Serway

1
Physics and Measurement
CHAPTER OUTLINE
1.1 1.2 1.3 1.4 1.5 1.6 1.7 Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures

ANSWERS TO QUESTIONS
Q1.1 Q1.2 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regularastronomical clocks. Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard. People have different size hands. Defining the unit precisely would be cumbersome. (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms (b) and (d). You cannot add or subtract quantities of differentdimension. 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.

Q1.3 Q1.4 Q1.5 Q1.6

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 to200 miles per day on vacation. On February 7, 2001, I am 55 years and 39 days old. 55 yr

Q1.8

F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10 GH 1 yr JK H 1d K

9

s ~ 10 9 s .

Many college students are just approaching 1 Gs. Q1.9 Q1.10 Q1.11 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10. The mass of the forty-six chapter textbook is onthe order of 10 0 kg . With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.

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Physics and Measurement

SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time

No problems in this section

Section 1.2 P1.1

Matter and Model-Building

From the figure, we may see that the spacing between diagonal planes is half thedistance 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 + L2 = 0.141 nm . 2

L = 0.200 nm , the diagonal planes are separated by

Section 1.3 *P1.2

Density and Atomic Mass 4 3 4 π r = π 6.37 × 10 6 m 3 3

Modeling the Earth as a sphere, we find itsvolume as density is then ρ =

e

j

3

= 1.08 × 10 21 m 3 . Its

m 5.98 × 10 24 kg = = 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. P1.3With V = base area height V = π r 2 h and ρ =

a

fb

g

e j

m , we have V

ρ=

1 kg m = 2 π r h π 19.5 mm 2 39.0 mm

a

fa
.

F 10 mm I f GH 1 m JK
9 3 3

ρ = 2.15 × 10 kg m
*P1.4

4

3

Let V represent the volume of the model, the same in ρ =

ρ gold =

m gold V

. Next,

ρ gold ρ iron

=

m gold 9.35 kg

and m gold

m for both. Then ρ iron = 9.35 kgV and V 19.3 × 10 3 kg / m3 = 23.0 kg . = 9.35 kg 7.86 × 10 3 kg / m3

F GH

I JK

P1.5

V = Vo − Vi =

4 3 π r2 − r13 3

e

j
j e j
.

ρ=

3 4π ρ r2 − r13 m 4 3 , so m = ρV = ρ π r2 − r13 = V 3 3

FG IJ e H K

Chapter 1

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P1.6

For either sphere the volume is V =

4 4 3 π r and the mass is m = ρV = ρ π r 3 . We divide this equation 3 3 for the larger sphere bythe same equation for the smaller:

ρ 4π r 3 3 r 3 m = = = 5. m s ρ 4π rs3 3 rs3
Then r = rs 3 5 = 4.50 cm 1.71 = 7.69 cm . P1.7 Use 1 u = 1.66 × 10 −24 g . (a) For He, m 0 = 4.00 u

a f

(b)

For Fe, m 0

(c) *P1.8 (a)

For Pb, m 0

F 1.66 × 10 GH 1 u F 1.66 × 10 = 55.9 uG H 1u F 1.66 × 10 = 207 uG H 1u

-24

-24

−24

I = 6.64 × 10 JK gI JK = 9.29 × 10 gI JK = 3.44 × 10...