Electromagnetismo

1
Physics and Measurement
CHAPTER OUTLINE
1.1

ANSWERS TO QUESTIONS

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

Atomic clocks are based on electromagnetic waves which atoms
emit. Also, pulsars are highly regular astronomical clocks.Q1.2

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.

Q1.4

(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

Q1.5

(b) and (d). You cannot add or subtract quantities ofdifferent
dimension.

Q1.6

1.7

Q1.1

Q1.3

1.2
1.3
1.4
1.5
1.6

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 perday. 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.
55 yr

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

Zero digits. An order-of-magnitude calculation is accurate only within a factorof 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
P1.1

Matter and Model-BuildingFrom 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

L = 0.200 nm , the diagonal planes are separated by

Section 1.3
*P1.2

12
L + L2 = 0.141 nm .
2

Density andAtomic Mass

Modeling the Earth as a sphere, we find its volume as

434
π r = π 6.37 × 10 6 m
3
3

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 issignificantly higher, so higher-density material
must be down below the surface.
density is then ρ =

P1.3

a

fb

g

ej

With V = base area height V = π r 2 h and ρ =

ρ=

a

fa

ρ = 2.15 × 10 kg m

3

3

9

3

.

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

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

ρ gold =

P1.5

F 10 mm I
f GH 1 m JK

1 kg
m
=
2
π r h π 19.5 mm 2 39.0 mm
4

*P1.4

m
, we have
V

m gold
V

V = Vo − Vi =

ρ=

. Next,

ρ gold
ρ iron

4
3
π r2 − r13
3

e

=

m gold
9.35 kg

and m gold

F
GH

j

FG IJ e
HK

e

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

j

j

.

I
JK

Chapter1

P1.6

3

4
43
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:

For either sphere the volume is V =

ρ 4π r 3 3 r 3
m
=
=
= 5.
m s ρ 4π rs3 3 rs3

af

Then r = 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
GH 1 u
F 1.66 × 10
= 55.9 uG
H 1u
F 1.66...