OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
CHAPTER 1: INTRODUCTION: THE NATURE
OF SCIENCE AND PHYSICS
1.2 PHYSICAL QUANTITIES AND UNITS
1. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in
meters per second? (b) How many miles per hour is this?
Solution 100 km 1000m 1h
(a) × × =27.77 m/ s=27.8 m/ s
h 1 km 3600 s
100 km 1 mi
(b) × =62 mi /h
h 1.609 km
2. A car is traveling at a speed of 33 m/ s . (a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km /h speed limit?
Solution 33 m 1 km 3600 s 2
(a) × × =118.8 km/h=1.2× 1 0 km/h
s 1000 m 1h
(b) At 120 km/h, the car is travelling faster than the speed limit.
3. Show that 1.0 m/ s=3.6 km/h . Hint: Show the explicit steps involved in converting
1.0 m/s=3.6 km /h.
Solution 1.0 m 1.0 m 3600 s 1 km
= × × =3.6 km/h
s s hr 1000 m
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
4. American football is played on a 100-yd-long field, excluding the end zones. How
long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
Solution 3 ft 1m
100 yd × × =91.44 m=91.4 m
1 yd 3.281 ft
5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its
dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
Solution 1 ft
115 m× =377.3 ft=377 ft long
0.3048 m
12∈. 3
377.3 ft × =4528∈.=4.53 ×1 0 ∈.long
1.0 ft
1 ft 2
85 m× =278.9 ft=2.8 ×1 0 ft wide
0.3048 m
12∈.
278.9 ft × =3346∈.=3.3× 1 03 ∈. wide
1.0 ft
6. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1
meter equals 39.37 in.)
Solution
(
6 ft , 1.0∈.= 6 ft ×
12∈.
ft )+1.0∈.=73.0∈.; 73.0∈.×
1m
39.37 ∈.
=1.85 m
7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its
height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
Solution 1 km
29,028 ft × =8.847 km
3281 ft
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
8. The speed of sound is measured to be 342 m/s on a certain day. What is this in
km/h?
Solution 342m 1 km 3600 s 3
× × =1.23 ×1 0 km/h
s 1000 m 1h
9. Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose
that one such plate has an average speed of 4.0 cm/year. (a) What distance does it
move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
Solution
(a) ( 4.0ycm × 1 s) × 1001 mcm × 365.25
1y
×
1d
×
1h
d 24.0 h 3600 s
−9
=1.3× 10 m
(b) 4.0 cm 1m 1 km 10 6 y
× × × =40 km/My
y 100 cm 1000 m 1 My
10. (a) Refer to Table 1.3 to determine the average distance between the Earth and the
Sun. Then calculate the average speed of the Earth in its orbit in kilometers per
second. (b) What is this in meters per second?
8
Solution (a) d 2 πr 2 π (1 0 km) 1 d 1h
v= = = × × =20 km /s
t t 365.25 d 24 h 3600 s
20 km 1000 m 4
(b) v= × =2.0 ×1 0 m/ s
s 1 km
1.3 ACCURACY, PRECISION, AND SIGNIFICANT FIGURES
11. Suppose that your bathroom scale reads your mass as 65 kg with a 3 % uncertainty.
What is the uncertainty in your mass (in kilograms)?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 3%
δ m= ×65 kg=2kg
100 %
12. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What
is its percent uncertainty?
Solution δl 0.50 cm 1m
% unc l= × 100 %= × × 100 %=2.5 ×1 0−2 %
l 20 m 100 cm
13. (a) A car speedometer has a 5.0 % uncertainty. What is the range of possible speeds
when it reads 90 km/h ? (b) Convert this range to miles per hour.(1 km=0.6214 mi)
Solution 5.0 %
(a) δv= × 90.0 km/h=4.5 km/h
100 % .¿
Thus , the range=90.0 ±5 km/h=85 ¿ 95 km/ h
(b) 85.5 km 0.6214 mi 94.5 km 0.6214 mi
× =53.1 mi / h; × =58.7 mi/h
h 1 km 1h 1 km
So the range is53.1 ¿ 58.7 mi/h ¿ .
14. An infant’s pulse rate is measured to be 130 ±5 beats/min. What is the percent
uncertainty in this measurement?
Solution δA 5 beats/min
% unc= ×100 %= ×100 %=3.84 %=4 %
A 130 beats/min
15. (a) Suppose that a person has an average heart rate of 72.0 beats/min. How many
beats do they have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 72.0 beats 60.0 min 24.0 h 365.25 d
× × × ×2.0 y =7.5738× 10 7 beats
1min 1.00 h 1.00 d 1.00 y
a) (limited by 2.0 y)
7
7.6 ×1 0 beats
(
(b) 7.57 ×1 07 beats (limited by 2.00 y)
(c) 7.57 ×1 07 beats (limited by 72.0 beats/min)
16. A can contains 375 mL of soda. How much is left after 308 mL is removed?
Solution 375 mL−308 mL=67 mL (uncertainty in the 1’s column)
17. State how many significant figures are proper in the results of the following
calculations: (a) ( 106.7 ) ( 98.2 ) / ( 46.210 )( 1.01 ) (b)( 18.7 )2 (c) ( 1.60 ×1 0−19 ) ( 3712 ).
Solution (a) 3 (limited by 98.2 and 1.01)
(b) 3 (limited by 18.7)
(c) 3 (limited by 1.60)
18. (a) How many significant figures are in the numbers 99 and 100? (b) If the
uncertainty in each number is 1, what is the percent uncertainty in each? (c)
Which is a more meaningful way to express the accuracy of these two numbers,
significant figures or percent uncertainties?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution (a) 99 has 2 sig. figs. ; 100 has 3 sig. figs. at most
1
(b) × 100=1.01 %=1.0 %;
99
1
×100=1.00 %(if all zeros are significant)
100
(c) percent uncertainties
19. (a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h , what
is the percent uncertainty? (b) If it has the same percent uncertainty when it reads
60 km/h, what is the range of speeds you could be going?
Solution 2.0 km/h
(a) % unc= × 100 %=2.2 %
90 km/h
2.2 %
(b) δ v= ×60 km/h=1km/h
100 %
So the range is 60 ± 1km/h or 59 ¿ 61 km/h¿ .
20. (a) A person’s blood pressure is measured to be 120 ±2 mm Hg . What is its percent
uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in
a blood pressure measurement of 80 mm Hg?
Solution 2 mm Hg
(a) % unc= ×100 %=1.7 %=2 % (1 sig . fig because of 2 mm Hg)
120 mm Hg
1.7 %
(b) δ bp= ×80 mm Hg=1.3 mm Hg=1 mm Hg(1 sig . fig because of 2 mm Hg)
100 %
21. A person measures their heart rate by counting the number of beats in 30 s . If
40 ± 1 beats are counted in 30.0 ± 0.5 s, what is the heart rate and its uncertainty in
beats per minute?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution beats 40 beats 60.0 s
= × =80 beats/min .
minute 30.0 s 1.00 min
1 beat 0.5 s
% unc= ×100 %+ × 100 %=2.5 %+1.7 %=4.2 %=4 %
40 beats 30.0 s
% unc 4.2 %
δ A= × A= ×80 beats /min=3.3 beats/min=3 beats /min
100 % 100 %
The heart rate is 80 ± 3 beats/min .
22. What is the area of a circle 3.102 cm in diameter?
Solution 2
A=π r =π
d 2
2 () (
=π
3.102cm 2
2 )
=7.557 c m2
23. If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a
26.22-mi marathon?
Solution 26.22mi
=2.8 h
9.5 mi/h
24. A marathon runner completes a 42.188−km course in 2 h, 30 min, and 12 s. There
is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the
elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the
uncertainty in the elapsed time. (c) What is the average speed in meters per
second? (d) What is the uncertainty in the average speed?
Solution 25 m 1km
(a) % unc distance= × ×100 %=0.0593 %=0.059 %
42.188 km 1000 m
1s
(b) % unc time= ×100 %=0.0111 %=0.01 %
9012 s
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
42.188 km 1000 m
(c) average speed = × =4.681 m/ s
9012 s 1 km
% unc speed =% unc distance+ % unc time
(d)
¿ 0.0593 % +0.0111 %=0.0704 %=0.07 %
0.07 %
δ speed = × 4.681 m/s=0.003 m/s
100 %
25. The sides of a small rectangular box are measured to be 1.80 ± 0.01cm ,
2.05 ± 0.02cm , and 3.1 ±0.1 cm long. Calculate its volume and uncertainty in cubic
centimeters.
Solution V =1.80 cm×2.05 cm ×3.1 cm=11.4 c m3 . Use the methods of adding percents.
0.01 cm
1.80 ± 0.01cm → ×100 %=0.556 %
1.80 cm
0.02 cm
2.05 ± 0.02cm → ×100 %=0.976 %
2.05 cm
0.1 cm
3.1 ± 0.1cm → ×100 %=3.226 %
3.1 cm
Adding these values and rounding to 1 sig. fig., the percent uncertainty of the
volume is 5%. The uncertainty in the volume is therefore (11.4)(0.05) = 0.6. The
volume is thus 11.4 ±0.6 c m3 .
26. When non-metric units were used in the United Kingdom, a unit of mass called
the pound-mass (lbm) was employed, where 1 lbm=0.4539 kg . (a) If there is an
uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty?
(b) Based on that percent uncertainty, what mass in pound-mass has an
uncertainty of 1 kg when converted to kilograms?
, OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 0.0001 kg
(a) % unc lbm= 0.4539 kg ×100 %=0.022 %=0.02 %
(b)
δ lbm 1 kg 1lbm 4
lbm= ×100 %= × × 100 %=1× 10 lbm ,∨10,000 lbm.
% unc lbm 0.02 % 0.4539 kg
27. The length and width of a rectangular room are measured to be 3.955 ± 0.005m
and 3.050 ± 0.005 m. Calculate the area of the room and its uncertainty in square
meters.
Solution The area is 3.995 m× 3.050 m=12.06 m2 . Now use method of adding percents to get
uncertainty in the area.
0.005 m
% unc width= ×100 %=0.16 %
3.050 m
0.005 m
% unc length= ×100 %=0.13 %
3.955 m
% unc area=0.13 % +0.16 %=0.29 %=0.3 %
0.29 % 2 2 2
δ area= ×12.06 m =0.035 m =0.04 m
100 %
So the area is 2
12.06 ± 0.04 m .
28. A car engine moves a piston with a circular cross section of 7.500 ± 0.002cm
diameter a distance of 3.250 ± 0.001cm to compress the gas in the cylinder. (a) By
what amount is the gas decreased in volume in cubic centimeters? (b) Find the
uncertainty in this volume.
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
CHAPTER 1: INTRODUCTION: THE NATURE
OF SCIENCE AND PHYSICS
1.2 PHYSICAL QUANTITIES AND UNITS
1. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in
meters per second? (b) How many miles per hour is this?
Solution 100 km 1000m 1h
(a) × × =27.77 m/ s=27.8 m/ s
h 1 km 3600 s
100 km 1 mi
(b) × =62 mi /h
h 1.609 km
2. A car is traveling at a speed of 33 m/ s . (a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km /h speed limit?
Solution 33 m 1 km 3600 s 2
(a) × × =118.8 km/h=1.2× 1 0 km/h
s 1000 m 1h
(b) At 120 km/h, the car is travelling faster than the speed limit.
3. Show that 1.0 m/ s=3.6 km/h . Hint: Show the explicit steps involved in converting
1.0 m/s=3.6 km /h.
Solution 1.0 m 1.0 m 3600 s 1 km
= × × =3.6 km/h
s s hr 1000 m
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
4. American football is played on a 100-yd-long field, excluding the end zones. How
long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
Solution 3 ft 1m
100 yd × × =91.44 m=91.4 m
1 yd 3.281 ft
5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its
dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
Solution 1 ft
115 m× =377.3 ft=377 ft long
0.3048 m
12∈. 3
377.3 ft × =4528∈.=4.53 ×1 0 ∈.long
1.0 ft
1 ft 2
85 m× =278.9 ft=2.8 ×1 0 ft wide
0.3048 m
12∈.
278.9 ft × =3346∈.=3.3× 1 03 ∈. wide
1.0 ft
6. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1
meter equals 39.37 in.)
Solution
(
6 ft , 1.0∈.= 6 ft ×
12∈.
ft )+1.0∈.=73.0∈.; 73.0∈.×
1m
39.37 ∈.
=1.85 m
7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its
height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
Solution 1 km
29,028 ft × =8.847 km
3281 ft
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
8. The speed of sound is measured to be 342 m/s on a certain day. What is this in
km/h?
Solution 342m 1 km 3600 s 3
× × =1.23 ×1 0 km/h
s 1000 m 1h
9. Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose
that one such plate has an average speed of 4.0 cm/year. (a) What distance does it
move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
Solution
(a) ( 4.0ycm × 1 s) × 1001 mcm × 365.25
1y
×
1d
×
1h
d 24.0 h 3600 s
−9
=1.3× 10 m
(b) 4.0 cm 1m 1 km 10 6 y
× × × =40 km/My
y 100 cm 1000 m 1 My
10. (a) Refer to Table 1.3 to determine the average distance between the Earth and the
Sun. Then calculate the average speed of the Earth in its orbit in kilometers per
second. (b) What is this in meters per second?
8
Solution (a) d 2 πr 2 π (1 0 km) 1 d 1h
v= = = × × =20 km /s
t t 365.25 d 24 h 3600 s
20 km 1000 m 4
(b) v= × =2.0 ×1 0 m/ s
s 1 km
1.3 ACCURACY, PRECISION, AND SIGNIFICANT FIGURES
11. Suppose that your bathroom scale reads your mass as 65 kg with a 3 % uncertainty.
What is the uncertainty in your mass (in kilograms)?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 3%
δ m= ×65 kg=2kg
100 %
12. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What
is its percent uncertainty?
Solution δl 0.50 cm 1m
% unc l= × 100 %= × × 100 %=2.5 ×1 0−2 %
l 20 m 100 cm
13. (a) A car speedometer has a 5.0 % uncertainty. What is the range of possible speeds
when it reads 90 km/h ? (b) Convert this range to miles per hour.(1 km=0.6214 mi)
Solution 5.0 %
(a) δv= × 90.0 km/h=4.5 km/h
100 % .¿
Thus , the range=90.0 ±5 km/h=85 ¿ 95 km/ h
(b) 85.5 km 0.6214 mi 94.5 km 0.6214 mi
× =53.1 mi / h; × =58.7 mi/h
h 1 km 1h 1 km
So the range is53.1 ¿ 58.7 mi/h ¿ .
14. An infant’s pulse rate is measured to be 130 ±5 beats/min. What is the percent
uncertainty in this measurement?
Solution δA 5 beats/min
% unc= ×100 %= ×100 %=3.84 %=4 %
A 130 beats/min
15. (a) Suppose that a person has an average heart rate of 72.0 beats/min. How many
beats do they have in 2.0 y? (b) In 2.00 y? (c) In 2.000 y?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 72.0 beats 60.0 min 24.0 h 365.25 d
× × × ×2.0 y =7.5738× 10 7 beats
1min 1.00 h 1.00 d 1.00 y
a) (limited by 2.0 y)
7
7.6 ×1 0 beats
(
(b) 7.57 ×1 07 beats (limited by 2.00 y)
(c) 7.57 ×1 07 beats (limited by 72.0 beats/min)
16. A can contains 375 mL of soda. How much is left after 308 mL is removed?
Solution 375 mL−308 mL=67 mL (uncertainty in the 1’s column)
17. State how many significant figures are proper in the results of the following
calculations: (a) ( 106.7 ) ( 98.2 ) / ( 46.210 )( 1.01 ) (b)( 18.7 )2 (c) ( 1.60 ×1 0−19 ) ( 3712 ).
Solution (a) 3 (limited by 98.2 and 1.01)
(b) 3 (limited by 18.7)
(c) 3 (limited by 1.60)
18. (a) How many significant figures are in the numbers 99 and 100? (b) If the
uncertainty in each number is 1, what is the percent uncertainty in each? (c)
Which is a more meaningful way to express the accuracy of these two numbers,
significant figures or percent uncertainties?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution (a) 99 has 2 sig. figs. ; 100 has 3 sig. figs. at most
1
(b) × 100=1.01 %=1.0 %;
99
1
×100=1.00 %(if all zeros are significant)
100
(c) percent uncertainties
19. (a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h , what
is the percent uncertainty? (b) If it has the same percent uncertainty when it reads
60 km/h, what is the range of speeds you could be going?
Solution 2.0 km/h
(a) % unc= × 100 %=2.2 %
90 km/h
2.2 %
(b) δ v= ×60 km/h=1km/h
100 %
So the range is 60 ± 1km/h or 59 ¿ 61 km/h¿ .
20. (a) A person’s blood pressure is measured to be 120 ±2 mm Hg . What is its percent
uncertainty? (b) Assuming the same percent uncertainty, what is the uncertainty in
a blood pressure measurement of 80 mm Hg?
Solution 2 mm Hg
(a) % unc= ×100 %=1.7 %=2 % (1 sig . fig because of 2 mm Hg)
120 mm Hg
1.7 %
(b) δ bp= ×80 mm Hg=1.3 mm Hg=1 mm Hg(1 sig . fig because of 2 mm Hg)
100 %
21. A person measures their heart rate by counting the number of beats in 30 s . If
40 ± 1 beats are counted in 30.0 ± 0.5 s, what is the heart rate and its uncertainty in
beats per minute?
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution beats 40 beats 60.0 s
= × =80 beats/min .
minute 30.0 s 1.00 min
1 beat 0.5 s
% unc= ×100 %+ × 100 %=2.5 %+1.7 %=4.2 %=4 %
40 beats 30.0 s
% unc 4.2 %
δ A= × A= ×80 beats /min=3.3 beats/min=3 beats /min
100 % 100 %
The heart rate is 80 ± 3 beats/min .
22. What is the area of a circle 3.102 cm in diameter?
Solution 2
A=π r =π
d 2
2 () (
=π
3.102cm 2
2 )
=7.557 c m2
23. If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a
26.22-mi marathon?
Solution 26.22mi
=2.8 h
9.5 mi/h
24. A marathon runner completes a 42.188−km course in 2 h, 30 min, and 12 s. There
is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the
elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the
uncertainty in the elapsed time. (c) What is the average speed in meters per
second? (d) What is the uncertainty in the average speed?
Solution 25 m 1km
(a) % unc distance= × ×100 %=0.0593 %=0.059 %
42.188 km 1000 m
1s
(b) % unc time= ×100 %=0.0111 %=0.01 %
9012 s
,OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
42.188 km 1000 m
(c) average speed = × =4.681 m/ s
9012 s 1 km
% unc speed =% unc distance+ % unc time
(d)
¿ 0.0593 % +0.0111 %=0.0704 %=0.07 %
0.07 %
δ speed = × 4.681 m/s=0.003 m/s
100 %
25. The sides of a small rectangular box are measured to be 1.80 ± 0.01cm ,
2.05 ± 0.02cm , and 3.1 ±0.1 cm long. Calculate its volume and uncertainty in cubic
centimeters.
Solution V =1.80 cm×2.05 cm ×3.1 cm=11.4 c m3 . Use the methods of adding percents.
0.01 cm
1.80 ± 0.01cm → ×100 %=0.556 %
1.80 cm
0.02 cm
2.05 ± 0.02cm → ×100 %=0.976 %
2.05 cm
0.1 cm
3.1 ± 0.1cm → ×100 %=3.226 %
3.1 cm
Adding these values and rounding to 1 sig. fig., the percent uncertainty of the
volume is 5%. The uncertainty in the volume is therefore (11.4)(0.05) = 0.6. The
volume is thus 11.4 ±0.6 c m3 .
26. When non-metric units were used in the United Kingdom, a unit of mass called
the pound-mass (lbm) was employed, where 1 lbm=0.4539 kg . (a) If there is an
uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty?
(b) Based on that percent uncertainty, what mass in pound-mass has an
uncertainty of 1 kg when converted to kilograms?
, OpenStax College Physics 2e Instructor Solutions Manual Chapter 1
Solution 0.0001 kg
(a) % unc lbm= 0.4539 kg ×100 %=0.022 %=0.02 %
(b)
δ lbm 1 kg 1lbm 4
lbm= ×100 %= × × 100 %=1× 10 lbm ,∨10,000 lbm.
% unc lbm 0.02 % 0.4539 kg
27. The length and width of a rectangular room are measured to be 3.955 ± 0.005m
and 3.050 ± 0.005 m. Calculate the area of the room and its uncertainty in square
meters.
Solution The area is 3.995 m× 3.050 m=12.06 m2 . Now use method of adding percents to get
uncertainty in the area.
0.005 m
% unc width= ×100 %=0.16 %
3.050 m
0.005 m
% unc length= ×100 %=0.13 %
3.955 m
% unc area=0.13 % +0.16 %=0.29 %=0.3 %
0.29 % 2 2 2
δ area= ×12.06 m =0.035 m =0.04 m
100 %
So the area is 2
12.06 ± 0.04 m .
28. A car engine moves a piston with a circular cross section of 7.500 ± 0.002cm
diameter a distance of 3.250 ± 0.001cm to compress the gas in the cylinder. (a) By
what amount is the gas decreased in volume in cubic centimeters? (b) Find the
uncertainty in this volume.
