Solutions manual to Tipler -Physics for Scientists and Engineers with Modern Physics(5th Edit)

2958 Pages·1990·23.49 MB·English

by  Paul A. Tipler, Gene Mosca

Chapter 1
Systems of Measurement

Conceptual Problems

*1 •
Determine the Concept The fundamental physical quantities in the SI system include
mass, length, and time. Force, being the product of mass and acceleration, is not a
fundamental quantity. (c) is correct.

2 •
2
Picture the Problem We can express and simplify the ratio of m/s to m/s to determine
the final units.

Express and simplify the ratio of m
m/s to m/s2: s m ⋅ s2
= = s and (d) is correct.
m m ⋅s
2
s

3 •
Determine the Concept Consulting Table 1-1 we note that the prefix giga
9
means 10 . (c) is correct.

4 •
Determine the Concept Consulting Table 1-1 we note that the prefix mega
6
means 10 . (d ) is correct.

*5 •
Determine the Concept Consulting Table 1-1 we note that the prefix pico
−12
means 10 . (a) is correct.

6 •
Determine the Concept Counting from left to right and ignoring zeros to the left
of the first nonzero digit, the last significant figure is the first digit that is in doubt.
Applying this criterion, the three zeros after the decimal point are not significant figures,
but the last zero is significant. Hence, there are four significant figures in this number.
(c) is correct.

1
2 Chapter 1

7 •
Determine the Concept Counting from left to right, the last significant figure is
the first digit that is in doubt. Applying this criterion, there are six significant
figures in this number. (e) is correct.

8 •
Determine the Concept The advantage is that the length measure is always with you. The
disadvantage is that arm lengths are not uniform; if you wish to purchase a board of ″two
arm lengths″ it may be longer or shorter than you wish, or else you may have to physically
go to the lumberyard to use your own arm as a measure of length.

9 •
(a) True. You cannot add ″apples to oranges″ or a length (distance traveled) to a volume
(liters of milk).

(b) False. The distance traveled is the product of speed (length/time) multiplied by the
time of travel (time).

(c) True. Multiplying by any conversion factor is equivalent to multiplying by 1.
Doing so does not change the value of a quantity; it changes its units.

Estimation and Approximation

*10 ••
Picture the Problem Because θ is small, we can approximate it by θ ≈ D/rm
provided that it is in radian measure. We can solve this relationship for the diameter of
the moon.

Express the moon’s diameter D in D = θ rm
terms of the angle it subtends at the
earth θ and the earth-moon distance
rm:

Find θ in radians: 2π rad
θ = 0.524°× = 0.00915rad
360°

Substitute and evaluate D: D = (0.00915rad)(384Mm)

6
= 3.51× 10 m
Systems of Measurement 3

*11 ••
Picture the Problem We’ll assume that the sun is made up entirely of hydrogen. Then we
can relate the mass of the sun to the number of hydrogen atoms and the mass of each.

Express the mass of the sun MS as M S = NHM H
the product of the number of
hydrogen atoms NH and the mass of
each atom MH:

Solve for NH: MS
NH =
MH

30
Substitute numerical values and 1.99 × 10 kg 57
evaluate NH: NH = 1.67× 10−27 kg = 1.19 × 10

12 ••
Picture the Problem Let P represent the population of the United States, r the rate of
consumption and N the number of aluminum cans used annually. The population of the
8
United States is roughly 3×10 people. Let’s assume that, on average, each person drinks
one can of soft drink every day. The mass of a soft-drink can is approximately
−2
1.8 ×10 kg.

(a) Express the number of cans N N = rP∆t
used annually in terms of the daily
rate of consumption of soft drinks r
and the population P:

Substitute numerical values and ⎛ 1can ⎞ 8
N = ⎜ ⎟(3 × 10 people)
approximate N: ⎜
⎝ person ⋅d ⎟⎠
⎛ d ⎞
×(1y)⎜365.24 ⎟
⎜ ⎟
⎝ y ⎠
11
≈ 10 cans

(b) Express the total mass of M = Nm
aluminum used per year for soft
drink cans M as a function of the
number of cans consumed and the
mass m per can:
4 Chapter 1

11 −2
Substitute numerical values and M = (10 cans/y)(1.8 × 10 kg/can)

evaluate M: 9
≈ 2 × 10 kg/y

(c) Express the value of the Value = ($1/ kg)M
aluminum as the product of M and = ($1/ kg)(2 × 109 kg/y)
the value at recycling centers: 9
= $2 × 10 / y
= 2 billion dollars/y

13 ••
Picture the Problem We can estimate the number of words in Encyclopedia Britannica
by counting the number of volumes, estimating the average number of pages per volume,
estimating the number of words per page, and finding the product of these measurements
and estimates. Doing so in Encyclopedia Britannica leads to an estimate of
approximately 200 million for the number of words. If we assume an average word
length of five letters, then our estimate of the number of letters in Encyclopedia
9
Britannica becomes 10 .

(a) Relate the area available for one 2 π 2
Ns = d where d is the diameter of the
2
letter s and the number of letters N 4
to be written on the pinhead to the pinhead.
area of the pinhead:

Solve for s to obtain: πd 2
s =
4N

Substitute numerical values and 2
evaluate s: π ⎢⎡(116 in)⎛⎜2.54 cm ⎞⎟⎥⎤
⎣ ⎝ in ⎠⎦ −8
s = 9 ≈ 10 m
4(10 )

(b) Express the number of atoms per s
n =
letter n in terms of s and the atomic datomic
spacing in a metal datomic:

−8
Substitute numerical values and 10 m
n = ≈ 20atoms
evaluate n: 5 × 10−10 atoms/m

*14 ••
8
Picture the Problem The population of the United States is roughly 3 × 10 people.
Assuming that the average family has four people, with an average of two cars per
Systems of Measurement 5

8
family, there are about 1.5 × 10 cars in the United States. If we double that number to
8
include trucks, cabs, etc., we have 3 × 10 vehicles. Let’s assume that each vehicle uses,
on average, about 12 gallons of gasoline per week.

8
(a) Find the daily consumption of G = (3×10 vehicles)(2gal/d)

gasoline G: 8
= 6×10 gal/d

8
Assuming a price per gallon C = GP = (6 × 10 gal/d)($1.50 / gal)

P = $1.50, find the daily cost C of 8
= $9 × 10 / d ≈ $1 billion dollars/d
gasoline:

(b) Relate the number of barrels N Y G∆t
N = =
of crude oil required annually to the n n
yearly consumption of gasoline Y
and the number of gallons of
gasoline n that can be made from
one barrel of crude oil:

8
Substitute numerical values and (6×10 gal/d) (365.24d/y)
N =
estimate N: 19.4 gal/barrel

10
≈ 10 barrels/y

15 ••
Picture the Problem We’ll assume a population of 300 million (fairly accurate as of
September, 2002) and a life expectancy of 76 y. We’ll also assume that a diaper has a
volume of about half a liter. In (c) we’ll assume the disposal site is a rectangular hole in
the ground and use the formula for the volume of such an opening to estimate the surface
area required.

(a) Express the total number N of N = nD
disposable diapers used in the
United States per year in terms of
the number of children n in diapers
and the number of diapers D used
by each child in 2.5 y:

Use the daily consumption, the 3diapers 365.24d
D = × ×2.5y
number of days in a year, and the d y
estimated length of time a child is in 3
≈3×10 diapers/child
diapers to estimate the number of
diapers D required per child:
6 Chapter 1

Use the assumed life expectancy to ⎛ 2.5y ⎞ 6
n = ⎜ ⎟(300 ×10 children)
estimate the number of children n in
⎝⎜ 76y ⎟⎠
diapers: 7
≈10 children

7
Substitute to obtain: N = (10 children)
3
× (3 × 10 diapers/child)
10
≈ 3 × 10 diapers

(b) Express the required landfill V = NVone diaper
volume V in terms of the volume of
diapers to be buried:

10
Substitute numerical values and V = (3 × 10 diapers)(0.5L/diaper)

evaluate V: 7 3
≈ 1.5 × 10 m

(c) Express the required volume in V = Ah
terms of the volume of a rectangular
parallelepiped:

7 3
Solve and evaluate h: V 1.5 × 10 m 6 2
A = = =1.5 × 10 m
h 10m

2
Use a conversion factor to express 6 2 1 mi
A =1.5 × 10 m ×
this area in square miles: 2.590km2

2
≈ 0.6mi

16 •••
Picture the Problem The number of bits that can be stored on the disk can be found
from the product of the capacity of the disk and the number of bits per byte. In part (b)
we’ll need to estimate (i) the number of bits required for the alphabet, (ii) the average
number of letters per word, (iii) an average number of words per line, (iv) an average
number of lines per page, and (v) a book length in pages.

(a) Express the number of bits Nbits Nbits = Nbytes (8bits/byte)
as a function of the number of bits 9
= (2 × 10 bytes)(8bits/byte)
per byte and the capacity of the hard
10
= 1.60 × 10 bits
disk Nbytes:
Systems of Measurement 7

(b) Assume an average of 8 bits characters bits
8 ×8 = 64
letters/word and 8 bits/character to character word word

estimate the number of bytes bytes
= 8
required per word: word

Assume 10 words/line and 60 words bytes bytes
600 ×8 = 4800
lines/page: page word page

Assume a book length of 300 pages bytes 6
300pages× 4800 =1.44×10 bytes
and approximate the number bytes page
required:

9
Divide the number of bytes per disk 2 × 10 bytes
N =
books 6
by our estimated number of bytes 1.44 × 10 bytes/book

required per book to obtain an
≈ 1400books
estimate of the number of books the
2-gigabyte hard disk can hold:

*17 ••
Picture the Problem Assume that, on average, four cars go through each toll station per
minute. Let R represent the yearly revenue from the tolls. We can estimate the yearly
revenue from the number of lanes N, the number of cars per minute n, and the $6 toll per
car C.
cars min h d $6
R = NnC =14lanes×4 ×60 ×24 ×365.24 × = $177M
min h d y car

Units

18 •
Picture the Problem We can use the metric prefixes listed in Table 1-1 and the
abbreviations on page EP-1 to express each of these quantities.

(a) (c)
6 −6
1,000,000 watts = 10 watts 3×10 meter = 3µm

= 1MW
(b) (d)
−3 3
0.002gram = 2 × 10 g = 2mg 30,000seconds = 30 × 10 s = 30ks
8 Chapter 1

19 •
Picture the Problem We can use the definitions of the metric prefixes listed in
Table 1-1 to express each of these quantities without prefixes.

(a) (c)
−6 6
40µW = 40 × 10 W = 0.000040 W 3MW = 3 × 10 W = 3,000,000 W
(b) (d)
−9 3
4ns = 4 × 10 s = 0.000000004s 25km = 25 × 10 m = 25,000m

*20 •
Picture the Problem We can use the definitions of the metric prefixes listed in
Table 1-1 to express each of these quantities without abbreviations.

−12 6
(a) 10 boo = 1picoboo (e) 10 phone = 1megaphone

9 −9
(b) 10 low = 1gigalow (f) 10 goat = 1nanogoat

−6 12
(c) 10 phone = 1microphone (g) 10 bull = 1terabull

−18
(d) 10 boy = 1attoboy

21 ••
Picture the Problem We can determine the SI units of each term on the right-hand side
of the equations from the units of the physical quantity on the left-hand side.

(a) Because x is in meters, C1 and C is in m; C is in m/s
1 2
C2t must be in meters:

2 2
(b) Because x is in meters, ½C1t C is in m/s
1
must be in meters:

2 2 2 2
(c) Because v is in m /s , 2C1x must C is in m/s
1
2 2
be in m /s :

−1
(d) The argument of trigonometric C is in m; C is in s
1 2
function must be dimensionless; i.e.
without units. Therefore, because x
Systems of Measurement 9

is in meters:

−1
(e) The argument of an exponential C is in m/s; C is in s
1 2
function must be dimensionless; i.e.
without units. Therefore, because v
is in m/s:

22 ••
Picture the Problem We can determine the US customary units of each term on the
right-hand side of the equations from the units of the physical quantity on the left-hand
side.

(a) Because x is in feet, C1 and C2t C is in ft; C is in ft/s
1 2
must be in feet:

2 2
(b) Because x is in feet, ½C1t must C is in ft/s
1
be in feet:

2 2 2 2
(c) Because v is in ft /s , 2C1x must C is in ft/s
1
2 2
be in ft /s :

−1
(d) The argument of trigonometric C is in ft; C is in s
1 2
function must be dimensionless; i.e.
without units. Therefore, because x
is in feet:

−1
(e) The argument of an exponential C is in ft/s; C is in s
1 2
function must be dimensionless; i.e.
without units. Therefore, because v
is in ft/s:

Conversion of Units

23 •
Picture the Problem We can use the formula for the circumference of a circle to find the
radius of the earth and the conversion factor 1 mi = 1.61 km to convert distances in meters
into distances in miles.

7
(a) The Pole-Equator distance is c = 4 × 10 m
one-fourth of the circumference:
10 Chapter 1

−7
(b) Use the formula for the c 4×10 m 6
R = = = 6.37 × 10 m
circumference of a circle to obtain: 2π 2π

(c) Use the conversion factors 7 1 km 1mi
c = 4 × 10 m× ×
3
1 km = 1000 m and 1 mi = 1.61 km: 10 m 1.61km

4
= 2.48 × 10 mi
and
6 1km 1mi
R = 6.37 × 10 m× ×
3
10 m 1.61km

3
= 3.96 × 10 mi

24 •
Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert
speeds in km/h into mi/h.

Find the speed of the plane in km/s: v = 2(340m/s) = 680m/s
⎛ m ⎞⎛ 1km ⎞⎛ s ⎞
= ⎜680 ⎟
⎜ 3 ⎟⎜3600 ⎟
⎝ s ⎠⎝10 m ⎠⎝ h ⎠
= 2450km/h

Convert v into mi/h: ⎛ km ⎞⎛ 1mi ⎞
v = ⎜2450 ⎟⎜ ⎟
⎜ ⎟
⎝ h ⎠⎝1.61km ⎠
= 1520mi/h

*25 •
Picture the Problem We’ll first express his height in inches and then use the
conversion factor 1 in = 2.54 cm.

Express the player’s height into inches: 12in
h = 6ft × +10.5in = 82.5in
ft

Convert h into cm: 2.54cm
h = 82.5in× = 210cm
in

26 •
Picture the Problem We can use the conversion factors 1 mi = 1.61 km,
1 in = 2.54 cm, and 1 m = 1.094 yd to complete these conversions.