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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 )iscorrect.
2 •
Picture the Problem We can express and simplify the ratio of m/s to m/s
2 to determine
the final units.
Express and simplify the ratio of
m/s to m/s
2 : (^) s
m s
m s
s
m
s
m 2
2
= and ( d )iscorrect.
3 •
Determine the Concept Consulting Table 1-1 we note that the prefix giga
means 10
9
. ( c )iscorrect.
4 •
Determine the Concept Consulting Table 1-1 we note that the prefix mega
means 10
6
. ( d )iscorrect.
*5 •
Determine the Concept Consulting Table 1-1 we note that the prefix pico
means 10
− 12
. ( a )iscorrect.
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 )iscorrect.
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 )iscorrect.
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/r m
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
terms of the angle it subtends at the
earth θ and the earth-moon distance
r m:
D = θ r m
Find θ in radians:
- 00915 rad 360
2 rad
- 524 = °
= °×
Substitute and evaluate D : (^ )(^ )
- 51 10 m
0.00915rad 384 Mm
6 = ×
D =
4 Chapter 1
Substitute numerical values and
evaluate M :
2 10 kg/y
10 cans/y 1. 8 10 kg/can
9
11 2
≈ ×
= ×
− M
( c ) Express the value of the
aluminum as the product of M and
the value at recycling centers:
2 billion dollars/y
$ 2 10 /y
$ 1 /kg 2 10 kg/y
Value $ 1 /kg
9
9
= ×
= ×
= M
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
Britannica becomes 10
9 .
( a ) Relate the area available for one
letter s
2 and the number of letters N
to be written on the pinhead to the
area of the pinhead:
2 2
4
Ns d
= where d is the diameter of the
pinhead.
Solve for s to obtain:
N
d s 4
2
Substitute numerical values and
evaluate s : (^ )
10 m 410
in
cm in 2. 54 8 9
2
16
1
− ≈
s
( b ) Express the number of atoms per
letter n in terms of s and the atomic
spacing in a metal d atomic:
d atomic
s n =
Substitute numerical values and
evaluate n :
20 atoms 5 10 atoms/m
10 m 10
8 ≈ ×
− n
*14 ••
Picture the Problem The population of the United States is roughly 3 × 10
8 people.
Assuming that the average family has four people, with an average of two cars per
Systems of Measurement 5
family, there are about 1.5 × 10
8 cars in the United States. If we double that number to
include trucks, cabs, etc., we have 3 × 10
8 vehicles. Let’s assume that each vehicle uses,
on average, about 12 gallons of gasoline per week.
( a ) Find the daily consumption of
gasoline G :
6 10 gal/d
3 10 vehicles 2 gal/d
8
8
= ×
G = ×
Assuming a price per gallon
P = $1.50, find the daily cost C of
gasoline:
$ 9 10 /d $1billion dollars/d
6 10 gal/d $ 1. 50 /gal
8
8
= × ≈
C = GP = ×
( b ) Relate the number of barrels N
of crude oil required annually to the
yearly consumption of gasoline Y
and the number of gallons of
gasoline n that can be made from
one barrel of crude oil:
n
G t
n
Y
N
Substitute numerical values and
estimate N :
10 barrels/y
19.4gal/barrel
6 10 gal/d 365. 24 d/y
10
8
×
N =
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
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:
N = nD
Use the daily consumption, the
number of days in a year, and the
estimated length of time a child is in
diapers to estimate the number of
diapers D required per child:
3 10 diapers/child
5 y y
24 d
d
3 diapers
3 ≈ ×
D = × ×
Systems of Measurement 7
( b ) Assume an average of 8
letters/word and 8 bits/character to
estimate the number of bytes
required per word: (^) word
bytes 8
word
bits 64 word
characters 8 character
bits 8
× =
Assume 10 words/line and 60
lines/page: page
bytes 4800 word
bytes 8 page
words 600 × =
Assume a book length of 300 pages
and approximate the number bytes
required:
- 44 10 bytes page
bytes 300 pages 4800
6 × = ×
Divide the number of bytes per disk
by our estimated number of bytes
required per book to obtain an
estimate of the number of books the
2-gigabyte hard disk can hold:
1400 books
- 44 10 bytes/book
2 10 bytes 6
9
books
×
×
N =
*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.
$ 177 M
car
y
d
- 24 d
h 24 h
min 60 min
cars R = NnC = 14 lanes× 4 × × × × =
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 )
1 MW
1 , 000 , 000 watts 10 watts
6
( c )
3 10 meter 3 m
6
× = μ
−
( b )
- 002 gram 2 10 g 2 mg
3 = × =
−
( d )
30 , 000 seconds 30 10 s 30 ks
3 = × =
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 )
40 W 40 10 W 0.000040W
6 = × =
−
( c )
3 MW 3 10 W 3 , 000 , 000 W
6 = × =
( b )
4 ns 4 10 s 0.00000000 4 s
9 = × =
−
( d )
25 km 25 10 m 25 , 000 m
3 = × =
*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.
( a ) 10 boo 1 picoboo
12
− ( e ) 10 phone 1 megaphone
6
( b ) 10 low 1 gigalow
9 = ( f ) 10 goat 1 nanogoat
9
−
( c ) 10 phone 1 microphone
6
− ( g ) 10 bull 1 terabull
12
( d ) 10 boy 1 attoboy
18
−
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, C 1 and
C 2 t must be in meters:
C 1 is inm; C 2 isinm/s
( b ) Because x is in meters, ½ C 1 t
2
must be in meters:
2 C 1 isinm/s
( c ) Because v
2 is in m
2 /s
2 , 2 C 1 x must
be in m
2 /s
2 :
2 C 1 isinm/s
( d ) The argument of trigonometric
function must be dimensionless; i.e.
without units. Therefore, because x
1 1 is^ inm; 2 isins
− C C
10 Chapter 1
( b ) Use the formula for the
circumference of a circle to obtain:
- 37 10 m 2
4 10 m
2
6
7 = ×
×
−
c R
( c ) Use the conversion factors
1 km = 1000 m and 1 mi = 1.61 km:
2.48 10 mi
1.61km
1 mi
10 m
1 km 4 10 m
4
3
7
= ×
c = × × ×
and
- 96 10 mi
1.61km
1 mi
10 m
1 km
- 37 10 m
3
3
6
= ×
R = × × ×
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: (^ )
2450 km/h
h
s 3600 10 m
1 km
s
m 680
2340 m/s 680 m/s
3
v = =
Convert v into mi/h:
1520 mi/h
1.61km
1 mi
h
km 2450
v =
*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: 10.5in 82.5in ft
12 in h = 6 ft× + =
Convert h into cm: 210 cm in
2.54cm h = 8 2.5 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.
Systems of Measurement 11
( a )
h
mi
1.61km
1 mi
h
km 100 h
km 100 = × =
( b ) 23.6in 2.54cm
1 in 60 cm= 60 cm× =
( c ) 91.4m 1.094yd
1 m 100 yd= 100 yd× =
27 •
Picture the Problem We can use the conversion factor 1.609 km = 5280 ft to convert the
length of the main span of the Golden Gate Bridge into kilometers.
Convert 4200 ft into km: 1.28km 5280 ft
1.609km 4200 ft= 4200 ft× =
*28 •
Picture the Problem Let v be the speed of an object in mi/h. We can use the conversion
factor 1 mi = 1.61 km to convert this speed to km/h.
Multiply v mi/h by 1.61 km/mi to
convert v to km/h:
- 61 km/h mi
1.61km
h
mi
h
mi v = v × = v
29 •
Picture the Problem Use the conversion factors 1 h = 3600 s, 1.609 km = 1 mi,
and 1 mi = 5280 ft to make these conversions.
( a ) h s
km
3600 s
1 h
h
km
- 296 10 h
km
- (^296102)
5 2
5
⋅
× = ×
( b ) (^2)
(^23)
2
5 2
5
s
m
km
10 m
3600 s
1 h
h
km
- 296 10 h
km
- (^29610) ⎟⎟= ⎠
× = ×
( c ) s
ft
- 0 3600 s
1 h
1 mi
5280 ft
h
mi 60 h
mi 60 = ⎟
( d ) s
m
- 8 3600 s
1 h
km
10 m
1 mi
1.609km
h
mi 60 h
mi 60
3
⎟⎟^ = ⎠
Systems of Measurement 13
*33 ••
Picture the Problem We can treat the SI units as though they are algebraic
quantities to simplify each of these combinations of physical quantities and
constants.
( a ) Express and simplify the units of
v
2 /x :
2 2
(^22)
s
m
m s
m
m
m s
⋅
( b ) Express and simplify the units of
x a :
s s m/s
m (^2) 2 = =
( c ) Noting that the constant factor
2
(^1) has no units, express and simplify
the units of
2 2
(^1) at :
( ) (s ) m
s
m s s
m (^2) 2
2 2 ⎟ = ⎠
Dimensions of Physical Quantities
34 •
Picture the Problem We can use the facts that each term in an equation must have the
same dimensions and that the arguments of a trigonometric or exponential function must
be dimensionless to determine the dimensions of the constants.
( a )
x = C 1 + C 2 t
T
T
L
L L
( d )
x = C 1 cos C 2 t
T
T
L L
( b ) 2 2 1 x =^1 C t
2 2 T T
L
L
( e )
v = C 1 exp( − C 2 t )
T
T T
L
T
L 1
( c )
v C 1 x
2 = 2
L
T
L
T
L
2 2
2
35 ••
Picture the Problem Because the exponent of the exponential function must be dimensionl
the dimension of λ must be.
− 1 T
14 Chapter 1
*36 ••
Picture the Problem We can solve Newton’s law of gravitation for G and
substitute the dimensions of the variables. Treating them as algebraic quantities
will allow us to express the dimensions in their simplest form. Finally, we can
substitute the SI units for the dimensions to find the units of G.
Solve Newton’s law of gravitation
for G to obtain: (^12)
2
m m
Fr G =
Substitute the dimensions of the
variables: 2
3
2
2 2
MT
L
M
L
T
ML
G =
×
Use the SI units for L, M, and T :
2
3
kg s
m Units of are ⋅
G
37 ••
Picture the Problem Let m represent the mass of the object, v its speed, and r the
radius of the circle in which it moves. We can express the force as the product of
m , v , and r (each raised to a power) and then use the dimensions of force F, mass m,
speed v, and radius r to obtain three equations in the assumed powers. Solving these
equations simultaneously will give us the dependence of F on m, v, and r.
Express the force in terms of
powers of the variables:
a b c F = mvr
Substitute the dimensions of the
physical quantities:
c
b a L T
L
MLT M ⎟
− 2
Simplify to obtain:
a bc b MLT M L T
− + −
2
Equate the exponents to obtain: a = 1,
b + c = 1, and
− b = − 2
Solve this system of equations to
obtain:
a = 1, b = 2, and c = − 1
Substitute in equation (1):
r
v F mvr m
2 2 1 = =
−
16 Chapter 1
Substitute the dimensions of force
and power and simplify to obtain: [ ] T
L
T
ML
T
ML
X = =
2
3
2
Because the dimensions of velocity
are L / T, we can conclude that:
[ P ] =[ F ][ ] v
Remarks: While it is true that P = Fv , dimensional analysis does not reveal the
presence of dimensionless constants. For example, if P = πFv, the analysis shown
above would fail to establish the factor of π.
*41 ••
Picture the Problem We can find the dimensions of C by solving the drag force
equation for C and substituting the dimensions of force, area, and velocity.
Solve the drag force equation for
the constant C :
2
air
Av
F
C =
Express this equation
dimensionally:
[ ]
[ ]
[ ][ ]
2
air
A v
F
C =
Substitute the dimensions of force,
area, and velocity and simplify to
obtain:
[ ] 2 3 2
2
L
M
T
L
L
T
ML
C =
42 ••
Picture the Problem We can express the period of a planet as the product of these
factors (each raised to a power) and then perform dimensional analysis to
determine the values of the exponents.
Express the period T of a planet as
the product of
a b c r , G ,and M S:
a b c T = CrGM S (1)
where C is a dimensionless constant.
Solve the law of gravitation for the
constant G : 1 2
2
m m
Fr G =
Express this equation dimensionally: [ ]
[ ][ ]
[ 1 ][ 2 ]
2
m m
F r G =
Systems of Measurement 17
Substitute the dimensions of F , r ,
and m : [ ]
( )
2
3
2 2
MT
L
M M
L
T
ML
G =
×
×
Noting that the dimension of time is
represented by the same letter as is
the period of a planet, substitute the
dimensions in equation (1) to
obtain:
( ) ( )
c
b a M MT
L
T L
3
Introduce the product of M
0 and L
0
in the left hand side of the equation
and simplify to obtain:
c b a b b M LT M L T
0 0 1 − + 3 − 2
Equate the exponents on the two
sides of the equation to obtain:
0 = c – b ,
0 = a + 3 b , and
1 = –2 b
Solve these equations
simultaneously to obtain:
2
1 2
1 2 a = 3 , b =− ,and c =−
Substitute in equation (1): (^32)
S
12 S
3 2 12 r GM
C
T = Cr G M =
− −
Scientific Notation and Significant Figures
*43 •
Picture the Problem We can use the rules governing scientific notation to express each
of these numbers as a decimal number.
( a ) 3 10 30 , 000
4 × = ( c ) 4 10 0. 000004
6 × =
−
( b ) 6. 2 10 0. 0062
3 × =
− ( d ) 2. 17 10 217 , 000
5 × =
44 •
Picture the Problem We can use the rules governing scientific notation to express each
of these measurements in scientific notation.
( a ) 3.1GW 3.1 10 W
9 = × ( c ) 2. 3 fs 2.3 10 s
− 15 = ×
Systems of Measurement 19
( c ) Express both terms in scientific
notation and note that the second
has only three significant figures.
Hence the result will have only
three significant figures.
4
4
4 4
4
= ×
= + ×
= × + ×
+ ×
( d ) Because the divisor has three
significant figures, the result will
have three significant figures.
4 3 1.^5210
- 17 10
= ×
×
−
*47 •
Picture the Problem Let N represent the required number of membranes and
express N in terms of the thickness of each cell membrane.
Express N in terms of the thickness
of a single membrane: 7 nm
1 in N =
Convert the units into SI units and
simplify to obtain:
6
9
10 m
1 nm
100 cm
1 m
in
2.54cm
7 nm
1 in
= ×
N = × × × −
48 •
Picture the Problem Apply the general rules concerning the multiplication,
division, addition, and subtraction of measurements to evaluate each of the
given expressions.
( a ) Both factors and the result have
three significant figures:
4 2 3
- 00 × 10 6. 10 × 10 = 1. 22 × 10
−
( b ) Because the second factor has
three significant figures, the result
will have three significant figures:
5 6
- 141592 4. 00 × 10 = 1. 26 × 10
( c ) Both factors and the result have
three significant figures:
5 8
3
- 00 10
- 16 10
= ×
×
×
( d ) Write both terms using the same
power of 10. Note that the result
will have only three significant
figures:
3
3
3 3
3 2
= ×
= + ×
= × + ×
× + ×
20 Chapter 1
( e ) Follow the same procedure used
in ( d ):
2
2 2
2 5
= ×
= × + ×
× + ×
−
*49 •
Picture the Problem Apply the general rules concerning the multiplication,
division, addition, and subtraction of measurements to evaluate each of the
given expressions.
( a ) The second factor and the
result have three significant figures:
2 3
- 141592654 × 23. 2 = 1. 69 × 10
( b ) We’ll assume that 2 is exact.
Therefore, the result will have two
significant figures:
2 × 3. 141592654 × 0. 76 = 4. 8
( c ) We’ll assume that 4/3 is exact.
Therefore the result will have two
significant figures:
π× =
( d ) Because 2.0 has two significant
figures, the result has two significant
figures:
5
=
General Problems
50 •
Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert 100
km/h into mi/h.
Multiply 100 km/h by 1 mi/1.61 km
to obtain:
- 1 mi/h
1.61km
1 mi
h
km 100 h
km 100
= ×
*51 •
Picture the Problem We can use a series of conversion factors to convert 1 billion
seconds into years.
Multiply 1 billion seconds by the appropriate conversion factors to convert into years:
