Solucionario física de serway 7ma edición

c1
Physics and Measurement
CHAPTER OUTLINE
1.1
1.2
1.3
1.4
1.5

Standards of Length, Mass, and
Time
Matter and Model-Building
Dimensional Analysis
Conversion of Units
Estimates and Order-ofMagnitude Calculations
Significant Figures

ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition.
Density varies with temperature and pressure. It would
be necessary tomeasure both mass and volume very
accurately in order to use the density of water as a
standard.

Q1.2

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

*Q1.3

In the base unit we have (a) 0.032 kg (b) 0.015 kg
(c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is
c=e>d>a>b

Q1.4

1.6

Q1.1

No: A dimensionally correct equation need not be true.
Example: 1chimpanzee = 2 chimpanzee is dimensionally correct.
Yes: If an equation is not dimensionally correct, it cannot
be correct.

*Q1.5

The answer is yes for (a), (c), and (f ). You cannot add or subtract a number of apples and a
number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kg 2 m,
or the volume of a cube, (2 m)3. Thus we have (a) yes (b) no (c) yes(d) no (e) no (f ) yes

*Q1.6

41 € ≈ 41 € (1 L 1.3 €)(1 qt 1 L)(1 gal 4 qt) ≈ (10 1.3) gal ≈ 8 gallons, answer (c)

*Q1.7

The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and
(c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the
meterstick measurement.

*Q1.8

0.02(1.365) = 0.03. The result is (1.37 ±0.03) × 107 kg. So (d) 3 digits are significant.

SOLUTIONS TO PROBLEMS
Section 1.1
P1.1

Standards of Length, Mass, and Time

3
4
4
Modeling the Earth as a sphere, we find its volume as π r 3 = π ( 6.37 × 10 6 m ) = 1.08 × 10 21 m 3.
3
3
m 5.98 × 10 24 kg
Its density is then ρ = =
= 5.52 × 10 3 kg m 3 . This value is intermediate
21
3
V 1.08 × 10 m
between the tabulated densitiesof 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.

1

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P1.2

Chapter 1

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

ρ=

m
, we have
V

⎛ 10 9 mm 3 ⎞
m
1 kg
=
2
2
⎝⎠
π r h π (19.5 mm ) ( 39.0 mm ) ⎜ 1 m 3 ⎟

ρ = 2.15 × 10 4 kg m 3 .
P1.3

and ρgold =

*P1.4

m
for both. Then ρiron = 9.35 kg V
V
⎛ 19.3 × 10 3 kg/m 3 ⎞
= 23.0 kg .
= 9.35 kg ⎜
⎠
⎝ 7.86 × 10 3 kg/m 3 ⎟

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

. Next,

mgold
ρgold
and mgold
=
ρiron 9.35 kg

ρ = m / V and V = (4 / 3)π r 3 = (4 / 3)π (d /2)3 = π d 3 / 6 where d is the diameter.
Then ρ = 6 m / π d 3 =

6(1.67 × 10 −27 kg)
= 2.3 × 1017 kg/m 3
π (2.4 × 10 −15 m)3

2.3 × 1017 kg/m 3 /(11.3 × 10 3 kg/m 3 ) = it is 20 × 1012 times the density of lead .
P1.5

4
43
π r and the mass is m = ρV = ρ π r 3. We divide
3
3
this equation for the larger sphere by the same equation for the smaller:
For either sphere the volume is V=

m
ρ 4π r 3 3 r 3
=
=
= 5.
ms ρ 4π rs3 3 rs3
Then r = rs 3 5 = 4.50 cm (1.71) = 7.69 cm .

Section 1.2
P1.6

Matter and Model-Building

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 theatoms are separated by a distance
12
L = 0.200 nm , the diagonal planes are separated by
L + L2 = 0.141 nm .
2

Section 1.3

Dimensional Analysis

P1.7

(a)

This is incorrect since the units of [ ax ] are m 2 s 2 , while the units of [ v ] are m s .

(b)

−1
This is correct since the units of [ y ] are m, and cos ( kx ) is dimensionless if [ k ] is in m .

P1.8

(a)...