 000 000 0600 s
(e) 22.5 fm 22.5 10 15 m 0.000 000 000 000 022 5 m
(f) 2.50 gigavolts 2.5 109 volts 2, 500, 000, 000 volts
13. (a) 1 106 volts 1 megavolt 1 Mvolt
(b) 2 10 6 meters 2 micrometers 2 m
3
(c) 6 10 days 6 kilodays 6 kdays
(d) 18 102 bucks 18 hectobucks 18 hbucks
(e) 8 10 9 pieces 8 nanopieces 8 npieces
14. (a) Assuming a height of 5 feet 10 inches, then 5 '10" 70 in 1 m 39.37 in 1.8 m
(b) 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 9.8 m/s2.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
4
, Giancoli Physics: Principles with Applications, 6th Edition
15. (a) 93 million miles 93 10 6 miles 1610 m 1 mile 1.5 1011 m
11 9
(b) 1.5 10 m 150 10 m 150 gigameters or 1.5 1011 m 0.15 1012 m 0.15 terameters
2
16. (a) 1 ft 2 1 ft 2 1 yd 3 ft 0.111 yd 2
2
(b) 1 m 2 1 m2 3.28 ft 1 m 10.8 ft 2
17. Use the speed of the airplane to convert the travel distance into a time.
1h 3600 s
1.00 km 3.8 s
950 km 1h
10 10
18. (a) 1.0 10 m 1.0 10 m 39.37 in 1 m 3.9 10 9 in
1m 1 atom
(b) 1.0 cm 10
1.0 108 atoms
100 cm 1.0 10 m
19. To add values with significant figures, adjust all values to be added so that their units are all the
same.
1.80 m 142.5 cm 5.34 105 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 m
When adding, the final result is to be no more accurate than the least accurate number used.
In this case, that is the first measurement, which is accurate to the hundredths place.
0.621 mi
20. (a) 1k h 0.621mi h
1 km
3.28 ft
(b) 1m s 3.28 ft s
1m
1000 m 1h
(c) 1km h 0.278 m s
1 km 3600 s
21. One mile is 1.61 103 m . It is 110 m longer than a 1500-m race. The percentage difference is
110 m
100% 7.3%
1500 m
22. (a) 1.00 ly 2.998 108 m s 3.156 10 7 s 9.46 1015 m
9.462 1015 m 1 AU
(b) 1.00 ly 11
6.31 104 AU
1.00 ly 1.50 10 m
1 AU 3600 s
(c) 2.998 108 m s 11
7.20 AU h
1.50 10 m 1 hr
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
5
Answers to Questions
1. (a) Fundamental standards should be accessible, invariable, indestructible, and reproducible. A
particular person’s foot would not be very accessible, since the person could not be at more than
one place at a time. The standard would be somewhat invariable if the person were an adult, but
even then, due to swelling or injury, the length of the standard foot could change. The standard
would not be indestructible – the foot would not last forever. The standard could be
reproducible – tracings or plaster casts could be made as secondary standards.
(b) If any person’s foot were to be used as a standard, “standard” would vary significantly
depending on the person whose foot happened to be used most recently for a measurement. The
standard would be very accessible, because wherever a measurement was needed, it would be
very easy to find someone with feet. The standard would be extremely variable – perhaps by a
factor of 2. That also renders the standard as not reproducible, because there could be many
reproductions that were quite different from each other. The standard would be almost
indestructible in that there is essentially a limitless supply of feet to be used.
2. There are various ways to alter the signs. The number of meters could be expressed in one
significant figure, as “900 m (3000 ft)”. Or, the number of feet could be expressed with the same
precision as the number of meters, as “914 m (2999 ft)”. The signs could also be moved to different
locations, where the number of meters was more exact. For example, if a sign was placed where the
elevation was really 1000 m to the nearest meter, then the sign could read “1000 m (3280 ft)”.
3. Including more digits in an answer does not necessarily increase its accuracy. The accuracy of an
answer is determined by the accuracy of the physical measurement on which the answer is based. If
you draw a circle, measure its diameter to be 168 mm and its circumference to be 527 mm, their
quotient, representing , is 3.136904762. The last seven digits are meaningless – they imply a
greater accuracy than is possible with the measurements.
4. The problem is that the precision of the two measurements are quite different. It would be more
appropriate to give the metric distance as 11 km, so that the numbers are given to about the same
precision (nearest mile or nearest km).
5. A measurement must be measured against a scale, and the units provide that scale. Units must be
specified or the answer is meaningless – the answer could mean a variety of quantities, and could be
interpreted in a variety of ways. Some units are understood, such as when you ask someone how old
they are. You assume their answer is in years. But if you ask someone how long it will be until they
are done with their task, and they answer “five”, does that mean five minutes or five hours or five
days? If you are in an international airport, and you ask the price of some object, what does the
answer “ten” mean? Ten dollars, or ten pounds, or ten marks, or ten euros?
6. If the jar is rectangular, for example, you could count the number of marbles along each dimension,
and then multiply those three numbers together for an estimate of the total number of marbles. If the
jar is cylindrical, you could count the marbles in one cross section, and then multiply by the number
of layers of marbles. Another approach would be to estimate the volume of one marble. If we
assume that the marbles are stacked such that their centers are all on vertical and horizontal lines,
then each marble would require a cube of edge 2R, or a volume of 8R3, where R is the radius of a
marble. The number of marbles would then be the volume of the container divided by 8R3.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
1
,Chapter 1 Introduction, Measurement, Estimating
7. The result should be written as 8.32 cm. The factor of 2 used to convert radius to diameter is exact –
it has no uncertainty, and so does not change the number of significant figures.
8. sin 30.0o 0.500
9. Since the size of large eggs can vary by 10%, the random large egg used in a recipe has a size with
an uncertainty of about 5% . Thus the amount of the other ingredients can also vary by about 5%
and not adversely affect the recipe.
10. In estimating the number of car mechanics, the assumptions and estimates needed are:
the population of the city
the number of cars per person in the city
the number of cars that a mechanic can repair in a day
the number of days that a mechanic works in a year
the number of times that a car is taken to a mechanic, per year
We estimate that there is 1 car for every 2 people, that a mechanic can repair 3 cars per day, that a
mechanic works 250 days a year, and that a car needs to be repaired twice per year.
(a) For San Francisco, we estimate the population at one million people. The number of mechanics
is found by the following calculation.
repairs
2
1 car year 1 yr 1 mechanic
1 106 people 1300 mechanics
2 people 1 car 250 workdays repairs
3
workday
(b) For Upland, Indiana, the population is about 4000. The number of mechanics is found by a
similar calculation, and would be 5 mechanics . There are actually two repair shops in Upland,
employing a total of 6 mechanics.
Solutions to Problems
1. (a) 14 billion years 1.4 1010 years
(b) 1.4 1010 y 3.156 107 s 1 y 4.4 1017 s
2. (a) 214 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) 3236 4 significant figures
(g) 8700 2 significant figures
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
2
,Giancoli Physics: Principles with Applications, 6th Edition
3. (a) 1.156 1.156 100
(b) 21.8 2.18 101
3
(c) 0.0068 6.8 10
(d) 27.635 2.7635 101
1
(e) 0.219 2.19 10
(f) 444 4.44 102
4. (a) 8.69 10 4 86, 900
(b) 9.1 103 9,100
1
(c) 8.8 10 0.88
(d) 4.76 10 2 476
5
(e) 3.62 10 0.0000362
5. The uncertainty is taken to be 0.01 m.
0.01 m
% uncertainty 100% 1%
1.57 m
0.25 m
6. % uncertainty 100% 6.6%
3.76 m
0.2 s
7. (a) % uncertainty 100% 4%
5s
0.2 s
(b) % uncertainty 100% 0.4%
50 s
0.2 s
(c) % uncertainty 100% 0.07%
300 s
8. To add values with significant figures, adjust all values to be added so that their exponents are all the
same.
9.2 103 s 8.3 10 4 s 0.008 106 s 9.2 103 s 83 103 s 8 103 s 9.2 83 8 103 s
100 103 s 1.00 105 s
When adding, keep the least accurate value, and so keep to the “ones” place in the parentheses.
9. 2.079 10 2 m 0.082 10 1
1.7 m . When multiplying, the result should have as many digits as
the number with the least number of significant digits used in the calculation.
10. 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 .
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
3
,Chapter 1 Introduction, Measurement, Estimating
2 2
Aspecified rspecified 3.8 10 4 cm 4.5 109 cm 2
2 2
Amin rmin 3.7 104 cm 4.30 109 cm 2
2 2
Amax rmax 3.9 10 4 cm 4.78 109 cm 2
A 1
2
Amax Amin 1
2
4.78 109 cm 2 4.30 109 cm 2 0.24 109 cm 2
Thus the area should be quoted as A 4.5 0.2 109 cm 2
11. To find the approximate uncertainty in the volume, calculate the volume for the specified radius, the
minimum radius, and the maximum radius. Subtract the extreme volumes. The uncertainty in the
volume is then half this variation in volume.
3 3
Vspecified 4
3
rspecified 4
3
2.86 m 9.80 101 m 3
3 3
Vmin 4
3
rmin 4
3
2.77 m 8.903 101 m 3
3 3
Vmax 4
3
rmax 4
3
2.95 m 10.754 101 m 3
V 1
2
Vmax Vmin 1
2
10.754 101 m 3 8.903 101 m 3 0.926 101 m 3
V 0.923 101 m 3
The percent uncertainty is 100 0.09444 9%
Vspecified 9.80 101 m 3
12. (a) 286.6 mm 286.6 10 3 m 0.286 6 m
(b) 85 V 85 10 6 V 0.000 085 V
(c) 760 mg 760 10 6 kg 0.000 760 kg (if last zero is significant)
(d) 60.0 ps 60.0 10 12 s 0.000 000 000 0600 s
(e) 22.5 fm 22.5 10 15 m 0.000 000 000 000 022 5 m
(f) 2.50 gigavolts 2.5 109 volts 2, 500, 000, 000 volts
13. (a) 1 106 volts 1 megavolt 1 Mvolt
(b) 2 10 6 meters 2 micrometers 2 m
3
(c) 6 10 days 6 kilodays 6 kdays
(d) 18 102 bucks 18 hectobucks 18 hbucks
(e) 8 10 9 pieces 8 nanopieces 8 npieces
14. (a) Assuming a height of 5 feet 10 inches, then 5 '10" 70 in 1 m 39.37 in 1.8 m
(b) 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 9.8 m/s2.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
4
, Giancoli Physics: Principles with Applications, 6th Edition
15. (a) 93 million miles 93 10 6 miles 1610 m 1 mile 1.5 1011 m
11 9
(b) 1.5 10 m 150 10 m 150 gigameters or 1.5 1011 m 0.15 1012 m 0.15 terameters
2
16. (a) 1 ft 2 1 ft 2 1 yd 3 ft 0.111 yd 2
2
(b) 1 m 2 1 m2 3.28 ft 1 m 10.8 ft 2
17. Use the speed of the airplane to convert the travel distance into a time.
1h 3600 s
1.00 km 3.8 s
950 km 1h
10 10
18. (a) 1.0 10 m 1.0 10 m 39.37 in 1 m 3.9 10 9 in
1m 1 atom
(b) 1.0 cm 10
1.0 108 atoms
100 cm 1.0 10 m
19. To add values with significant figures, adjust all values to be added so that their units are all the
same.
1.80 m 142.5 cm 5.34 105 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 m
When adding, the final result is to be no more accurate than the least accurate number used.
In this case, that is the first measurement, which is accurate to the hundredths place.
0.621 mi
20. (a) 1k h 0.621mi h
1 km
3.28 ft
(b) 1m s 3.28 ft s
1m
1000 m 1h
(c) 1km h 0.278 m s
1 km 3600 s
21. One mile is 1.61 103 m . It is 110 m longer than a 1500-m race. The percentage difference is
110 m
100% 7.3%
1500 m
22. (a) 1.00 ly 2.998 108 m s 3.156 10 7 s 9.46 1015 m
9.462 1015 m 1 AU
(b) 1.00 ly 11
6.31 104 AU
1.00 ly 1.50 10 m
1 AU 3600 s
(c) 2.998 108 m s 11
7.20 AU h
1.50 10 m 1 hr
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
5
