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Exam (elaborations) PHYS 195 (PHYS195) Serway Physics 6th Edition Solutions
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Serway Physics 6th Edition Solutions Exam (elaborations) PHYS 195 (PHYS195) Serway Physics 6th Edition Solutions
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1Physics and Measurement
CHAPTER OUTLINE ANSWERS TO QUESTIONS
1.1 Standards of Length, Mass,
and Time Q1.1 Atomic clocks are based on electromagnetic waves which atoms
1.2 Matter and Model-Building
emit. Also, pulsars are highly regular astronomical clocks.
1.3 Density and Atomic Mass
1.4 Dimensional Analysis
1.5 Conversion of Units Q1.2 Density varies with temperature and pressure. It would be
1.6 Estimates and Order-of- necessary to measure both mass and volume very accurately in
Magnitude Calculations
order to use the density of water as a standard.
1.7 Significant Figures
Q1.3 People have different size hands. Defining the unit precisely
would be cumbersome.
Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms
Q1.5 (b) and (d). You cannot add or subtract quantities of different
dimension.
Q1.6 A dimensionally correct equation need not be true. Example:
1 chimpanzee = 2 chimpanzee is dimensionally correct. If an
equation is not dimensionally correct, it cannot be correct.
Q1.7 If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about
10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on
vacation.
Q1.8 On February 7, 2001, I am 55 years and 39 days old.
F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10
55 yr GH 1 yr JK H 1d K
9
s ~ 10 9 s .
Many college students are just approaching 1 Gs.
Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.
Q1.10 The mass of the forty-six chapter textbook is on the order of 10 0 kg .
Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
1
, 2 Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
No problems in this section
Section 1.2 Matter and Model-Building
P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between
diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the
Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance
1 2
L = 0.200 nm , the diagonal planes are separated by L + L2 = 0.141 nm .
2
Section 1.3 Density and Atomic Mass
4 3 4
*P1.2 Modeling the Earth as a sphere, we find its volume as
3 3
e
π r = π 6.37 × 10 6 m j 3
= 1.08 × 10 21 m 3 . Its
m 5.98 × 10 24 kg
density is then ρ = = = 5.52 × 10 3 kg m3 . This value is intermediate between the
V 1.08 × 10 21 m 3
tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to
3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density material
must be down below the surface.
P1.3 a fb g
With V = base area height V = π r 2 h and ρ = e j m
V
, we have
m 1 kg F 10 mm I
9 3
ρ= =
a
π r h π 19.5 mm 2 39.0 mm
2
fa f GH 1 m JK
3
4 3
ρ = 2.15 × 10 kg m .
m
*P1.4 Let V represent the volume of the model, the same in ρ = for both. Then ρ iron = 9.35 kg V and
V
m gold ρ gold m gold F
19.3 × 10 3 kg / m3 I
ρ gold =
V
. Next,
ρ iron
=
9.35 kg
and m gold = 9.35 kg GH
7.86 × 10 3 kg / m3 JK
= 23.0 kg .
4
P1.5 V = Vo − Vi =
3
e
π r23 − r13 j
ρ=
m 4 FG IJ e
, so m = ρV = ρ π r23 − r13 =
4π ρ r23 − r13
j e j
V 3 H K 3
.
, Chapter 1 3
4 3 4
P1.6 For either sphere the volume is V = π r and the mass is m = ρV = ρ π r 3 . We divide this equation
3 3
for the larger sphere by the same equation for the smaller:
m A ρ 4π rA3 3 rA3
= = = 5.
m s ρ 4π rs3 3 rs3
a f
Then rA = rs 3 5 = 4.50 cm 1.71 = 7.69 cm .
P1.7 Use 1 u = 1.66 × 10 −24 g .
F 1.66 × 10 -24
g I = 6.64 × 10
(a) For He, m 0 = 4.00 u GH 1 u JK −24
g .
F 1.66 × 10
= 55.9 uG
-24
gI
(b) For Fe, m 0
H 1u JK = 9.29 × 10 −23
g .
= 207 uG
F 1.66 × 10 −24
gI
(c) For Pb, m 0
H 1u JK = 3.44 × 10 −22
g .
*P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m 0 of one
atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is
m 0 = 27.0 u = 27.0 u × 1.66 × 10 −27 kg 1 u = 4.48 × 10 −26 kg .
Then the mass of 6.02 × 10 23 atoms is
m = Nm 0 = 6.02 × 10 23 × 4.48 × 10 −26 kg = 0.027 0 kg = 27.0 g .
Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be
written m = Nm 0 .
0.027 kg
0.027 0 kg = 6.02 × 10 23 m 0 , so m 0 = = 4.48 × 10 −26 kg ,
6.02 × 10 23
in agreement with the first assertion.
(b) The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u ,
where M is the numerical value of the atomic mass. It divides out exactly for all substances,
giving 1.000 000 0 × 10 −3 kg = N 1.660 540 2 × 10 −27 kg . With eight-digit data, we can be quite
sure of the result to seven digits. For one mole the number of atoms is
F 1 I 10
N= GH 1.660 540 2 JK −3 + 27
= 6.022 137 × 10 23 .
(c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one
b g
molecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .
(d) b g
For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.
, 4 Physics and Measurement
P1.9 Mass of gold abraded: ∆m = 3.80 g − 3.35 g = 0.45 g = 0.45 g b gFGH 101 kgg IJK = 4.5 × 10
3
−4
kg .
F 1.66 × 10 −27
kg I
Each atom has mass m 0 = 197 u = 197 u GH 1 u JK = 3.27 × 10 kg . −25
Now, ∆m = ∆N m 0 , and the number of atoms missing is
∆m 4.5 × 10 −4 kg
∆N = = = 1.38 × 10 21 atoms .
m0 3.27 × 10 −25 kg
The rate of loss is
∆N 1.38 × 10 21 atoms 1 yr FG IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ
∆t
=
50 yr 365.25 d H K H 24 h K H 60 min K H 60 s K
∆N
= 8.72 × 10 11 atoms s .
∆t
P1.10 (a) e
m = ρ L3 = 7.86 g cm 3 5.00 × 10 −6 cm je j 3
= 9.83 × 10 −16 g = 9.83 × 10 −19 kg
m 9.83 × 10 −19 kg
(b) N= = = 1.06 × 10 7 atoms
e
m 0 55.9 u 1.66 × 10 −27 kg 1 u j
P1.11 (a) The cross-sectional area is
a fa
A = 2 0.150 m 0.010 m + 0.340 m 0.010 m f a fa f.
−3 2
= 6.40 × 10 m .
The volume of the beam is
e ja
V = AL = 6.40 × 10 −3 m 2 1.50 m = 9.60 × 10 −3 m3 . f
Thus, its mass is
FIG. P1.11
e
m = ρV = 7.56 × 10 kg / m je9.60 × 10 m j = 72.6 kg .
3 3 −3 3
F 1.66 × 10 kg I = 9.28 × 10
The mass of one typical atom is m = a55.9 ufG
−27
H 1 u JK
−26
(b) 0 kg . Now
m 72.6 kg
m = Nm 0 and the number of atoms is N = = −26
= 7.82 × 10 26 atoms .
m 0 9.28 × 10 kg
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