Solution Manual for Physics, 5th Edition

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1–1Chapter 1: Introduction to PhysicsAnswers to Even-Numbered Conceptual Questions2.The quantityT+ddoesnot make sense physically, because it adds together variables that have different physicaldimensions. The quantityd/Tdoes make sense, however; it could represent the distancedtraveled by an object in thetimeT.4.The frequency is a scalar quantity. It has a numerical value, but no associated direction.6.(a)107s;(b)10,000 s;(c)1 s;(d)1017s;(e)108s to 109s.Solutions to Problems and Conceptual Exercises1.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:(a)Convert the units:91 gigadollars$152, 000, 0000.152 gigadollars1 10dollars=(b)Convert the units again:4121 teradollars$152, 000, 0001.5210teradollars1 10dollars−=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.2.Picture the Problem:This problem is about the conversion of units.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:(a)Convert the units:651.010m85m8.510mm−−=(b)Convert the units again:61.010m1000 mm85m0.085 mmm1 m−=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.3.Picture the Problem:This problem is about the conversion of units.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:Convert the units:98Gm1 10m0.3310m/ssGm=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–24.Picture the Problem:This problem is about the conversion of units.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:Convert the units:1295teracalculation1 10calculations1 10s136.8steracalculationns136,800 calculations/ns1.36810calculations/ns−==Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.5.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution: 1. (a)Substitutedimensions for the variables:222121[L][L][T][L]The equation is dimensionally consistent.2[T]xat===\æö÷ç÷ç÷çèø2. (b)Substitute dimensionsfor the variables: LT1TNot dimensionally consistentLTvtx==3. (c)Substitute dimensionsfor the variables: 222LTTTDimensionally consistentLTxta====Insight:The number 2 does not contribute any dimensions to the problem.6.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:1. (a)Substitute dimensionsfor the variables: L1TYesLT1 Txv===2. (b)Substitute dimensions for the variables:   ()21TTLT1NoLT1TTav===3. (c)Substitute dimensions for the variables: 222L21TTYesLT1Txa====4. (d)Substitute dimensions for the variables: ()  2222222LTLTLLNoLLTLTva====Insight:When squaring the velocity you must remember to square the dimensions of both thenumerator (meters) andthe denominator (seconds).

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–37.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:1. (a)Substitute dimensionsfor the variables: LTLYesTv t==2. (b)Substitute dimensions for the variables:2211222LTLYesTa t==3. (c)Substitute dimensions for the variables:2LL22TNoTTa t==4. (d)Substitute dimensions for the variables: ()  2222222LTLTLLYesLLTLTva====Insight:When squaring the velocity you must remember to square the dimensions of both the numerator (meters) andthe denominator (seconds).8.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:1. (a)Substitute dimensionsfor the variables:2211222LTLNoTa t==2. (b)Substitute dimensions for the variables:2LLTYesTTa t==3. (c)Substitute dimensions for the variables: 22 L2TNoLTxa==4. (d)Substitute dimensions for the variables:222LLL22LYesTTTa x===Insight:Whentaking the square root of dimensions you need not worry about the positive andnegative roots; only thepositive root is physical.9.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:Substitute dimensions for thevariables:222212LLLTTLLtherefore1pppva xp+====Insight:Thenumber 2 does not contribute any dimensions to the problem.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–410.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:Substitute dimensionsfor the variables:222[L][L][T][T][T][T]therefore2pppax tp−==== −Insight:The number 2 does not contribute any dimensions to the problem.11.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:Substitute dimensionsfor the variables: 1221 2122T1T[L][L]LTLTLTthereforeppppthgp−=====Insight:We conclude thehbelongs inside the square root, and the time to fall from rest a distancehis2.th g=12.Picture the Problem: This is a dimensional analysis question.Strategy:Rearrange the expression to solve for the forceF, and then substitutetheappropriatedimensionsfor thecorresponding variables.Solution:Substitute dimensions for the variables,using[M]torepresent the dimension of mass:2[L][M][T]Fm a==Insight:This unit,kg·m/s2,will later be given the name “Newton”and abbreviated as N.13.Picture the Problem: This is a dimensional analysis question.Strategy:Rearrange the expression to solve for the force constantk, and then substitutetheappropriatedimensionsforthe corresponding variables.Solution:1.Solve fork:222242square both sides:4ormmmTTkkkT===2.Substitute the dimensions,using[M]torepresent the dimension of mass:2[M][T]k=Insight:This unit will later be renamed “Newton/meter.”The 42does not contribute any dimensions.14.Picture the Problem: This is a significant figures question.Strategy:Follow the given rules regarding the calculation and display of significant figures.Solution:Round to the 3rddigit:882.997910m/s3.0010m/sInsight:It is important not to round numbers off too early when solving a problem because excessive rounding cancause your answer to significantly differ from the true answer, especially when two large values are subtracted to find asmall difference between them.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–515.Picture the Problem: The parking lot is a rectangle.Strategy:The perimeter of the parking lot is the sum of the lengths ofits four sides. Apply the rule for addition of numbers: the number ofdecimal places after addition equals the smallest number ofdecimalplaces in any of the individual terms.Solution:1.Add the numbers:124.3 +41.06 + 124.3 +41.06 m = 330.72 m2.Round to the smallest number ofdecimalplaces in any of the individual terms:330.72m330.7mInsight:Even if you changed the problem to()()2124.3 m241.06 m ,+you’d still have to report 330.7m as theanswer; the 2 is considered an exact number so it’s the“124.3 m”valuethat limits the number of significant digits.16.Picture the Problem:The weights of the fish are added.Strategy:Apply the rule for addition of numbers, which states thatthe number of decimal places after addition equalsthe smallest number of decimal places in any of the individual terms.Solution:1.Add the numbers:2.77+ 14.3+13.43 lb =30.50lb2.Round to the smallest number of decimalplaces in any of the individual terms:30.50lb30.5lbInsight:The 14.3-lb rock cod is the limiting figure in this case; itis only measured to within an accuracy of 0.1 lb.17.Picture the Problem: This is a significant figures question.Strategy:Follow the given rules regarding the calculation and display of significant figures.Solution:1.(a)The leading zeros are not significant:0.0000303has3significantfigures2.(b)The middle zeros are significant:6.201×105has4significantfiguresInsight:Zeros are the hardest part of determining significant figures. Scientific notation can remove the ambiguity ofwhether a zero is significant because any zerowrittento the right of the decimal point is significant.18.Picture the Problem: This is asignificant figures question.Strategy:Apply the rule for multiplication of numbers, which states thatthe number ofsignificant figuresaftermultiplication equals the number of significant figures in theleastaccurately known quantity.Solution:1.(a)Calculate the area andround to four significant figures:()222211.37 m406.13536 m406.1 mAr===2.(b)Calculate the area and round totwo significant figures:()222226.8 m145.2672443 m1.5 10mAr===Insight:The numberis considered exact so it will never limit the number of significant digits you report in an answer.If we present the answer to part (b) as 150 m the number of significant figures is ambiguous, so we present the result inscientific notation to clarify that there are only two significant figures.41.06 m41.06 m124.3 m124.3 m

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–619.Picture the Problem: This is a significant figures question.Strategy:Follow the given rules regarding the calculation and display of significant figures.Solution:(a)Round to the 3rddigit:3.141592653589793.14(b)Round to the 5thdigit:3.141592653589793.1416(c)Round to the 7thdigit:3.141592653589793.141593Insight:It is important not to round numbers off too early when solving a problem because excessive rounding cancause your answer to significantly differ from the true answer.20.Picture the Problem: This problemis about the conversion of units.Strategy:Convert each speed to m/s units to compare their magnitudes.Solution:1.(a)The speed is already in m/s units:a0.25 m/sv=2.(b)Convert the speed to m/s units:bkm0.75v=h1000 m1 km1 h0.21 m/s3600 s =3.(c)Convert the speed to m/s units:cft12v=1 ms3.281 ft3.7 m/s =4.(d)Convert the speed to m/s units:dcm16v=1 ms100 cm0.16 m/s =5.Rank the four speeds:dbacvvvvInsight:To onesignificant digit the speeds in (b) and (d) are identical(0.2 m/s), but it is ambiguous how to round the0.25 m/s of (a) to one significant digit (either 0.2 or 0.3 m/s).Notice that it is impossible to compare these speedswithout converting to the same unit of measure.21.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.Find the length in feet:()17.7 in1 ft2.5 cubit3.68 ft1 cubit12 in =2.Find the width and height in feet:()17.7 in1 ft1.5 cubit2.21 ft1 cubit12 in =3.Find the volume in cubic feet:()()()33.68 ft2.21 ft2.21 ft18 ftVLWH===Insight:Conversion factors areconceptually equal to one, even though numerically they often equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.22.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertmi/hto km/h:2mi1.609 km68109 km/h1.1 10km/hh1 mi ==Insight:The given 68 mi/h has only two significantfigures, thus the answer is limited to two significant figures. If wepresent the answer as 110 km/h the zero is ambiguous, thus we use scientific notation to remove the ambiguity.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–723.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertfeetto kilometers:()1 mi1.609 km3212 ft0.9788 km5280 ft1 mi =Insight:Conversion factors are conceptually equal to one, even thoughnumerically they often equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.24.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert seconds to weeks:41 msg3600 s24 h7 dmsgmsg67, 2007109 shdwkwkwk ==Insight:In this problemthere is only one significant figure associated with thephrase,“every 9seconds.”25.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertfeetto meters:()1 m108 ft32.9 m3.281 ft =Insight:Conversion factors are conceptually equal to one, even though numerically they often equalsomething otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.26.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertcaratsto pounds:()0.20 g1 kg2.21 lb530.2 ct0.23 lbct1000 gkg=Insight:Conversion factors are conceptually equal to one, even though numerically they often equalsomething otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.27.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertm/s2to feet per second per second:22m3.28 ftft98.13221 mss =Insight:Conversion factors are conceptually equal to one, even though numerically they often equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.28.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)The speed must begreaterthan55 km/hbecause1 mi/h = 1.609 km/h.2.(b)Convert the miles tokilometers:mi1.609 kmkm5588hmih =Insight:Conversion factors are conceptually equal to one, even though numerically they oftenare equal tosomethingother than one. They often helptodisplay a number in a convenient, useful, oreasy-to-comprehend fashion.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–829.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution: 1.(a)Convert to feet per second:m3.28 ftft2375s1 ms =2. (b)Convert to miles per hour:m1 mi3600 smi2351s1609 m1 hrh =Insight:Mantis shrimp havebeen known to shatter the glass wallsof the aquarium in which they are kept.30.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.In this problem, one“jiffy” corresponds to the time in seconds that it takes light to travel one centimeter.Solution:1.(a):Determine the magnitude of a jiffy:118111 s1 msjiffy3.3357101100 cmcmcm2.997910m1 jiffy3.335710s−− ===2.(b)Convert minutes to jiffys:()121160 s1 jiffy1 minute1.798710jiffy1 min3.335710s− =Insight:A jiffy is 33.357 billionths of a second. In other terms 1 jiffy = 33.357 picosecond (ps).31.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Convertcubic feet to mutchkins:()3328.3 L1 mutchkin1 ft67 mutchkin0.42 Lft =2.(b)Convert noggins to gallons:()0.28 mutchkin0.42 L1 gal1 noggin0.031 galnogginmutchkin3.785 L =Insight:To convert noggins to gallons, multiply the number of noggins by 0.031 gal/noggin. Conversely,there are 1noggin/0.031 gal = 32 noggins/gallon. That means a noggin is about half a cup.A mutchkin is about 1.8 cups.32.Picture the Problem:A cubic meterof oil is spread out into a slickthat isone molecule thick.Strategy:The volume of the slick equals its area times its thickness. Use this fact to find the area.Solution:Calculate the area forthe known volume and thickness:36261.0 m1m2.010m0.50m1 10mVAh−===Insight:Two million square meters is about 772 square miles!33.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert meters to feet:22m3.28 ft9.8132.2 ft/s1 ms =Insight:Conversion factors are conceptually equal to one, even though numerically they often areequal to somethingother than one. They often help to display a number in a convenient, useful, or easy-to-comprehend fashion.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–934.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution: 1.(a)Convert m/s to ft/s:m3.281 ft25.082.0 ft/ssm =2. (b)Convert m/s to mi/h:m1 mi3600 s25.055.9 mi/hs1609 m1 h =Insight:Conversion factors are conceptually equal to one, even though numerically they often equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.35.Picture the Problem: Therows of seatsin a ballparkare arranged into roughly a circle.Strategy:Estimate thata baseball field is a circle around 300 ft in diameter, with 100 rows of seats around outside ofthe field, arranged in circles that have perhaps anaverage diameterof 500 feet. The length of each row is then thecircumference of the circle, ord=(500 ft).Suppose there is a seat every 3 feet.Solution:Multiply the quantitiesto make an estimate:()5ft1 seat100 rows50052, 400 seats10seatsrow3 ftN==Insight:Some college football stadiums can hold as many as 100,000 spectators, but most less than that.Regardless,for an order of magnitudeestimatewe round to the nearest factor of ten, in this case105.36.Picture the Problem:Hair grows at a steady rate.Strategy:Estimate thatyour hair grows about a centimeter a month, or 0.010 m in 30 days.Solution:Multiply the quantitiesto make an estimate:990.010 m1 d1 h3.910m/s3.9 nm/s10m/s30 d24 h3600 sv−−===Insight:This rate corresponds to about 40 atomic diameters persecond. The length of human hair accumulates0.12 m or about 5 inches per year.37.Picture the Problem:Suppose all milk is purchased by the gallon in plastic containers.Strategy:There are about 300 million people in the United States, and if each of these were to drink a half gallon ofmilk every week, that’s about 25 gallons per person per year.Each plastic container is estimated to weigh about anounce.Solution:1.(a)Multiply thequantities to make an estimate:()()691030010people25 gal/y/person7.510gal/y10gal/y=2.(b)Multiply the gallons bythe weight of the plastic:()()10891 lb1 10gal/y1 oz/gal6.2510 lb/y10lb/y16 oz=Insight:Abouthalf a billion pounds of plastic! Concerted recycling can prevent much of these containers fromclogging up our landfills.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1038.Picture the Problem: The Earth is roughly a sphere rotating about its axis.Strategy:Use the fact the Earth spins once about its axis every 24 hours to find the estimated quantities.Solution:1.(a)Divide distance by time:33000 mi1000 mi/h10mi/h3 hdvt===2.(b)Multiply speed by 24 hours:()()4circumference3000 mi/h24 h24,000 mi10mivt===3.(c)Circumference equals 2r:3circumference24, 000 mi3800 mi10mi22r====Insight:These estimates are “in the ballpark.” The speed of a point on the equator is 1038 mi/h, the circumference ofthe equator is 24,900 mi, and the equatorial radius of the Earth is 3963 mi.39.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:1. (a)Substitutedimensions forthe variables:( )2mmmsThe equation is dimensionally consistent.sssva t===2. (b)Substitute dimensionsfor the variables:( )2122122mmsmdimensionally consistentssva t==NOT3. (c)Substitute dimensionsfor the variables:2m s1sdimensionally consistentm ssatv==NOT4. (d)Substitute dimensionsfor the variables:()2222222mmm2mdimensionally consistentsssva x===Insight:The number 2 does not contribute any dimensions to the problem.40.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:1. (a)Substitute dimensionsfor the variables:()( )222msm sNox t==2. (b)Substitute dimensions for the variables:2222msmYesmsvx==3. (c)Substitute dimensions for the variables:22mYessxt=4. (d)Substitute dimensions for the variables:2m sm=Yesssvt=Insight:One of the equations to be discussed later is for calculating centripetal acceleration, where we’ll note that2centripetalavr=has units of acceleration, as we verified in part (b).

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1141.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Convert nm to mm:()9431 10m1 mm675 nm6.7510mm1 nm1 10m−−− =2.(b)Convertnm to in:()951 10m39.4 in675 nm2.6610in1 nm1 m−− =Insight:Conversion factors are conceptually equal to one, even though numerically they often are equal to somethingother than one. They often help to display a number in a convenient, useful, or easy-to-comprehend fashion.42.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert ft/day to m/s:4ft1 m1 day2107.41 10m/sday3.281 ft86, 400 s−=Insight:This is amuch slower speed than the 1.3 m/s average walking speed of a human being.43.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertcubic feet of gold to pounds:()33328.3 L42.5 lb1.0 ft1200 lb1.210lb1 L1 ft ==Insight:A cube of solid gold one foot on a side weighs over half a ton! You will need help moving your valuablediscovery.44.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert m/s to miles per hour:85m1 mimi3.00101.8610s1609 ms=Insight:The equatorial circumference of the Earth is 40,075 km or 24,907 mi.Thus a beam of light, traveling at186,000 miles per second, can travel around the globe 7.5 times every second.45.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertshakes perminute to shakes per second:shakes1 min330055 shakes/secondmin60 s =Insight:When analyzing the characteristic shake frequencies ofrattlesnakes, it is advisable to work from a distance.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1246.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1. (a)Convertpg to kg:()121410g1 kg27 pg2.710kgpg1000 g−− =2. (b)Convertpg to ng:()12910g1 ng27 pg0.027 ngpg10g−− =Insight:The inside back cover of thetextbook has a helpful chart of the metric prefixes.47.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1. (a)Convertcm/day to mm/s:4cm10 mm1 d1 h4.14.710mm/sdcm24 h3600 s− =2. (b)Convertcm/day to ft/week:cm1 ft7 d4.10.94 ft/weekd30.5 cm1 week =Insight:Given that the average width of a human hair is 0.10 mm,acornplant that grows at thisrategains theadditional height ofthe width of a human hair every 3.5 minutes.48.Picture the Problem: This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution: 1.(a)Convert seconds to minutes:4605 beats60 s3.6310beats/minsmin =2. (b)Convert beats to cycles:1 s9,192,631,770 cycles15,194, 433 cycles/beat605 beatss =Insight:Conversion factors are conceptually equal to one, even though numerically they often are equal to somethingother than one. They often help to display a number in a convenient, useful, or easy-to-comprehend fashion.49.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)The acceleration must begreaterthan14ft/s2becausethere are about 3 ft per meter.2.(b)Convert m/s2to ft/s2:22m3.281 ftft1446mss =Insight:Conversion factors are conceptually equal to one, even though numerically they often are equal to somethingother than one. They often help todisplay a number in a convenient, useful, or easy-to-comprehend fashion.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1350.Picture the Problem:A speeding bullet covers a large distance in a small interval of time.Strategy:Useconversion factors to change the unitsfrom ft/s to mi/h.Then multiply the speed of the bullet by the timeinterval to find the distance traveled.Solution: 1.(a)Convertft/s to mi/h:ft1 mi3600 s42252881 mi/hs5280 ft1 h =2. (b)Multiply the speed by thetime to find the distanced:ft1 m0.001 s42255.0 ms6.4 ms3.281 ft1 msd==Insight:The bullet covers 21 feet in 5.0 milliseconds. Because the normal length of a blink is 300 milliseconds, thebullet can cover 1270 ft (nearly a quarter mile) in a blink of an eye.51.Picture the Problem:Nerve impulses cover a large distance in a small interval of time.Strategy:Useconversion factors to change the unitsfrom m/s to mi/h.Then multiply the speed of the nerve impulsesby the time interval to find the distance traveled.Solution:1.(a)Convert m/s to mi/h:m1 mi3600 smi140310s1609 m1 hh =2. (b)Multiply the speed by thetime to find the distanced:3m1 10s1405.0 ms0.70 ms1 msd−==Insight:The nerve impulses travel more than two feet in 5.0 milliseconds. Because the normal length of a blink is300 milliseconds, the nerve impulses can cover 42 ft in a blink of an eye.52.Picture the Problem:This problemis about the conversion of units.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Convert mg/min to g/day:3mg110g1440 ming1.62.3minmg1 dayday−=2.(b)Divide the mass gain by the rate:0.0075 kg1000 g/kg3.3 days2.3 g/daymtrate===Insight:The rate of brain growth slows down considerably as thechildmatures, and stops growingat around 10 yearsof age. Brain weight decreases a small amount, and very slowly, after age 20.53.Picture theProblem: TheHuygens spaceprobe rotates many times per minute.Strategy:Find the time it takes the probe to travel 150 yards and then determine how many rotations occurred duringthat time interval.Convert units to figure out the distance moved per revolution.Solution:1.(a)Find the timeto travel 150 yards:1 s30.5 cm3 fts2.95150 yd443 s31 cmftydyd==2.Find the number ofrotations in that time:()7 rev1 min443 s51.6 rev51 complete revolutionsmin60 s ==3.(b)Convert min/rev to ft/rev:1 min60 s31 cm1 ft8.7 ft/rev7 rev1 mins30.5 cm =Insight:In later chapters the rotation rate will be represented by the symbolωand we will discover that the total anglethrough which the probe rotated is given by.t=

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1454.Picture the Problem:A dragonfly spins rapidly whilebeing recorded by a high-speed video camera.Strategy:Convert the spin revolution per frame value into units of revolutions per minute.Solution:Convert rev/frame to rev/min:31 rev240 frames60 s1029 rev/min1.010rpm14 frames1 s1 min ==Insight:The value of one spin revolution every14frames contains onlytwosignificant figures, so our answer isaccurate to onlytwosignificant figures. Greater precision can be achieved by averaging the rotation rate over manyframes.55.Picture theProblem: This is a dimensional analysis question.Strategy:Findpto make the length dimensions match andqto make the time dimensions match.Solution:1.Make the length dimensions match:2[L][L][T]implies1[T][T]pqp==2.Now make the time units match:21211[T]or[T][T] [T]implies1[T][T]qqq−−=== −Insight:Sometimes you can determine whether you’ve made a mistake in your calculations simply by checking toensure the dimensions work out correctly on both sides of your equations.56.Picture the Problem: This is a dimensional analysis question.Strategy:Findqto make the time dimensions match and thenpto make the distance dimensions match.RecallLmusthave dimensions of meters andgdimensions of m/s2.Solution:1.Makethe time dimensions match:()2122L[T]L=LL[T]implies[T]qqppq−== −2.Now make the distance units match:12122[L][T]Limplies[T]pp−==Insight:Sometimes you can determine whether you’ve made a mistake in your calculations simply by checking toensure the dimensions work out correctly on both sides of your equations.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,5thEdition1–1557.Picture the Problem: Your car travels 1.0mile in each situation, but the speed and times are different in the secondcase than the first.Strategy:Set the distances traveled equal to each other, then mathematically solve for the initial speedv0. The knownquantities are that the change in speed is7.9 mi/hv=and the change in time is13 s.t = −Solution:1.Set the distances equal:12dd=2.Substituteforthedistances:()()00v tvvtt=+ + 3.Multiply the terms on the right side:000v tv tvttvv t=+ + +  4.Subtract0v tfrom both sides and substitute0dtv=:000dvvtvtv= + +  5.Multiply both sides by0vand rearrange:()200vtv t vvd = +  + 6.Solve the quadratic equation for0v:()()22042vtvttvdvt−  −=7.Substitute in the numbers:()()mi1 h7.913 s0.0285 miandh3600 s1 h13 s0.00361 h,and1 mi3600 svttd  =+−= − =−= −=8.Find0v:()()()()()2000.0285 mi0.0285 mi40.0285 mi1mi20.00361 h43 mi/h ,51 mi/hvv− −−−−=−=−Insight:This was a very complex problem, but it does illustrate that it is necessary to know how to convert units inorder to properly solve problems. Theunits must be consistent with each other in order for the math to succeed.58.Picture the Problem: Thesnowycricket chirps at a rate that is linearly dependent upon the temperature.Strategy:Take note of the given mathematical relationship between the number of chirpsNin 13 seconds and thetemperatureTin Fahrenheit. Use the relationship to determine the appropriate graph ofNvs.T.Solution:The given formula,N=T− 40, is a linear equation of the formy=mx+b. By comparing the twoexpressions we see thatNis akin toy,Tis akin tox, the slopem= 1.00 chirps °F−1, andb= −40chirps. In the displayedgraphs ofNvs.T, onlythreeof the plots are linearwith nonzero slope, plots A, C,andE, so we consider only those. Ofthosethree, onlytwohave positive slopes, A and C, so we rule out plot E. Using the formula at 70°F, we expect thenumber of chirps to beN= (1.00 chirps °F−1)(70°F)–40 chirps = 30 chirps, and bynoting the values of plots A and C at70°Fwe conclude that the correct plot isplotC.Insight:Plot Bis quadratic and corresponds totheformula()24030.NT=−59.Picture the Problem:The snowycricket chirps at a rate that is linearly dependent upon the temperature.Strategy:Use the given formula to determine the number of chirpsNin 13 seconds, and then use that rate to find thetime elapsed for the snowy cricket to chirp 12 times.Solution: 1.Find thenumber ofchirpsper second:4043400.23 chirps13 s13 ssNTt−−===2.Find thetime elapsed for 12 chirps:1 s12 chirps52 s0.23 chirp=Insight:Notice thatwe can employ either the ratio0.23 chirp 1 sor the ratio1 s 0.23 chirp,whichever is most usefulfor answering the particular question that is posed.

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