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Chapter 1
Measurement and Vectors
Conceptual Problems
1 • [SSM] Which of the following is not one of the base quantities in the SI system? ( a ) mass, ( b ) length, ( c ) energy, ( d ) time, ( e ) All of the above are base quantities.
Determine the Concept The base quantities in the SI system include mass, length,
and time. Force is not a base quantity. ( c ) is correct.
2 • In doing a calculation, you end up with m/s in the numerator and m/s^2 in the denominator. What are your final units? ( a ) m^2 /s 3 , ( b ) 1/s, ( c ) s 3 /m^2 , ( d ) s, ( e ) m/s.
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 ) is correct.
3 • The prefix giga means ( a ) 10^3 , ( b ) 10^6 , ( c ) 10^9 , ( d ) 10^12 , ( e ) 10^15.
Determine the Concept Consulting Table 1-1 we note that the prefix giga
means 10^9. ( c ) is correct.
4 • The prefix mega means ( a ) 10–9^ , ( b ) 10–6^ , ( c ) 10 –3^ , ( d ) 10 6 , ( e ) 10 9.
Determine the Concept Consulting Table 1-1 we note that the prefix mega
means 10^6. ( d ) is correct.
5 • [SSM] Show that there are 30.48 cm per foot. How many centimeters are there in one mile?
Picture the Problem We can use the facts that there are 2.540 centimeters in 1 inch and 12 inches in 1 foot to show that there are 30.48 cm per ft. We can then use the fact that there are 5280 feet in 1 mile to find the number of centimeters in one mile.
2 Chapter 1
Multiply 2.540 cm/in by 12 in/ft to find the number of cm per ft:
- 48 cm/ft ft
in 12 in
cm
- (^540) ⎟= ⎠
Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one mile:
- 609 10 cm/mi mi
5280 ft ft
- 48 cm = ×^5 ⎟ ⎠
Remarks: Because there are exactly 2.54 cm in 1 in and exactly 12 inches in 1 ft, we are justified in reporting four significant figures in these results.
6 • The number 0.000 513 0 has significant figures. ( a ) one, ( b ) three, ( c ) four, ( d ) seven, ( e ) eight.
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.
7 • The number 23.0040 has significant figures. ( a ) two, ( b ) three, ( c ) four, ( d ) five, ( e ) six.
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 • Force has dimensions of mass times acceleration. Acceleration has dimensions of speed divided by time. Pressure is defined as force divided by area. What are the dimensions of pressure? Express pressure in terms of the SI base units kilogram, meter and second.
Determine the Concept We can use the definitions of force and pressure, together with the dimensions of mass, acceleration, and length, to find the dimensions of pressure. We can express pressure in terms of the SI base units by substituting the base units for mass, acceleration, and length in the definition of pressure.
4 Chapter 1
Determine the Concept Let the + x direction be to the right and the + y direction be upward. The figure shows the vector A
r − pointing in the − y direction and three unlabeled possibilities for vector B.
r Note that the choices for B
r start at the end of vector A
r − rather than at its initial point.
A
r −
x
y
Several B choices
r
13 • [SSM] Is it possible for three equal magnitude vectors to add to zero? If so, sketch a graphical answer. If not, explain why not.
Determine the Concept In order for the three equal magnitude vectors to add to zero, the sum of the three vectors must form a triangle. The equilateral triangle shown to the right satisfies this condition for the vectors A
r , B
r , and C
r for which it is true that A = B = C, whereas A + B + C = 0.
r r r A
r
B
r C
r
Estimation and Approximation
14 • The angle subtended by the moon’s diameter at a point on Earth is about 0.524º (Fig. 1-2). Use this and the fact that the moon is about 384 Mm away to find the diameter of the moon. HINT: The angle can be determined from the diameter of the moon and the distance to the moon.
Picture the Problem Let θ represent the angle subtended by the moon’s
diameter, D represent the diameter of the moon, and r m the distance to the moon.
Because θ is small, we can approximate it by θ ≈ D/r m where θ 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 Earth
θ and the Earth-moon distance r m:
D = θ r m
Substitute numerical values and evaluate D : ( )
- 51 10 m
384 Mm 360
2 rad
- 524
= ×^6
= °×
D
Measurement and Vectors 5
15 • [SSM] Some good estimates about the human body can be made if it is assumed that we are made mostly of water. The mass of a water molecule is 29.9 × 10 −^27 kg. If the mass of a person is 60 kg, estimate the number of water molecules in that person.
Picture the Problem We can estimate the number of water molecules in a person whose mass is 60 kg by dividing this mass by the mass of a single water molecule.
Letting N represent the number of water molecules in a person of mass m human body, express N in terms of m human body and the mass of a water molecule m water molecule:
watermolecule
humanbody m
m N =
Substitute numerical values and evaluate N :
- 0 10 molecules
molecule
- 9 10 kg
60 kg
27
27
= ×
×
−
N
16 •• In 1989, IBM scientists figured out how to move atoms with a scanning tunneling microscope (STM). One of the first STM pictures seen by the general public was of the letters IBM spelled with xenon atoms on a nickel surface. The letters IBM were 15 xenon atoms across. If the space between the centers of adjacent xenon atoms is 5 nm (5 × 10 −^9 m), estimate how many times could ″IBM″ could be written across this 8.5 inch page.
Picture the Problem We can estimate the number of times N that ″IBM″ could be written across this 8.5-inch page by dividing the width w of the page by the distance d required by each writing of ″IBM.″
Express N in terms of the width w of the page and the distance d required by each writing of ″IBM″:
d
N = w
Express d in terms of the separation s of the centers of adjacent xenon atoms and the number n of xenon atoms in each writing of ″IBM″:
d = sn
Substitute for d in the expression for N to obtain: sn
N = w
Measurement and Vectors 7
( b ) Express the required landfill volume V in terms of the volume of diapers to be buried:
V = NV onediaper
Substitute numerical values and evaluate V :
( ) 6 3 10 3 3
- 5 10 m L
10 m diaper
0. 5 L
- 1 10 diapers ⎟⎟≈ × ⎠
= × ×
− V
( c ) Express the required volume in terms of the volume of a rectangular parallelepiped:
V = Ah ⇒ h
A = V
Substitute numerical values evaluate A :
(^635). 5 105 m 2 10 m
A =^5.^5 ×^10 m = ×
Use a conversion factor (see Appendix A) to express this area in square miles: (^2)
2
5 2 2
- 2 mi
1 km
- 3861 mi
- 5 10 m
≈
A = × ×
18 •• ( a ) Estimate the number of gallons of gasoline used per day by automobiles in the United States and the total amount of money spent on it. ( b ) If 19.4 gal of gasoline can be made from one barrel of crude oil, estimate the total number of barrels of oil imported into the United States per year to make gasoline. How many barrels per day is this?
Picture the Problem The population of the United States is roughly 3 × 108 people. Assuming that the average family has four people, with an average of two cars per family, there are about 1.5 × 108 cars in the United States. If we double that number to include trucks, cabs, etc., we have 3 × 108 vehicles. Let’s assume that each vehicle uses, on average, 14 gallons of gasoline per week and that the United States imports half its oil.
( 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 = $3.00, find the daily cost C of gasoline:
( )( )
2 billion dollars/d
$ 18 10 /d
6 10 gal/d $ 3. 00 /gal 8
8
= ×
C = GP = ×
8 Chapter 1
( b ) Relate the number of barrels N of crude oil imported 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
fG t n
fY N
where f is the fraction of the oil that is imported.
Substitute numerical values and estimate N : (^ )
y
barrels 6 10
barrel
gallons
y
24 d d
5 6 10 gallons
9
8
≈ ×
⎛ ×
N =
Convert barrels/y to barrels/d to obtain:
d
barrels 2 10
365.24d
1 y y
6 10 barrels
7
9
≈ ×
N = × ×
19 •• [SSM] A megabyte (MB) is a unit of computer memory storage. A CD has a storage capacity of 700 MB and can store approximately 70 min of high-quality music. ( a ) If a typical song is 5 min long, how many megabytes are required for each song? ( b ) If a page of printed text takes approximately 5 kilobytes, estimate the number of novels that could be saved on a CD.
Picture the Problem We can set up a proportion to relate the storage capacity of a CD to its playing time, the length of a typical song, and the storage capacity required for each song. In ( b ) we can relate the number of novels that can be stored on a CD to the number of megabytes required per novel and the storage capacity of the CD.
( a ) Set up a proportion relating the ratio of the number of megabytes on a CD to its playing time to the ratio of the number of megabytes N required for each song:
70 min 5 min
700 MB N
Solve this proportion for N to obtain:
( 5 min) 50 MB 70 min
700 MB =
N =⎛
10 Chapter 1
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 μW= 40 × 10 −^6 W= 0.000040W
( c ) 3 MW= 3 × 106 W= 3 , 000 , 000 W
( b )
4 ns= 4 × 10 −^9 s= 0.00000000 4 s
( d ) 25 km= 25 × 103 m= 25 , 000 m
22 • Write the following (which are not SI units) using prefixes (but not their abbreviations). For example, 10^3 meters = 1 kilometer: ( a ) 10–12^ boo, ( b ) 10 9 low, ( c ) 10–6^ phone, ( d ) 10–18^ boy, ( e ) 10^6 phone, ( f ) 10–9^ goat, ( g ) 10^12 bull.
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 −^12 boo= 1 picoboo ( e ) 10 6 phone= 1 megaphone
( b ) 10 9 low= 1 gigalow ( f ) 10 −^9 goat= 1 nanogoat
( c ) 10 −^6 phone= 1 microphone ( g ) 1012 bull= 1 terabull
( d ) 10 −^18 boy= 1 attoboy
23 •• [SSM] In the following equations, the distance x is in meters, the time t is in seconds, and the velocity v is in meters per second. What are the SI units of the constants C 1 and C 2? ( a ) x = C 1 + C 2 t , ( b ) x = 12 C 1 t^2 , ( c ) v^2 = 2 C 1 x , ( d ) x = C 1 cos C 2 t , ( e ) v^2 = 2 C 1 v – ( C 2 x ) 2.
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
Measurement and Vectors 11
( 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 a trigonometric function must be dimensionless; i.e. without units. Therefore, because x is in meters:
1 1 is^ inm; 2 isins
C C −
( e ) All of the terms in the expression must have the same units. Therefore, because v is in m/s:
1 1 is^ inm/s; 2 isins
C C −
24 •• If x is in feet, t is in milliseconds, and v is in feet per second, what are the units of the constants C 1 and C 2 in each part of Problem 23?
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, C 1 and C 2 t must be in feet:
C 1 is inft; C 2 isinft/ms
( b ) Because x is in feet, 21 C 1 t^2
must be in feet:
C 1 isinft/ ( ms)^2
( c ) Because v^2 is in ft 2 /(ms) 2 , 2 C 1 x must be in ft 2 /s^2 :
C 1 isinft/ ( ms)^2
( d ) The argument of a trigonometric function must be dimensionless; that is, without units. Therefore, because x is in feet:
C 1 is inft; C 2 isin ( ms)− 1
( e ) The argument of an exponential function must be dimensionless; that is, without units. Therefore, because v is in ft/s:
C 1 is inft/ms; C 2 isin (ms ) −^1
Conversion of Units
Measurement and Vectors 13
Convert v into mi/h:
1.53 10 mi/h
1.609km
1 mi h
2.47 10 km
3
3
= ×
v =⎛^ ×
27 • A basketball player is 6 ft 10 12 in tall. What is his height in
centimeters?
Picture the Problem We’ll first express his height in inches and then use the conversion factor 1 in = 2.540 cm.
Express the player’s height in inches: (^) 10.5in 82.5in ft
12 in h = 6 ft× + =
Convert h into cm: (^210) cm in
h = 8 2.5 in×2.540cm=
28 • Complete the following: ( a ) 100 km/h = mi/h, ( b ) 60 cm = in, ( c ) 100 yd = m.
Picture the Problem We can use the conversion factors 1 mi = 1.609 km, 1 in = 2.540 cm, and 1 m = 1.094 yd to perform these conversions.
( a ) Convert h
km 100 to : h
mi
62.2mi/h
1.609km
1 mi h
100 km h
100 km
=
= ×
( b ) Convert 60 cm to in:
24 in
23.6in 2.540cm
1 in 60 cm 60 cm
=
= × =
( c ) Convert 100 yd to m:
91.4m
1.094yd
100 yd 100 yd^1 m
=
= ×
29 • The main span of the Golden Gate Bridge is 4200 ft. Express this distance in kilometers.
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.
14 Chapter 1
Convert 4200 ft into km by multiplying by 1 in the form
5280 ft
- 609 km: 1.28km
5280 ft
4200 ft 4200 ft 1.609km
=
= ×
30 • Find the conversion factor to convert from miles per hour into kilometers per hour.
Picture the Problem Let v be the speed of an object in mi/h. We can use the conversion factor 1 mi = 1.609 km to convert this speed to km/h.
Multiply v mi/h by 1.609 km/mi to convert v to km/h:
- 61 km/h mi
1.609km h
mi h
v mi^ = v × = v
31 • Complete the following: ( a ) 1.296 × 10 5 km/h^2 = km/(h⋅s), ( b ) 1.296 × 10 5 km/h^2 = m/s^2 , ( c ) 60 mi/h = ft/s, ( d ) 60 mi/h = m/s.
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 ) 36.00km/h s 3600 s
1 h h
km
- 296 10 h
km
- (^296105 252) ⎟⎟= ⋅ ⎠
× =⎛^ ×
( b ) 2
(^23) 2
5 2
(^5) 10.00m/s km
10 m 3600 s
1 h h
296 10 km h
296 10 km = ⎟⎟⎠
× =⎛^ ×
( c ) 88 ft/s 3600 s
1 h 1 mi
5280 ft h
mi 60 h
mi (^60) ⎟⎟= ⎠
( d ) 26. 8 m/s 27 m/s 3600 s
1 h km
10 m 1 mi
1.609km h
60 mi h
60 mi^3 = = ⎟⎟⎠
32 • There are 640 acres in a square mile. How many square meters are there in one acre?
Picture the Problem We can use the conversion factor given in the problem statement and the fact that 1 mi = 1.609 km to express the number of square meters in one acre. Note that, because there are exactly 640 acres in a square mile, 640 acres has as many significant figures as we may wish to associate with it.
16 Chapter 1
34 •• A right circular cylinder has a diameter of 6.8 in and a height of 2.0 ft. What is the volume of the cylinder in ( a ) cubic feet, ( b ) cubic meters, ( c ) liters?
Picture the Problem The volume of a right circular cylinder is the area of its base multiplied by its height. Let d represent the diameter and h the height of the right circular cylinder; use conversion factors to express the volume V in the given units.
( a ) Express the volume of the cylinder:
V = 41 π d^2 h
Substitute numerical values and evaluate V :
( ) ( )
( ) ( )
3 3
2 2 41
2 41
- 504 ft 0. 50 ft
12 in
1 ft 6.8in 2.0ft
6.8in 2.0ft
V π
( b ) Use the fact that 1 m = 3.281 ft to convert the volume in cubic feet into cubic meters:
( )
3
3
3 3
0.014m
0.0143m 3.281ft
0.504ft^1 m
=
V =
( c ) Because 1 L = 10−^3 m^3 : ( ) 14 L 10 m
1 L
0.0143m (^3 33) ⎟⎟= ⎠
V = −
35 •• [SSM] In the following, x is in meters, t is in seconds, v is in meters per second, and the acceleration a is in meters per second squared. Find the SI units of each combination: ( a ) v^2 / x , ( b ) x a , ( c ) 12 at^2.
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 : (^) m/s s s
m (^2) 2 = =
Measurement and Vectors 17
( c ) Noting that the constant factor 21 has no units, express and simplify the units of 21 at^2 :
( )s m s
m (^2) 2 ⎟⎠ =
Dimensions of Physical Quantities
36 • What are the dimensions of the constants in each part of Problem 23?
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 x = 21 C 1 t 2 T^2 T
L
L
( e ) v^2 = 2 C 1 v − ( C 2 ) 2 x^2 2
2 L T
T
L
T
L
T
L
( c ) v^2 = 2 C 1 x
L T
L
T
L
2 2
2
37 • The law of radioactive decay is N ( t ) = N 0 e −^ λ t , where N 0 is the number
of radioactive nuclei at t = 0, N ( t ) is the number remaining at time t , and λ is a
quantity known as the decay constant. What is the dimension of λ?
Picture the Problem Because the exponent of the exponential function must be
dimensionless, the dimension of λ must be T −^1.
38 •• The SI unit of force, the kilogram-meter per second squared (kg⋅m/s 2 ) is called the newton (N). Find the dimensions and the SI units of the constant G in Newton’s law of gravitation F = Gm 1 m 2 / r^2.
Measurement and Vectors 19
40 •• Show that the product of mass, acceleration, and speed has the dimensions of power.
Picture the Problem We note from Table 1-2 that the dimensions of power are ML^2 / T^3. The dimensions of mass, acceleration, and speed are M, L / T^2 , and L / T respectively.
Express the dimensions of mav : [ ] (^3)
2 (^2) T
ML
T
L
T
mav =M ×L× =
From Table 1-2: [ ] (^3)
2 T
ML
P =
Comparing these results, we see that the product of mass, acceleration, and speed has the dimensions of power.
41 •• [SSM] The momentum of an object is the product of its velocity and mass. Show that momentum has the dimensions of force multiplied by time.
Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T^2.
Express the dimensions of momentum:
[ ] T
ML
T
L
mv =M × =
From Table 1-2: (^) [ ] T^2
F =ML
Express the dimensions of force multiplied by time:
[ ] T
ML
T
T
ML
Ft = 2 × =
Comparing these results, we see that momentum has the dimensions of force multiplied by time.
42 •• What combination of force and one other physical quantity has the dimensions of power?
Picture the Problem Let X represent the physical quantity of interest. Then we can express the dimensional relationship between F, X, and P and solve this relationship for the dimensions of X.
20 Chapter 1
Express the relationship of X to force and power dimensionally:
[ F ][ X ] =[ P ]
Solve for [ X ]: (^) [ ] [ ] [ (^) F ]
X = P
Substitute the dimensions of force and power and simplify to obtain: (^) [ ] T
L
T
ML
T
ML
2
3
2
X = =
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.
43 •• [SSM] When an object falls through air, there is a drag force that depends on the product of the cross sectional area of the object and the square of its velocity, that is, F air = CAv^2 , where C is a constant. Determine the dimensions of C.
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
C = F
Substitute the dimensions of force, area, and velocity and simplify to obtain:
[ ] (^23) 2
2 L
M
T
L
L
T
ML
C =
44 •• Kepler’s third law relates the period of a planet to its orbital radius r , the constant G in Newton’s law of gravitation ( F = Gm 1 m 2 / r^2 ), and the mass of the sun M s. What combination of these factors gives the correct dimensions for the period of a planet?
