physics for scientists and engineers 6th edition: CHAPTER 1

8/11/2019 physics for scientists and engineers 6th edition: CHAPTER 1

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1

Chapter 1

Measurement and Vectors

Conceptual Problems

1 [SSM] Which of the following is not one of the base quantities in theSI 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/s2in 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/s2to

determine the final units.

Express and simplify the ratio of

m/s to m/s2: ssm

sm

s

ms

m2

2

=

= and )(d is correct.

3 The prefix giga means (a) 103, (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 109. )(c is correct.

4 The prefix mega means (a) 109, (b) 10

6, (c) 103, (d) 10

6, (e) 109.

Determine the Concept Consulting Table 1-1 we note that the prefix mega

means 106. )(d is correct.

5 [SSM] Show that there are 30.48 cm per foot. How many centimetersare 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.


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Chapter 12

Multiply 2.540 cm/in by 12 in/ft to

find the number of cm per ft:cm/ft48.30

ft

in12

in

cm540.2 =

Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one

mile:

cm/mi10609.1mi

ft5280

ft

cm48.30 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 leftof 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 hasdimensions 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 baseunits 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 thedimensions 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.


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Measurement and Vectors 3

Use the definition of pressure and

the dimensions of force and area to

obtain:[ ]

[ ][ ] 22

2

LT

M

L

T

ML

===A

FP

Express pressure in terms of the SI

base units to obtain: 222

2 smkg

ms

mkg

mN =

=

9 True or false: Two quantities must have the same dimensions in orderto be multiplied.

False. For example, the distance traveled by an object is the product of its speed

(length/time) multiplied by its time of travel (time).

10 A vector has a negativex component and a positivey component. Itsangle measured counterclockwise from the positivexaxis is (a) between zero and90 degrees. (b) between 90 and 180 degrees. (c) More than 180 degrees.

Determine the ConceptBecause a vector with a negativex-component and a

positivey-component is in the second quadrant, its angle is between 90 and 180

degrees. ( ) correct.isb

11 [SSM]A vectorr

points in the +xdirection. Show graphically at

least three choices for a vector Br

such thatBrr

+ points in the +ydirection.

Determine the Concept The figureshows a vector

r pointing in the

positivex direction and three unlabeled

possibilities for vector .Br

Note that the

choices for Br

start at the end of vectorr

rather than at its initial point. Note

further that this configuration could be

in any quadrant of the reference system

shown. x

y

A

choicesSeveralB

12 A vector r points in the +ydirection. Show graphically at least three

choices for a vector Br

such that Brr

points in the +x direction.


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Chapter 14

Determine the Concept Let the +xdirection be to the right and the +ydirection be upward. The figure shows

the vectorr

pointing in the ydirection and three unlabeled

possibilities for vector .Br

Note that the

choices for Br start at the end of vectorr

rather than at its initial point.

A

x

y

choicesSeveralB

13 [SSM] Is it possible for three equal magnitude vectors to add tozero? 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 vectorsmust form a triangle. The equilateral

triangle shown to the right satisfies this

condition for the vectorsr

, Br

, and

Cr

for which it is true that A = B = C,

whereas .0=++ CBArrr

A

B

C

Estimation and Approximation

14 The angle subtended by the moons diameter at a point on Earth is

about 0.524 (Fig. 1-2). Use this and the fact that the moon is about 384 Mm awayto find the diameter of the moon.HINT: The angle can be determined from thediameter of the moon and the distance to the moon.

Picture the Problem Let represent the angle subtended by the moons

diameter,D represent the diameter of the moon, and rmthe distance to the moon.

Because is small, we can approximate it by D/rm where is in radianmeasure. We can solve this relationship for the diameter of the moon.

Express the moons diameterD in

terms of the angle it subtends at Earthand the Earth-moon distance rm:

mrD =

Substitute numerical values and

evaluateD:( )

m1051.3

Mm384360

rad2524.0

6=

=

D


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15 [SSM] Some good estimates about the human body can be made if itis assumed that we are made mostly of water. The mass of a water molecule is

29.9 1027 kg. If the mass of a person is 60 kg, estimate the number of watermolecules in that person.

Picture the ProblemWe 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

mhuman body, expressNin terms of

mhuman bodyand the mass of a water

molecule mwater molecule:

moleculewater

bodyhuman

m

mN=


evaluateN:

molecules100.2

molecule

kg109.29

kg60

27

27

=

= N

16 In 1989, IBM scientists figured out how to move atoms with ascanning tunneling microscope (STM). One of the first STM pictures seen by thegeneral public was of the letters IBM spelled with xenon atoms on a nickelsurface. The letters IBM were 15 xenon atoms across. If the space between the

centers of adjacent xenon atoms is 5 nm (5 109m), estimate how many timescould IBMcould be written across this 8.5 inch page.

Picture the ProblemWe can estimate the number of times N that IBM couldbe written across this 8.5-inch page by dividing the width wof the page by the

distance drequired by each writing of IBM.

ExpressN in terms of the width wof

the page and the distance d required

by each writing of IBM:d

wN=

Express din terms of the separations

of the centers of adjacent xenon

atoms and the number nof xenon

atoms in each writing of IBM:

snd=

Substitute for din the expression for

Nto obtain: sn

wN=


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Chapter 16


evaluateN:( )

( )

6

9103

15nm

m10nm5

in

cm540.2in5.8

=

=

N

17 There is an environmental debate over the use of cloth versusdisposable diapers. (a) If we assume that between birth and 2.5 y of age, a childuses 3 diapers per day, estimate the total number of disposable diapers used in theUnited States per year. (b) Estimate the total landfill volume due to these diapers,assuming that 1000 kg of waste fills about 1 m

3of landfill volume. (c) How many

square miles of landfill area at an average height of 10 m is needed for thedisposal of diapers each year?

Picture the Problem Well assume a population of 300 million and a life

expectancy of 76 y. Well also assume that a diaper has a volume of about half a

liter. In (c) well assume the disposal site is a rectangular hole in the ground anduse the formula for the volume of such an opening to estimate the surface area

required.

(a) Express the total numberN of

disposable diapers used in the United

States per year in terms of the

number of children nin diapers and

the number of diapersD used by

each child per year:

nDN= (1)

Use the estimated daily consumption

and the number of days in a year to

estimate the number of diapersD

required per child per year:yilddiapers/ch101.1

y

d24.365

dchild

diapers3

3

=D

Use the assumed life expectancy to

estimate the number of children n in

diapers yearly:

( )

children10

children10300y76

y5.2

7

6

=n

Substitute numerical values in equation

(1) to obtain:( )

diapers101.1

y

diapers103children10

10

37

=N


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(b) Express the required landfill

volume Vin terms of the volume of

diapers to be buried:

diaperoneNVV=

Substitute numerical values and evaluate V:

( ) 3633

10 m105.5L

m10

diaper

L5.0diapers101.1

=

V

(c) Express the required volume in

terms of the volume of a rectangular

parallelepiped:

AhV= h

VA =

Substitute numerical values evaluate

A:25

36

m105.5m10

m105.5

=

=A

Use a conversion factor (see

Appendix A) to express this area in

square miles: 2

2

225

mi2.0

km1

mi3861.0m105.5

=A

18 (a) Estimate the number of gallons of gasoline used per day byautomobiles in the United States and the total amount of money spent on it. (b) If19.4 gal of gasoline can be made from one barrel of crude oil, estimate the totalnumber of barrels of oil imported into the United States per year to makegasoline. How many barrels per day is this?

Picture the Problem The population of the United States is roughly 3 108people. Assuming that the average family has four people, with an average of two

cars per family, there are about 1.5 108cars in the United States. If we doublethat number to include trucks, cabs, etc., we have 3 108vehicles. Lets assumethat 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:( )( )

gal/d106

gal/d2vehicles103

8

8

=

=G

Assuming a price per gallon

P= $3.00, find the daily cost Cof

gasoline:

( )( )

dollars/dbillion2

d/1018$

gal/00.3$gal/d106

8

8

=

== GPC


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Chapter 18

(b) Relate the number of barrelsNof

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

tfG

n

fYN

==

wheref is the fraction of the oil that is

imported.


estimateN:( )

y

barrels106

barrel

gallons19.4

y

d24.365

d

gallons1065.0

9

8

=N

Convert barrels/y to barrels/d toobtain:

d

barrels102

d365.24y1

ybarrels106

7

9

=N

19 [SSM] A megabyte (MB) is a unit of computer memory storage. ACD has a storage capacity of 700 MB and can store approximately 70 min ofhigh-quality music. (a) If a typical song is 5 min long, how many megabytes arerequired for each song? (b) If a page of printed text takes approximately5 kilobytes, estimate the number of novels that could be saved on a CD.

Picture the ProblemWe 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 megabytesNrequired for each song:

min5min70

MB700 N=

Solve this proportion forNto

obtain:( ) MB50min5

min70

MB700=

=N


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(b) Letting nrepresent the number of

megabytes per novel, express the

number of novels novelsN that can be

stored on a CD in terms of the

storage capacity of the CD:

nN

MB700novels=

Assuming that a typical page in a

novel requires 5 kB of memory,

express nin terms of the number of

pagespin a typical novel:

pn

=

page

kB5

Substitute for n in the expression

for novelsN to obtain:p

N

=

page

kB5

MB700novels

Assuming that a typical novel has

200 pages:

novels107

novel

pages200

page

kB5

MB

kB10MB700

2

3

novels

=

=N

Units

20 Express the following quantities using the prefixes listed in Table 1-1and the unit abbreviations listed in the table Abbreviations for Units. For example,

10,000 meters = 10 km. (a) 1,000,000 watts, (b) 0.002 gram, (c) 3 106meter,(d) 30,000 seconds.

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)

MW1watts10watts000,000,1 6 ==

(c)

m3meter103 6 =

(b)

mg2g102gram002.0 3 ==

(d)

ks30s1003seconds000,30 3 ==

21 Write each of the following without using prefixes: (a) 40 W,(b) 4 ns, (c) 3 MW, (d) 25 km.


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Chapter 110

Picture the ProblemWe can use the definitions of the metric prefixes listed in

Table 1-1 to express each of these quantities without prefixes.

(a)

W0.000040W1040W40 6 ==

(c)

W000,000,3W103MW3 6 ==

(b)

s40.00000000s104ns4 9 ==

(d)

m000,52m1025km25 3 ==

22 Write the following (which are not SI units) using prefixes (but nottheir abbreviations). For example, 103meters = 1 kilometer: (a) 10

12boo,(b) 109low, (c) 106phone, (d) 1018boy, (e) 106phone, (f) 109goat,(g) 1012bull.

Picture the ProblemWe can use the definitions of the metric prefixes listed in

Table 1-1 to express each of these quantities without abbreviations.

(a) picoboo1boo10 12 = (e) megaphone1phone106 =

(b) gigalow1low109 = (f) nanogoat1goat10 9 =

(c) microphone1phone10 6 = (g) terabull1bull1012 =

(d) attoboy1boy10 18 =

23 [SSM] In the following equations, the distancex is in meters, thetime t is in seconds, and the velocityv is in meters per second. What are the SI

units of the constants C1and C2? (a)x = C1+ C2t, (b)x =12 C1t

2, (c) v2= 2C1x,

(d)x = C1cos C2t, (e) v2= 2C1v (C2x)

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) Becausexis in meters, C1and

C2tmust be in meters:m/sinism;inis 21 CC

(b) Becausexis in meters, C1t2

must be in meters:

2

1 m/sinisC


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(c) Because v2is in m2/s2, 2C1x

must be in m2/s

2:

2

1 m/sinisC

(d) The argument of a trigonometric

function must be dimensionless; i.e.

without units. Therefore, becausexis in meters:

1

21 sinism;inisCC

(e) All of the terms in the expression

must have the same units. Therefore,

because vis in m/s:

121 sinism/s;inis

CC

24 Ifx is in feet, t is in milliseconds, andv is in feet per second, what are

the units of the constants C1and C2in each part of Problem 23?

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

left-hand side.

(a) Becausexis in feet, C1and

C2tmust be in feet:ft/msinisft;inis 21 CC

(b) Becausexis in feet, 2121 tC

must be in feet:

( )21 msft/inisC

(c) Because v2is in ft2/(ms)2,

2C1xmust be in ft2/s

2:

( )21 msft/inisC

(d) The argument of a trigonometric

function must be dimensionless; that

is, without units. Therefore, because

xis in feet:

( ) 121 msinisft;inis

CC

(e) The argument of an exponential

function must be dimensionless; that

is, without units. Therefore, because

vis in ft/s:

( ) 121 msinisft/ms;inis

CC

Conversion of Units

25 From the original definition of the meter in terms of the distance alonga meridian from the equator to the North Pole, find in meters (a) the


12/54

Chapter 112

circumference of Earth and (b) the radius of Earth. (c) Convert your answers for(a) and (b) from meters into miles.

Picture the ProblemWe can use the formula for the circumference of a circle to

find the radius of Earth and the conversion factor 1 mi = 1.609 km to convert

distances in meters into distances in miles.

(a) The Pole-Equator distance is

one-fourth of the circumference:m1044 7equator-pole == DC

(b) The formula for the circumference

of a circle is:RDC 2==

2

CR=


evaluateR:

m106

m1037.62

m104

6

67

=

=

=

R

(c) Use the conversion factors

1 km = 1000 m and 1 mi = 1.609 km

to express Cin mi:mi102

km1.609

mi1

m10

km1m104

4

3

7

=

=C

Use the conversion factors

1 km = 1000 m and 1 mi = 1.61 km

to expressRin mi:mi104

km1.609

mi1

m10

km1m1037.6

3

3

6

=

=R

26 The speed of sound in air is 343 m/s at normal room temperature.What is the speed of a supersonic plane that travels at twice the speed of sound?Give your answer in kilometers per hour and miles per hour.

Picture the ProblemWe 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/h: ( )

km/h102.47

h

s3600

m10

km1

s

m686

m/s686m/s3432

3

3

=

=

==v


13/54


Convert vinto mi/h:

mi/h101.53

km1.609

mi1

h

km102.47

3

3

=

=v

27 A basketball player is 6 ft 10 1

2 in tall. What is his height incentimeters?

Picture the Problem Well first express his height in inches and then use the

conversion factor 1 in = 2.540 cm.

Express the players height in inches:in82.5in10.5

ft

in12ft6 =+=h

Convert hinto cm:cm210

in

cm2.540in2.58 ==h

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) Converth

km100 to :

h

mi

mi/h62.2

km1.609

mi1

h

km100

h

km100

=

=

(b) Convert 60 cm to in:

in24

in23.6cm2.540

in1cm60cm60

=

==

(c) Convert 100 yd to m:

m91.4

yd1.094

m1yd100yd100

=

=

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.


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Chapter 114

Convert 4200 ft into km by

multiplying by 1 in the form

ft5280

km609.1: km1.28

ft5280

km1.609ft4200ft4200

=

=

30 Find the conversion factor to convert from miles per hour into

kilometers per hour.

Picture the Problem Let vbe 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 vmi/h by 1.609 km/mi to

convert vto km/h:km/h61.1

mi

km1.609

h

mi

h

mivvv ==

31 Complete the following: (a) 1.296 105

km/h

2

= km/(hs),(b) 1.296 105km/h2= m/s2, (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) skm/h36.00s3600

h1

h

km10296.1

h

km10296.1

2

5

2

5 =

=

(b)2

32

2

5

2

5

m/s10.00km

m10

s3600

h1

h

km10296.1h

km10296.1 =

=

(c) ft/s88s3600

h1

mi1

ft5280

h

mi60

h

mi60 =

=

(d) m/s27m/s8.26s3600

h1

km

m10

mi1

km1.609

h

mi60

h

mi60

3

==

=

32 There are 640 acres in a square mile. How many square meters arethere 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.


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Multiply by 1 twice, properly chosen,

to convert one acre into square miles,

and then into square meters:

( )

2

22

m4045

mi

m1609

acres640

mi1acre1acre1

=

=

33 [SSM] You are a delivery person for the Fresh Aqua Spring WaterCompany. Your truck carries 4 pallets. Each pallet carries 60 cases of water. Eachcase of water has 24 one-liter bottles. You are to deliver 10 cases of water to eachconvenience store along your route. The dolly you use to carry the water into thestores has a weight limit of 250 lb. (a) If a milliliter of water has a mass of1 g, and a kilogram has a weight of 2.2 lb, what is the weight, in pounds, of all thewater in your truck? (b) How many full cases of water can you carry on the cart?

Picture the Problem The weight of the water in the truck is the product of the

volume of the water and its weight density of 2.2 lb/L.

(a) Relate the weight wof the water

on the truck to its volume V and

weight density (weight per unit

volume)D:

DVw=

Find the volume Vof the water:

L5760

)case

L24()

pallet

cases(60pallets)4(

=

=V

Substitute numerical values forD

andL and evaluate w:( )

lb103.1

lb10267.1L5760L

lb2.2

4

4

=

=

=w

(b) Express the number of cases of

water in terms of the weight limit of

the cart and the weight of each case

of water:

waterofcaseeachofweight

carttheoflimitweight=N


evaluateN:cases7.4

case

L24

L

lb2.2

lb250=

=N

cases.4carrycanYou


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Chapter 116

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 drepresent the diameter and hthe height of the right

circular cylinder; use conversion factors to express the volume Vin the givenunits.

(a) Express the volume of the

cylinder:

hdV 241 =


evaluate V:

( ) ( )

( ) ( )

33

2

2

41

2

41

ft50.0ft504.0

in12

ft1ft2.0in6.8

ft2.0in6.8

==

=

=

V

(b) Use the fact that 1 m = 3.281 ft

to convert the volume in cubic feet

into cubic meters:

( )

3

3

3

3

m0.014

m0.0143ft3.281

m1ft0.504

=

=

=V

(c) Because 1 L = 103m

3:

( ) L14m10

L10.0143m

33

3 =

=

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) v2/x, (b) ax , (c) 12 at2.

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 ofv

2/x: ( ) 22

22

sm

smm

msm ==

(b) Express and simplify the units of

ax :ss

m/s

m 22

==


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(c) Noting that the constant factor

21 has no units, express and simplify

the units of 221 at :

( ) mss

m 22

=

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 = C1 + C2 t

TT

LLL

(d)

x = C1 cos C2 t

TT

1LL

(b)2

121 tCx =

2

2T

T

LL

(e)

v2 = 2C1 v (C2)

2x2

2

2

LT

1

T

L

T

L

T

L

(c)

xCv 12 2=

LT

L

T

L22

2

37 The law of radioactive decay is ( ) teNtN = 0 , whereN0is the number

of radioactive nuclei at t = 0,N(t) is the number remaining at time t, and is a

quantity known as the decay constant. What is the dimension of ?

Picture the Problem Because the exponent of the exponential function must be

dimensionless, the dimension of must be .T 1

38 The SI unit of force, the kilogram-meter per second squared (kgm/s2)is called the newton (N). Find the dimensions and the SI units of the constant GinNewtons law of gravitationF= Gm1m2/r

2.


18/54

Chapter 118

Picture the Problem We can solve Newtons 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 Newtons law of gravitationfor Gto obtain:

21

2

mmFrG=

Substitute the dimensions of the

variables: [ ]2

3

2

2

2

MT

L

M

LT

ML

=

=G

Use the SI units forL,M,and Tto

obtain: 2

3

skg

m

39 The magnitude of the force (F) that a spring exerts when it is stretcheda distancex from its unstressed length is governed by Hookes law,F= kx.(a) What are the dimensions of theforce constant, k? (b) What are the dimensionsand SI units of the quantity kx2?

Picture the ProblemThe dimensions of mass and velocity areMandL/T,

respectively. We note from Table 1-2 that the dimensions of force areML/T2.

(a) We know, from Hookes law, that:

x

Fk=

Write the corresponding dimensional

equation:[ ] [ ]

[ ]xF

k =

Substitute the dimensions ofFandx

and simplify to obtain: [ ]2

2

T

M

L

T

LM

==k

(b) Substitute the dimensions of

kandx

2

and simplify to obtain:

[ ]2

22

2

2

T

MLL

T

M==kx

Substitute the units of kx2to obtain:

2

2

s

mkg


19/54


40 Show that the product of mass, acceleration, and speed has thedimensions of power.

Picture the ProblemWe note from Table 1-2 that the dimensions of power are

ML2/T3. The dimensions of mass, acceleration, and speed areM, L/T2, andL/T

respectively.

Express the dimensions of mav:[ ]

3

2

2 T

ML

T

L

T

LM ==mav

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 andmass. Show that momentum has the dimensions of force multiplied by time.

Picture the ProblemThe dimensions of mass and velocity are M and L/T,

respectively. We note from Table 1-2 that the dimensions of force are ML/T2.

Express the dimensions of

momentum:[ ]

T

ML

T

LM ==mv

From Table 1-2: [ ] 2TML

=F

Express the dimensions of force

multiplied by time:[ ]

T

MLT

T

ML2

==Ft

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 ProblemLetXrepresent the physical quantity of interest. Then we

can express the dimensional relationship betweenF,X,andPand solve this

relationship for the dimensions ofX.


20/54

Chapter 120

Express the relationship of X to

force and power dimensionally:

[ ][ ] [ ]PXF =

Solve for [ ]X : [ ] [ ][ ]FP

X =

Substitute the dimensions of force

and power and simplify to obtain: [ ]T

L

T

MLT

ML

2

3

2

==X

Because the dimensions of velocity

are L/T,we can conclude that:[ ] [ ][ ]vFP =

Remarks: While it is true thatP = Fv, dimensional analysis does not reveal

the presence of dimensionless constants. For example, if FvP= ,the analysisshown above would fail to establish the factor of .

43 [SSM] When an object falls through air, there is a drag force thatdepends on the product of the cross sectional area of the object and the square ofits velocity, that is,Fair= CAv

2, whereC is a constant. Determine the dimensionsof C.

Picture the ProblemWe can find the dimensions of Cby solving the drag force

equation for Cand substituting the dimensions of force, area, and velocity.

Solve the drag force equation for the

constant C: 2air

Av

FC=

Express this equation dimensionally: [ ] [ ][ ][ ]2

air

vA

FC =

Substitute the dimensions of force,

area, and velocity and simplify to

obtain:[ ]

32

2

2

L

M

T

L

L

T

ML

=

=C

44 Keplers third law relates the period of a planet to its orbital radius r,

the constant G in Newtons law of gravitation (F= Gm1m2/r2), and the mass of the

sunMs. What combination of these factors gives the correct dimensions for the

period of a planet?


21/54


Picture the ProblemWe can express the period of a planet as the product of the

factors r, G, and MS (each raised to a power) and then perform dimensional

analysis to determine the values of the exponents.

Express the period Tof a planet as

the product ofcba

MGr Sand,, :

cba MGCrT S= (1)

where Cis a dimensionless constant.

Solve the law of gravitation for the

constant G:21

2

mm

FrG=

Express this equation dimensionally:[ ] [ ][ ]

[ ][ ]21

2

mm

rFG =

Substitute the dimensions ofF, r,

and m: [ ]( )

2

32

2

MTL

MM

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:

( ) ( )cb

aT M

MT

LL

2

3

=

Introduce the product of M0and L0

in the left hand side of the equation

and simplify to obtain:

bbabc 23100 TLMTLM +=

Equating the exponents on the two

sides of the equation yields:

0 = c b, 0 = a + 3b, and 1 = 2b

Solve these equations

simultaneously to obtain:21

21

23 and,, === cba

Substitute for a, b, and c in equation

(1) and simplify to obtain:23

S

21

S

2123 rGM

CMGCrT ==

Scientific Notation and Significant Figures

45 [SSM] Express as a decimal number without using powers of 10

notation: (a) 3 104, (b) 6.2 103, (c) 4 106, (d) 2.17 105.


22/54

Chapter 122

Picture the Problem We can use the rules governing scientific notation to

express each of these numbers as a decimal number.

(a) 000,30103 4 = (c) 000004.0104 6 =

(b) 0062.0102.6 3 = (d) 000,2171017.2 5 =

46 Write the following in scientific notation: (a) 1345100 m = ____ km,(b) 12340. kW = ____ MW, (c) 54.32 ps = ____ s, (d) 3.0 m = ____ mm

Picture the Problem We can use the rules governing scientific notation to

express each of these numbers in scientific notation.

(a)

km103451.1m103451.1m13451003

6

==

(c)

s105.432s1054.32ps32.54

11

12

==

(b)

MW101.2340

W101.2340

kW101.2340kW.12340

1

7

4

=

=

=

(d)

mm100.3

m

mm10m3.0m0.3

3

3

=

=

47 [SSM] Calculate the following, round off to the correct number ofsignificant figures, and express your result in scientific notation:

(a) (1.14)(9.99 104), (b) (2.78 108) (5.31 109), (c) 12 /(4.56 103),(d) 27.6 + (5.99 102).

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 number of significant figures

in each factor is three; therefore theresult has three significant figures:

( )( ) 54 1014.11099.914.1 =


23/54


(b) Express both terms with the same

power of 10. Because the first

measurement has only two digits

after the decimal point, the result can

have only two digits after thedecimal point:

( ) ( )( )

8

8

98

1025.2

10531.078.2

1031.51078.2

=

=

(c) Well assume that 12 is exact.

Hence, the answer will have three

significant figures:

3

31027.8

1056.4

12=

(d) Proceed as in (b): ( )

2

2

1027.6

627

5996.271099.56.27

=

=

+=+

48 Calculate the following, round off to the correct number of significantfigures, and express your result in scientific notation: (a) (200.9)(569.3),

(b) (0.000000513)(62.3 107), (c) 28 401 + (5.78 104), (d) 63.25/(4.17 103).

Picture the Problem Apply the general rules concerning the multiplication,

division, addition, and subtraction of measurements to evaluate each of the

given expressions.

(a) Note that both factors have foursignificant figures. ( )( )5

10144.13.5699.200 =

(b) Express the first factor in scientific notation and note that both factors

have three significant figures.

( )( ) ( )( ) 2777 1020.3103.621013.5103.62000000513.0 ==

(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

444

1062.8

1078.5841.2

1078.510841.21078.528401

=

+=

+=+


24/54

Chapter 124

(d) Because the divisor has three

significant figures, the result will

have three significant figures.

4

31052.1

1017.4

25.63=

49 [SSM] A cell membrane has a thickness of 7.0 nm. How many cell

membranes would it take to make a stack 1.0 in high?

Picture the Problem LetNrepresent the required number of membranes and

expressNin terms of the thickness of each cell membrane.

ExpressNin terms of the thickness

of a single membrane: nm7.0

in1.0=N

Convert the units into SI units and simplify to obtain:

66

9 106.31063.3m10

nm1

cm100

m1

in

cm2.540

nm7.0

in1.0

=== N

50 A circular hole of radius 8.470 101cm must be cut into the frontpanel of a display unit. The toleranceis 1.0 103cm, which means the actualhole cannot differ by more than this much from the desired radius. If the actualhole is larger than the desired radius by the allowed tolerance, what is thedifference between the actual area and the desired area of the hole?

Picture the Problem Let r0represent the larger radius and r the desired radius of

the hole. We can find the difference between the actual and the desired area of the

hole by subtracting the smaller area from the larger area.

Express the difference between the

two areas in terms of rand r0:

( )220220 rrrrA ==

Factoring 220 rr to obtain: ( )( )rrrrA += 00

Substitute numerical values and evaluate A:

( ) ( )233113

cm103.5cm100.1cm10470.8cm10470.8cm100.1

=++= A

51 [SSM] A square peg must be made to fit through a square hole. Ifyou have a square peg that has an edge length of 42.9 mm, and the square holehas an edge length of 43.2 mm, (a) what is the area of the space available when


25/54


the peg is in the hole? (b) If the peg is made rectangular by removing 0.10 mm ofmaterial from one side, what is the area available now?

Picture the Problem Letshrepresent the side of the square hole and spthe side

of the square peg. We can find the area of the space available when the peg is in

the hole by subtracting the area of the peg from the area of the hole.

(a) Express the difference between

the two areas in terms ofshandsp:

2

p

2

h ssA =


evaluate A:( ) ( )

2

22

mm62

mm9.42mm2.43

=A

(b) Express the difference between

the area of the square hole and therectangular peg in terms ofsh ,spand

the new length of the peg lp:

pp2h lssA' =

Substitute numerical values and evaluate A:

( ) ( )( ) 22 mm03mm10.0mm9.42mm9.42mm2.43 =A'

Vectors and Their Properties

52 A vector 7.00 units long and a vector 5.50 units long are added. Theirsum is a vector 10.0 units long. (a) Show graphically at least one way that this canbe accomplished. (b) Using your sketch in Part (a), determine the angle betweenthe original two vectors.

Picture the Problem Letr

be the vector whose length is 7.00 units and let Brbe

the vector whose length is 5.50 units. BACrrr

+= is the sum ofr

and .Br

The fact

that their sum is 10.0 long tells us that the vectors are not collinear. We can use the

of the vectors to find the angle betweenr

and .Br

(a) A graphical representation of vectorsr

, Br

and Cr

is shown below. is the

angle betweenr

and .Br


26/54

Chapter 126

00.7=A

50.

5=

B

0.10=

C

x

(b) The components of the vectors

are related as follows:xxx CBA =+

and

yyy CBA =+

Substituting for the components

gives:

( ) ( ) cos0.10cos50.500.7 =+ and

( ) ( ) sin0.10sin50.5 =

Squaring and adding these equations yields:

( ) ( ) ( ) ( )[ ]2222 cos50.500.7sin50.5cos100sin100 ++=+ or

( )( ) ( ) ( )[ ]2222 cos50.500.7sin50.5cossin100 ++=+

Because 1cossin 22 =+ :

( ) ( )[ ]222 cos50.500.7sin50.5100 ++=

Solve this equation for (you can (1)use the same trigonometric identityused in the previous step to eliminate

either sin2or cos

2 in favor of the

other and then solve the resultingequation or (2) use your graphingcalculators SOLVER program) toobtain:

= 4.74

Remarks: You could also solve Part (b) of this problem by using the law of

cosines.

53 [SSM] Determine thex andy components of the following threevectors in thexyplane. (a) A 10-m displacement vector that makes an angle of

30clockwise from the +y direction. (b) A 25-m/s velocity vector that makes an


27/54


angle of 40counterclockwise from the xdirection. (c) A 40-lb force vectorthat makes an angle of 120counterclockwise from the ydirection.

Picture the ProblemThexandycomponents of these vectors are theirprojections onto thexandyaxes. Note that the components are calculated usingthe angle each vector makes with the +xaxis.

(a) Sketch the displacement vector

(call itr

) and note that it makes an

angle of 60with the +xaxis:

A

x

y

30

Ay

Ax

Find thexandycomponents ofr

: ( ) m0.560cosm10 ==xA

and

( ) m.7860sinm10 ==yA

(b) Sketch the velocity vector (call

it vr

) and note that it makes an

angle of 220with the +xaxis:

v

40

y

xxv

yv

Find thexandycomponents of vr

: ( ) m/s19220cosm/s52 ==xv and

( ) m/s16220sinm/s52 ==yv


28/54

Chapter 128

(c) Sketch the force vector (call itr

) and note that it makes an angle

of 30with the +x axis:

x

120

F

y

Find thexandycomponents ofr

:

( ) lb3530coslb04 ==xF and

( ) lb0230sinlb04 ==yF

54 Rewrite the following vectors in terms of their magnitude and angle(counterclockwise from the +xdirection). (a) A displacement vector with anx

component of +8.5 m and aycomponent of 5.5 m. (b) A velocity vector with anxcomponent of 75 m/s and aycomponent of +35 m/s. (c) A force vector with amagnitude of 50 lb that is in the third quadrant with anxcomponent whosemagnitude is 40 lb.

Picture the Problem We can use the Pythagorean Theorem to find magnitudes of

these vectors from their x and y components and trigonometry to find their

direction angles.

(a) Sketch the components of the

displacement vector (call it

r

) andshow their resultant:

y

x

A

m5.8=xA

m5.5

=y

A

Use the Pythagorean Theorem to

find the magnitude ofr

:( ) ( )

m10

m5.5m5.82222

=

+=+= yx AAA

Find the direction angle ofr

:

==

=

=

32733

m5.8

m5.5tantan 11

x

y

A

A


29/54


(b) Sketch the components of the

velocity vector (call it vr

) and

show their resultant:v

x

y

m/s

35

=

yv

m/s75=xv


find the magnitude of vr

and

trigonometry to find its direction

angle:

( ) ( )

m/s38

m/s53m/s752222

=

+=+= yx vvv

=

=

= 25

m/s57

m/s53tantan 11

x

y

v

v

and

=== 15525180180

(c) Sketch the force vector (call it Fr

)

and itsx-component and showFy:x

y

lb40=xF

yFlb

50=

F

Use trigonometry to find and,

hence, :=

=

= 37

lb50

lb40coscos 11

F

Fx

=+=+= 21737180180

Use the relationship between a

vector and its vertical component

to findFy:

( )

lb30

217sinlb50sin

=

== FFy

Remarks: In Part (c) we could have used the Pythagorean Theorem to find

the magnitude of Fy.

55 You walk 100 m in a straight line on a horizontal plane. If this walktook you 50 m east, what are your possible north or south movements? What arethe possible angles that your walk made with respect to due east?


30/54

Chapter 130

Picture the Problem There are two directions that you could have walked that

are consistent with your having walked 50 m to the east. One possibility is

walking in the north-east direction and the other is walking in the south-east

direction. We can use the Pythagorean Theorem to find the distance you walked

either north or south and then use trigonometry to find the two angles that

correspond to your having walked in either of these directions.

LettingArepresent the magnitude of

your displacement on the walk, use

the Pythagorean Theorem to relateA

to its horizontal (Ax) and vertical (Ay)

components:

22

xy AAA =


evaluateAy:( ) ( ) m87m50m100 22 ==yA

The plus-and-minus-signs mean that you could have gone 87 m north or 87 m

south.

Use trigonometry to relate the

directions you could have walked

to the distance you walked and its

easterly component:

=

= 60m100

m50cos 1

The plus-and-minus-signs mean that you could have walked 60north of east or

60south of east.

56 The final destination of your journey is 300 m due east of your startingpoint. The first leg of this journey is the walk described in Problem 55, and thesecond leg is also a walk along a single straight-line path. Estimate graphically thelength and heading for the second leg of your journey.

Picture the Problem Letr

be the vector whose length is 100 m and whose

direction is 60N of E. Let Br

be the vector whose direction and magnitude we

are to determine and assume that you initially walked in a direction north of east.

The graphical representation that we can use to estimate these quantities is shown

below. Knowing that the magnitude ofr is 100 m, use any convenient scale to

determine the length of Br

and a protractor to determine the value of .


31/54


32/54

Chapter 132


evaluateD:( ) ( ) 8.635.492.40 22 =+=D

Use trigonometry to express and

evaluate the angle :

=

=

=

51

2.40

5.49tantan 11

x

y

D

D

where the minus sign means thatDr

is

in the 4th quadrant.

58 Given the following force vectors:r

is 25 lb at an angle of 30clockwise from the +xaxis, and

rB is 42 lb at an angle of 50clockwise from the

+yaxis. (a) Make a sketch and visually estimate the magnitude and angle of the

vectorrC such that 2

r+

rC

rB results in a vector with a magnitude of

35 lb pointing in the +xdirection. (b) Repeat the calculation in Part (a) using themethod of components and compare your result to the estimate in (a).

Picture the Problem A diagram showing the condition that iBCA 352 =+rrr

and

from which one can scale the values of Cand is shown below. In (b) we can usethe two scalar equations corresponding to this vector equation to check ourgraphical results.

(a) A diagram showing the conditions

imposed by iBCA 352 =+rrr

approximately to scale is shown to the

right.

The magnitude of Cr

is approximately

lb57 and the angle is

approximately .68

B

B

A

2

C

50

50

30x

y

i

35

(b) Express the condition relating thevectors ,A

r ,Br

and :Cr

iBCA 352 =+

rrr

The corresponding scalar equations

are:

352 =+ xxx BCA

and

02 =+ yyy BCA


33/54


Solve these equations for Cxand Cy

to obtain:xxx BAC += 235

and

yyy BAC += 2

Substitute for thexandy

components ofr

and Br

to obtain:

( )[ ]

( )lb9.23

40coslb42

330coslb252lb35

=+

=xC

and

( )[ ] ( )

lb52

40sinlb42330sinlb252

=

+=yC


relate the magnitude of Cr

to its

components:

22yx CCC +=


evaluate C:( ) ( ) lb57lb0.52lb9.23 22 =+=C

Use trigonometry to find the

direction of Cr

:=

=

= 65

lb9.23

lb0.52tantan 11

x

y

C

C

Remarks: The analytical results for Cand are in excellent agreement with

the values determined graphically.

59 [SSM] Calculate the unit vector (in terms of iandj ) in the direction

opposite to the direction of each of the vectors in Problem 57.

Picture the ProblemThe unit vector in the direction opposite to the direction of avector is found by taking the negative of the unit vector in the direction of thevector. The unit vector in the direction of a given vector is found by dividing thegiven vector by its magnitude.

The unit vector in the direction ofr

is given by:22

yx AA +

==A

A

AA

r

r

r

Substitute for ,Ar

Ax, andAyand

evaluate :A ( ) ( )ji

jiA 81.059.0

7.44.3

7.44.322

+=+

+=


34/54

Chapter 134

The unit vector in the direction

opposite to that ofr

is given by:jiA 81.059.0 =

The unit vector in the direction of

Br

is given by: 22

yx BB +

==B

B

BB

r

r

r

Substitute for Br

,Bx, andByand

evaluate :B ( ) ( )

ji

jiB

38.092.0

2.37.7

2.37.722

+=

+

+=


opposite to that of Br

is given by:jiB 38.092.0 =

The unit vector in the direction ofCr

is given by: 22

yx CC +== C

CCC

r

r

r

Substitute for Cr

, Cx, and Cyand

evaluate :C ( ) ( )ji

jiC 86.051.0

1.94.5

1.94.522

=+

=


opposite that of Cr

is given by:jiC 86.051.0 +=

60 Unit vectors iandjare directed east and north, respectively. Calculate

the unit vector (in terms of iandj ) in the following directions. (a) northeast,

(b) 70clockwise from the yaxis, (c) southwest.

Picture the ProblemThe unit vector in a given direction is a vector pointing in

that direction whose magnitude is 1.

(a) The unit vector in the northeast

direction is given by:

ji

jiu

707.0

707.0

45sin)1(45cos)1(NE

+=

+=

(b) The unit vector 70clockwisefrom the yaxis is given by:

ji

jiu

342.0940.0

200sin)1(200cos)1(

=

+=


35/54


(c) The unit vector in the southwest

direction is given by:ji

jiu

707.0707.0

225sin)1(225cos)1(SW

=

+=

Remarks: One can confirm that a given vector is, in fact, a unit vector by

checking its magnitude.

General Problems

61 [SSM] The Apollo trips to the moon in the 1960's and 1970'stypically took 3 days to travel the Earth-moon distance once they left Earth orbit.Estimate the spacecraft's average speed in kilometers per hour, miles per hour,and meters per second.

Picture the ProblemAverage speed is defined to be the distance traveled divided

by the elapsed time. The Earth-moon distance and the distance and time

conversion factors can be found on the inside-front cover of the text. Well

assume that 3 days means exactly three days.

Express the average speed of Apollo

as it travels to the moon: timeelapsed

traveleddistanceav=v

Substitute numerical values to

obtain: d3

mi102.39

5

av

=v

Use the fact that there are 24 h in

1 d to convert 3 d into hours:

mi/h1032.3

mi/h10319.3

d

h24d3

mi102.39

3

3

5

av

=

=

=v

Use the fact that 1 mi is equal to 1.609 km to convert the spacecrafts average

speed to km/h:

km/h1034.5h

km10340.5mi

km609.1h

mi10319.3 333av ===v


36/54

Chapter 136

Use the facts that there are 3600 s in 1 h and 1000 m in 1 km to convert the

spacecrafts average speed to m/s:

m/s1049.1m/s10485.1s3600

h1

km

m10

h

km10340.5 33

36

av ===v

Remarks: An alternative to multiplying by 103m/km in the last step is to

replace the metric prefix k in kmby 103.

62 On many of the roads in Canada the speed limit is 100 km/h. What isthis speed limit in miles per hour?

Picture the Problem We can use the conversion factor 1 mi = 1.609 km to

convert 100 km/h into mi/h.

Multiply 100 km/h by 1 mi/1.609 kmto obtain:

mi/h2.62

km1.609mi1

hkm100

hkm100

=

=

63 If you could count $1.00 per second, how many years would it take tocount 1.00 billion dollars?

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:

y31.7days365.24

y1

h24

day1

s3600

h1s10s10 99 ==

64 (a) The speed of light in vacuum is 186,000 mi/s = 3.00 108m/s. Usethis fact to find the number of kilometers in a mile. (b) The weight of 1.00 ft3ofwater is 62.4 lb, and 1.00 ft = 30.5 cm. Use this and the fact that 1.00 cm

3of

water has a mass of 1.00 g to find the weight in pounds of a 1.00-kg mass.

Picture the Problem In both the examples cited we can equate expressions for

the physical quantities, expressed in different units, and then divide both sides of

the equation by one of the expressions to obtain the desired conversion factor.


37/54


(a) Divide both sides of the equation

expressing the speed of light in the

two systems of measurement by

186,000 mi/s to obtain:

km/mi61.1

m10

km1

mi

m1061.1

m/mi1061.1mi/h101.86

m/s103.001

3

3

3

5

8

=

=

=

=

(b) Find the volume of 1.00 kg of

water:

Volume of 1.00 kg = 103g is 10

3cm

3

Express 103cm

3in ft

3:

( ) 33

3ft0.03525

cm5.03

ft1.00cm10 =

Relate the weight of 1.00 ft3of water

to the volume occupied by 1.00 kg ofwater:

33

ft

lb62.4

ft0.03525

kg1.00=

Divide both sides of the equation

by the left-hand side to obtain: lb/kg20.2

ft0.03525

kg1.00ft

lb62.4

1

3

3

==

65 The mass of one uranium atom is 4.0 1026kg. How many uraniumatoms are there in 8.0 g of pure uranium?

Picture the Problem We can use the given information to equate the ratios of the

number of uranium atoms in 8 g of pure uranium and of 1 atom to its mass.

Express the proportion relating the

number of uranium atomsNUin

8.0 g of pure uranium to the mass of

1 atom:

kg104.0

atom1

g0.8 26U

=

N

Solve for and evaluateNU:

23

263U

100.2

kg104.0

atom1

g10

kg1

g0.8

=

= N


38/54

Chapter 138

66 During a thunderstorm, a total of 1.4 in of rain falls. How much waterfalls on one acre of land? (1 mi2= 640 acres.) Express your answer in (a) cubicinches, (b) cubic feet, (c) cubic meters, and (d) kilograms. Note that the density ofwater is 1000 kg/m

3.

Picture the Problem Assuming that the water is distributed uniformly over the

one acre of land, its volume is the product of the area over which it is distributedand its depth. The mass of the water is the product of its density and volume. Therequired conversion factors can be found in the front material of the text.

(a) Express the volume Vof water interms of its depth hand the areaAoverwhich it falls:

AhV=


( ) ( )36

36

222

in108.8

in10782.8in4.1ft

in12

mi

ft5280

acres640

mi1acre1

=

=

=V

(b) Convert in3into ft3:

3333

3

36

ft101.5ft10082.5

in12

ft1in10782.8

==

=V

(c) Convert ft3into m

3:

32

3

333

33

m104.1

m102439.1cm100

m1

in

cm540.2

ft

in12ft10082.5

=

=

=V

(d) The mass of the water is the

product of its densityand volume

V:

Vm =


evaluate m:( )

kg104.1

m10439.1m

kg10

5

32

3

3

=

=m


39/54


67 An iron nucleus has a radius of 5.4 1015m and a mass of9.3 1026kg. (a) What is its mass per unit volume in kg/ m3? (b) If Earth had thesame mass per unit volume, what would be its radius? (The mass of Earth is

5.98 1024kg.)

Picture the ProblemThe mass per unit volume of an object is its density.

(a) The densityof an object is its

mass mper unit volume V: V

m=

Assuming the iron nucleus to be

spherical, its volume as a function of

its radius ris given by:

3

3

4rV =

Substitute for V and simplify to

obtain: 33 4

3

3

4 r

m

r

m

== (1)


evaluate:

( )( )

317

317

315

26

kg/m104.1

kg/m10410.1

m104.54

kg103.93

=

=

=

(b) Solve equation (1) for rto

obtain:3

4

3

mr=


evaluate r:( )

m102.2

m

kg10410.14

kg1098.53

2

3

3

17

24

=

=

r

or about 200 m!

68 The Canadian Norman Wells Oil Pipeline extends from NormanWells, Northwest Territories, to Zama, Alberta. The 8.68 105-m-long pipelinehas an inside diameter of 12 in and can be supplied with oil at 35 L/s. (a) What is

the volume of oil in the pipeline if it is full at some instant in time? (b) How long

would it take to fill the pipeline with oil if it is initially empty?


40/54

Chapter 140

Picture the Problem The volume of a cylinder is the product of its cross-sectional area and its length. The time required to fill the pipeline with oil is theratio of its volume to the flow rateR of the oil. Well assume that the pipe has adiameter of exactly 12 in.

(a) Express the volume Vof the

cylindrical pipe in terms of its radius

rand its lengthL:

LrV 2=


( ) 3452

2

m103.6m1068.8in

m102.540in6 =

=

-

V

(b) Express the time tto fill the pipein terms of its volume V and the flow

rateRof the oil:

RVt=


evaluate t:s108.1

L

m10

s

L35

m103.6

6

33

34

=

=

t

or about 21 days!

69 The astronomical unit (AU) is defined as the mean center-to-center

distance from Earth to the sun, namely 1.496 1011m. The parsec is the radius ofa circle for which a central angle of 1 s intercepts an arc of length 1 AU. Thelight-year is the distance that light travels in 1 y. (a) How many parsecs are therein one astronomical unit? (b) How many meters are in a parsec? (c) How manymeters in a light-year? (d) How many astronomical units in a light-year? (e) Howmany light-years in a parsec?

Picture the ProblemWe can use the relationship between an angle ,measured

in radians, subtended at the center of a circle, the radius Rof the circle, and the

lengthLof the arc to answer these questions concerning the astronomical units of

measure. Well take the speed of light to be 2.998 108m/s.

(a) Relate the angle subtended by

an arc of length Sto the distanceR: R

S= RS= (1)


41/54


Substitute numerical values and evaluate S:

( )( ) parsec10848.4360

rad2

min60

1

s60

min1s1parsec1 6=

=

S

(b) Solving equation (1) forRyields:

SR=


evaluateR:( )

m10086.3

360

rad2

min60

1

s60

min1s1

m10496.1

16

11

=

=

R

(c) The distanceDlight travels in a

given interval of time tis given by:tcD =

Substitute numerical values and evaluateD:

m10461.9min

s60

h

min60

d

h24

y

d365.24

s

m102.998 158 =

=D

(d) Use the definition of 1 AU and

the result from Part (c) to obtain:( )

AU106.324

m101.496

AU1m109.461y1

4

11

15

=

=c

(e) Combine the results of Parts (b) and (c) to obtain:

( ) y262.3m109.461

y1m10086.3parsec1

15

16 =

= cc

70 If the average density of the universe is at least 6 1027kg/m3, thenthe universe will eventually stop expanding and begin contracting. (a) How many

electrons are needed in each cubic meter to produce the critical density? (b) Howmany protons per cubic meter would produce the critical density?

(me= 9.11 1031

kg; mp= 1.67 1027

kg.)


42/54


43/54


drop a penny, how long should it take to fall 1.0 m? (e) In the next chapter, we

will show that the expected relationship betweenyand tis ,221 aty= where ais

the acceleration of the object. What is the acceleration of objects dropped on themoon?

y(m) 10 20 30 40 50

t(s) 3.5 5.2 6.0 7.3 7.9

Picture the ProblemWe can plot logyversus log tand find the slope of the best-

fit line to determine the exponent C. The value of B can be determined from the

intercept of this graph. Once we know CandB,we can solve CBty= for tas a

function ofy and use this result to determine the time required for an object to fall

a given distance on the surface of the moon.

(a) The following graph of log y versus log t was created using a spreadsheet

program. The equation shown on the graph was obtained using Excels AddTrendline function. (Excels Add Trendline function uses regression analysisto generate the trendline.)

logy = 1.9637logt- 0.0762

0.80

0.90

1.00

1.10

1.20

1.30

1.40

1.50

1.60

1.70

1.80

0.50 0.60 0.70 0.80 0.90 1.00

log t

logy

(b) Taking the logarithm of both

sides of the equation CBty= yields:( )

tCB

tBBty CC

loglog

loglogloglog

+=

+==

(c) Note that this result is of the

form:

mXbY += where

Y= logy, b = logB, m= C, and

X= log t


44/54

Chapter 144

From the regression analysis

(trendline) we have:

log B = 0.076

Solving for B yields: 2076.0 m/s84.010 == B

where we have inferred the units from th

given forCBty= .

Also, from the regression analysis

we have:0.296.1 =C

(d) SolveCBty= for t to obtain: C

B

yt

1

=


evaluate t to determine how long it

would take a penny to fall 1.0 m: s1.1

s

m84.0

m0.1

2

1

2

=t

(e) Substituting forB andC inCBty= yields:

2

2s

m84.0 ty

=

Compare this equation to 221 aty= to

obtain: 221

s

m84.0=a

and

22 s

m7.1

s

m84.02 =

=a

Remarks: One could use a graphing calculator to obtain the results in Parts

(a) and (c).

72 A particular companys stock prices vary with the market and withthe companys type of business, and can be very unpredictable, but people often

try to look for mathematical patterns where they may not belong. Corning is amaterials-engineering company located in upstate New York. Below is a table ofthe price of Corning stock on August 3, for every 5 years from 1981 to 2001.Assume that the price follows a power law: price (in $) = BtC where tis expressedin years. (a) Evaluate the constantsBand C. (b) According to the power law, whatshould the price of Corning stock have been on August 3, 2000? (It was actually$82.83!)


45/54


Price (dollars) 2.10 4.19 9.14 10.82 16.85

Years since 1980 1 6 11 16 21

Picture the Problem We can plot log P versus log t and find the slope of the

best-fit line to determine the exponent C. The value ofB can be determined from

the intercept of this graph. Once we know CandB,we can use

C

BtP= to predictthe price of Corning stock as a function of time.

(a) The following graph of log P versus log was created using a spreadsheet

program. The equation shown on the graph was obtained using Excels AddTrendline function. (Excels Add Trendline function uses regression analysisto generate the trendline.)

logP= 0.658log t+ 0.2614

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40

log t

logP

Taking the logarithm of both sides

of the equation CBtP= yields:( )

tCB

tBBtP CC

loglog

loglogloglog

+=

+==

Note that this result is of the form: mXbY += where

Y= logP, b = logB, m= C, and

X= log t

From the regression analysis

(trendline) we have:

logB= 0.2614 .831$10 2614.0 ==B

Also, from the regression analysis

we have:658.0=C

(b) Substituting forB and C inCBtP= yields:

( ) 658.083.1$ tP=


46/54

Chapter 146

EvaluateP(20 y) to obtain: ( ) ( )( ) 14.13$2083.1$y20 658.0 ==P

Remarks: One could use a graphing calculator to obtain these results.

73 [SSM] The Super-Kamiokande neutrino detector in Japan is a large

transparent cylinder filled with ultra pure water. The height of the cylinder is41.4 m and the diameter is 39.3 m. Calculate the mass of the water in the cylinder.Does this match the claim posted on the official Super-K Web site that thedetector uses 50000 tons of water?

Picture the ProblemWe can use the definition of density to relate the mass ofthe water in the cylinder to its volume and the formula for the volume of acylinder to express the volume of water used in the detectors cylinder. To convertour answer in kg to lb, we can use the fact that 1 kg weighs about 2.205 lb.

Relate the mass of water contained in

the cylinder to its density andvolume:

Vm =

Express the volume of a cylinder interms of its diameter dand height h:

hdhAV 2base4

==

Substitute in the expression for m toobtain:

hdm 2

4

=

Substitute numerical values andevaluate m: ( ) ( ) ( )

kg10022.5

m4.41m3.394

kg/m10

7

233

=

=

m

Convert 5.02 107kg to tons:

ton104.55

lb2000

ton1

kg

lb2.205kg10022.5

3

7

=

=m

The 50,000-ton claim is conservative. The actual weight is closer to 55,000 tons.

74 You and a friend are out hiking across a large flat plain and decide to

determine the height of a distant mountain peak, and also the horizontal distancefrom you to the peak Figure 1-19). In order to do this, you stand in one spot and

determine that the sightline to the top of the peak is inclined at 7.5above thehorizontal. You also make note of the heading to the peak at that point: 13east ofnorth. You stand at the original position, and your friend hikes due west for1.5 km. He then sights the peak and determines that its sightline has a heading of

15east of north. How far is the mountain from your position, and how high is itssummit above your position?


47/54


Picture the Problem Vectorr

lies in the plane of the plain and locates the base

or the peak relative to you. VectorBr

also lies in the plane of the plain and locatesthe base of the peak relative to your friend when he/she has walked 1.5 km to thewest. We can use the geometry of the diagram and the E-W components of the

vectors dr

, Br

andr

to find the distance to the mountain from your position. Once

we know the distanceA,we can use a trigonometric relationship to find the heightof the peak above your position.

15

13

W E

N

d = 1.5 km

A

B

Your positionYour friend's

position

d

d

Peak

Referring to the diagram, note that: hB =sin and

hA =sin

Equating these expressions for hgives:

sinsin AB = AB

sin

sin=

Adding theE-Wcomponents of the

vectors dr

, Br

andr

yields:

coscoskm5.1 AB =

Substitute forB and simplify toobtain:

=

=

costan

sin

coscossin

sinkm5.1

A

AA

Solving forA yields:

costan

sin

km5.1

=A


48/54

Chapter 148

Substitute numerical values andevaluateA:

km42

km52.41

77cos75tan

77sin

km5.1

=

=

=A

Referring to the following diagram,we note that:

h

A5.7

( )

km5.5

5.7tankm52.41

5.7tan

=

==Ah

Remarks: One can also solve this problem using the law of sines.

75 The table below gives the periods T and orbit radii rfor the motions offour satellites orbiting a dense, heavy asteroid. (a) These data can be fitted by the

formula nCrT= . Find the values of the constantsC and n. (b) A fifth satellite isdiscovered to have a period of 6.20 y. Find the radius for the orbit of this satellite,which fits the same formula.

Period T, y 0.44 1.61 3.88 7.89

Radius r, Gm 0.088 0.208 0.374 0.600

Picture the ProblemWe can plot log Tversus log r and find the slope of thebest-fit line to determine the exponent n. We can then use any of the ordered pairs

to evaluate C. Once we know nand C,we can solve nCrT= for ras a function ofT.

(a) Take the logarithm (well

arbitrarily use base 10) of both sides

of nCrT= and simplify to obtain:

( ) ( )Crn

rCCrT nn

loglog

loglogloglog

+=

+==

Note that this equation is of the formbmxy += . Hence a graph of log Tvs.

log rshould be linear with a slope of nand a log T -intercept log C.


49/54


The following graph of log T versus log r was created using a spreadsheet

program. The equation shown on the graph was obtained using Excels AddTrendline function. (Excels Add Trendlinefunction uses regression analysisto generate the trendline.)

log T= 1.5036logr+ 1.2311

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

-1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2

log r

logT

From the regression analysis we

note that:50.1=n ,

( ) 232311.1 Gmy/0.1710 ==C ,

and

( )( )50.123

Gmy/0.17 rT= (1)

(b) Solve equation (1) for the radius

of the planets orbit: ( )

32

23Gm/y0.17

=

Tr


evaluate r: ( )Gm510.0

Gmy/0.17

y20.632

23 =

=r

76 The period T of a simple pendulum depends on the lengthL of thependulum and the acceleration of gravityg (dimensionsL/T2). (a) Find a simple

combination ofL andg that has the dimensions of time. (b) Check the dependence

of the period T on the lengthLby measuring the period (time for a complete

swing back and forth) of a pendulum for two different values of L. (c) The correct

formula relating T toL andg involves a constant that is a multiple of , and

cannot be obtained by the dimensional analysis of Part (a). It can be found by


50/54

Chapter 150

experiment as in Part (b) ifg is known. Using the valueg = 9.81 m/s2and your

experimental results from Part (b), find the formula relating T toL andg.

Picture the Problem We can express the relationship between the period Tof the

pendulum, its length L, and the acceleration of gravity g as bagCLT= and

perform dimensional analysis to find the values of a and b and, hence, thefunction relating these variables. Once weve performed the experiment called for

in Part (b), we can determine an experimental value for C.

(a) Express T as the product ofL

andg raised to powersa andb:

bagCLT= (1)

where Cis a dimensionless constant.

Write this equation in dimensional

form:

[ ] [ ] [ ]ba gLT =

Substituting the dimensions of the

physical quantities yields:

b

a

=2T

LLT

Because Ldoes not appear on the

left-hand side of the equation, we

can write this equation as:

bba 210 TLTL +=

Equate the exponents to obtain: 12and0 ==+ bba

Solve these equations simultaneouslyto find aand b:

21

21

and == ba

Substitute in equation (1) to obtain:

g

LCgCLT == 2121 (2)

(b) If you use pendulums of lengths

1.0 m and 0.50 m; the periods should

be about:

( )

( )s4.1m50.0

and

s0.2m0.1

=

=

T

T

(c) Solving equation (2) for Cyields:

L

gTC=


51/54


Evaluate CwithL = 1.0 m and

T = 2.0 s: ( ) 226.6m1.0m/s9.81

s2.02

==C

Substitute in equation (2) to obtain:

g

LT 2=

77 A sled at rest is suddenly pulled in three horizontal directions at thesame time but it goes nowhere. Paul pulls to the northeast with a force of 50 lb.

Johnny pulls at an angle of 35south of due west with a force of 65 lb. Conniepulls with a force to be determined. (a) Express the boys' two forces in terms ofthe usual unit vectors (b) Determine the third force (from Connie), expressing itfirst in component form and then as a magnitude and angle (direction).

Picture the ProblemA diagram showing the forces exerted by Paul, Johnny, and

Connie is shown below. Once weve expressed the forces exerted by Paul and

Johnny in vector form we can use them to find the force exerted by Connie.

N

S

EW 45

35

ConnieF

PaulF

JohnnyF

lb50

lb56

(a) The force that Paul exerts is:

( )[ ] ( )[ ] ( ) ( )

( ) ( )ji

jijiF

lb35lb35

lb4.35lb4.3545sinlb5045coslb50Paul

+=

+=+=r

The force that Johnny exerts is:

( )[ ] ( )[ ] ( ) ( )

( ) ( )ji

jijiF

lb37lb53

lb3.37lb53.2215sinlb56215coslb56Johnny

+=

=+=r


52/54

Chapter 152

The sum of the forces exerted by Paul and Johnny is:

( ) ( ) ( ) ( )

( ) ( )ji

jijiFF

lb9.1lb8.71

lb3.37lb53.2lb4.35lb4.35JohnnyPaul

=

+=+rr

(b) The condition that the three

forces must satisfy is:

0ConnieJohnnyPaul =++ FFFrrr

Solving for ConnieFr

yields: ( )JohnnyPaulConnie FFFrrr

+=

Substitute for JohnnyPaul FFrr

+ to

obtain:

( ) ( )[ ]( ) ( )ji

jiF

lb.91lb81

lb9.1lb8.71Connie

+=

=r

The magnitude of ConnieFr

is: ( ) ( ) lb18lb9.1lb8.17 22Connie =+=F

The direction that the force exerted

by Connie acts is given by:EofN1.6

lb17.8

lb9.1tan 1 =

=

78 You spot a plane that is 1.50 km North, 2.5 km East and at an altitudeof 5.0 km above your position. (a) How far from you is the plane? (b) At whatangle from due north (in the horizontal plane) are you looking? (c) Determine the

plane's position vector (from your location) in terms of the unit vectors, letting i

be toward the east direction, j be toward the north direction, and k be vertically

upward. (d) At what elevation angle (above the horizontal plane of Earth) is theairplane?

Picture the ProblemA diagram showing the given information is shown below.

We can use the Pythagorean Theorem, trigonometry, and vector algebra to find

the distance, angles, and expression called for in the problem statement.


53/54


h

d

km5.2

km5.1

N

E

hereareYou

Plane

(a) Use the Pythagorean Theorem to

express din terms of hand l:

22hd += l


evaluate d:( ) ( ) ( )

km8.5

km0.5km5.1km5.2222

=

++=d

(b) Use trigonometry to evaluate the

angle from due north at which you

are looking at the plane:

NofE59km1.5

km5.2tan 1 =

=

(c) We can use the coordinates of the

plane relative to your position to

express the vector dr

:

( ) ( ) ( )kjidkm.05km5.1km5.2

++=

r

(d) Express the elevation angle in

terms of hand l:

= l

h1tan


evaluate :( ) ( )

horizontheabove60

km5.1km5.2

km0.5tan

22

1

=

+=


54/54

Chapter 154

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physics for scientists and engineers 6th edition: CHAPTER 1