Solution Manual for Fundamental Managerial Accounting Concepts, 9th Edition

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1–1Chapter 1: Introduction to PhysicsAnswers to Even-Numbered Conceptual Questions2.The quantityT+ddoes not make sense physically, because it adds together variables that have differentphysical dimensions. The quantityd/Tdoes make sense, however; it could represent the distancedtraveledby an object in the timeT.4.(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 is simply a units conversion problem.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:(a)Convert the units:91 gigadollars$114,000,0000.114 gigadollars1 10dollars=(b)Convert the units again:4121 teradollars$114,000,0001.1410teradollars1 10dollars−=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.2.Picture the Problem: This is simply a units conversion problem.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:(a)Convert the units:651.010m70m7.010mm−−=(b)Convert the units again:681.010m1 km70m7.010kmm1000 m−−=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.3.Picture the Problem: This is simply a units conversion problem.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:Convert the units:98Gm110m0.3310m/ssGm=Insight:The inside back cover of the textbook has a helpful chart of the metric prefixes.4.Picture the Problem: This is simply a units conversion problem.Strategy:Multiply the given number by conversion factors to obtain the desired units.Solution:Convert the units:1267teracalculation110calculations110s70.72steracalculations7.07210calculations/s−=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,4thEdition1–25.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:( )mmsmThe equation is dimensionally consistent.sxvt===2. (b)Substitute dimensionsfor the variables:( )2122122mmsmdimensionally consistentsxat===3. (c)Substitute dimensionsfor the variables:222msssdimensionally consistentm sxta====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:( )msmYessvt==2. (b)Substitute dimensions for the variables:( )2211222msmYessat==3. (c)Substitute dimensions for the variables:( )2mm22sNossat==4. (d)Substitute dimensions for the variables:()222m smYesm sva==Insight:When squaring the velocity you must remember to square the dimensions of both the numerator (meters) andthe denominator (seconds).7.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:( )2211222msmNosat==2. (b)Substitute dimensions for the variables:( )2mmsYesssat==3. (c)Substitute dimensions for the variables:222msNom sxa==4. (d)Substitute dimensions for the variables:()2mm22mYesssax==Insight:Whentaking the square root of dimensions you need not worry about the positive and negative roots; only thepositive root is physical.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–38.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:Substitute dimensions for the variables:()222212mmmssmmtherefore1pppvaxp+====Insight:Thenumber 2 does not contribute any dimensions to the problem.9.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]therefore2pppaxtp−==== −Insight:The number 2 does not contribute any dimensions to the problem.10.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 on both sides of the equation:02[L][L][L] [T][T][T][T][L][L]It is dimensionally consistent![T][T]vvat=+=+=Insight:Two numbers must have the same dimensions in order to be added or subtracted.11.Picture the Problem: This is a dimensional analysis question.Strategy:Manipulate the dimensions in the same manner as algebraic expressions.Solution:Substitute dimensions for the variables,where [M] represents the dimension of mass:2[L][M][T]Fma==Insight:This unit(kgm/s2)will later be given the name “Newton.”12.Picture the Problem: This is a dimensional analysis question.Strategy:Solve the formula forkand substitute the units.Solution:1.Solve fork:222242square both sides:4ormmmTTkkkT===2.Substitute the dimensions, where [M]represents the dimension of mass:2[M][T]k=Insight:This unit will later be renamed “Newton/m.” The 42does not contribute any dimensions.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–413.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.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.15.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:144.3 +47.66 + 144.3 +47.66 m = 383.92 m2.Round to the smallest number ofdecimalplaces in any of the individual terms:383.92 m383.9mInsight:Even if you changed the problem to()()2144.3 m247.66 m+you’d still have to report 383.9 m as theanswer; the 2 is considered an exact number so it’s the 144.3 m that 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:2.35 + 12.1 + 12.13 lb = 26.58 lb2.Round to the smallest number of decimalplaces in any of the individual terms:26.58 lb26.6lbInsight:The 12.1 lb rock cod is the limiting figure in this case; it is 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:0.000054has2significantfigures2.(b)The middle zeros are significant:3.001×105has4significantfiguresInsight:Zeros are the hardest part of determining significant figures. Scientific notation can remove the ambiguity ofwhether a zerois significant because any trailing zero to the right of the decimal point is significant.47.66 m47.66 m144.3 m144.3 m

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–518.Picture the Problem: This is a significant 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:()222214.37 m648.729144 m648.7 mAr===2.(b)Calculate the area and round totwo significant figures:()22223.8 m45.3645979 m45 mAr===Insight:The numberis considered exact so it will never limit the number of significant digits you report in an answer.19.Picture the Problem:This is a units conversion problem.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: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.20.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Find the length in feet:()3.28 ft631 m2069 ft1 m =2.Find the width in feet:()3 ft707 yd2121 ft1 yd =3.Find the volume in cubic feet:()()()832069 ft2121 ft110 ft4.83 10ftVLWH===4.(b)Convert to cubic meters:()383731 m4.8310ft1.3710m3.28 ft=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.21.Picture the Problem:This is a units conversion problem.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 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.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–622.Picture the Problem:This is a units conversion problem.Strategy:Convert the frequency of cesium-133 given on page4 to units ofmicroseconds per megacycle, then multiplyby the number of megacycles to find the elapsed time.Solution:Convert tomicroseconds per megacycleandmultiply by1.5 megacycles:6641 s1 10cycles1ss108.78277579,192, 631, 770 cyclesMcycleMcycle1 10ss108.78277571.5 Mcycle160s1.610sMcycle−−===Insight:Only twosignificant figures remain in the answer becauseof the 1.5 Mcycle figure given in the problemstatement. The metric prefix conversionsare considered exact and have an unlimited number of significant figures, butmost other conversion factors have a limited number of significant figures.23.Picture the Problem:This is a units conversion problem.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 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.24.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert seconds to weeks:41 msg3600 s24 h7 dmsg9107 shdwkwk =Insight:In this problemthere is only one significant figure associated with thephrase,“7seconds.”25.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertfeetto meters:()1 m108 ft32.9 m3.28 ft =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.26.Picture the Problem:This is a units conversion problem.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 equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–727.Picture the Problem:This is a units conversion problem.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 to kilometers:mi1.609 kmkm5588hmih =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 is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convertm/sto miles per hour:88m1 mi3600 smi3.00106.71 10s1609 m1 hh=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.29.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert to ft per secondper 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.30.Picture the Problem:This is a units conversion problem.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: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.31.Picture the Problem:This is a units conversion problem.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.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–832.Picture the Problem: The volume of the 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 is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units. Then use a ratio to findthe factor change in part (b).Solution:1.(a)Convert square inches to square meters:()2221 m8.5 in11 in0.060 m1550 inA==2.(b)Calculate a ratio to find the new area:()()11oldold22newnewnewoldoldoldoldold1newold414LWALWALWLWAA====Insight:If you learn to use ratiosyoucan often make calculations like these very easily.Always put the new quantityin the numerator and the old quantity in the denominator to make the new quantity easier to calculate at the end.34.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.Convert m/s to ft/s:m3.28 ft20.065.6 ft/ssm =2.(b)Convert m/s to mi/h:m1 mi3600 s20.044.7 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:This is a units conversion problem.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 equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.36.Picture the Problem: The rows of seats are 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. Still, for anorder of magnitude we round to the nearest factor of ten, in this case it’s 105.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–937.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.38.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: The lottery winnings are represented either by quarters or paper dollars.Strategy:There are about 5 quarters and about 30 dollar bills per ounce.Solution:1.(a)Multiply byconversion factors:()661 oz1 lb41210quarters600, 000 lb10lb5 quarters16 oz=2.(b)Repeat for the dollar bills:()641 oz1 lb1210dollars25,000 lb10lb30 dollars16 oz=Insight:Better go with large denominations or perhaps a single check when you collect your lottery winnings! Eventhe dollar bills weigh overtentons!

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–1040.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 consistentssvat==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.41.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 sNoxt==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).42.Picture the Problem:This is a units conversion problem.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 nm2.6610mm1 nm1 m−− =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.43.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:Convert ft/day to m/s:4ft1 m1 day2107.4110m/sday3.28 ft86400 s− =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.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–1144.Picture the Problem:This is a units conversion problem.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 equal something otherthan one. They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion.45.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.Convertten feetto m:()1 m10.0 ft3.05 m above the water3.28 ft =2.Convert ten knots to m/s:()1.0 n.mi/hr1.852 km1000 m1 hr10.0 knot5.14 m/sknotn.mikm3600 s =Insight:If we were to describe the flying parameters of the helicopter in SI units, we would say it is flying “3 and 5”!46.Picture the Problem:This is a units conversion problem.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/s2:22m3.28 ftft1446mss =3.(c)Convert m/s2to km/h2:2522m1 km3600 skm141.8101000 mhsh=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.47.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Convert m/s to mi/h:m1 mi3600 smi140310s1609 m1 hh =2.(b)Convert m/s to m/ms3m1 10sm1400.145.0 ms0.70 ms1 msms−==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.48.Picture the Problem:This is a units conversion problem.Strategy:Multiply the known quantity by appropriate conversion factors to change the units.Solution:1.(a)Convert mg/min to g/day:3mg1 10g1440 ming1.62.3minmg1 dayday−=2.(b)Divide the mass gain by the rate:0.0075 kg1000 g/kg3.3 days2.3 g/daymtrate===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.

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Solution Manual for Fundamental Managerial Accounting Concepts, 9th Edition - Page 13 preview image

Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–1249.Picture the Problem: The probe 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: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.50.Picture the Problem: 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.51.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.Make the 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.52.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 = +  + 

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Solution Manual for Fundamental Managerial Accounting Concepts, 9th Edition - Page 14 preview image

Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–136.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. The units must be consistent with each other in order for the math to succeed.53.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= −40 °F. In the displayedgraphs ofNvs.T, only two of the plots are linear, plots A and C, so we consider only those. Of those two, only one hasan intercept of −40 °F, so we conclude that the correct plot isplotC.Insight:Plot B would be an appropriate depiction of a formula likeN=T2− 40.54.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:Note that we can employ either the ratio0.23 chirp 1 sor the ratio1 s 0.23 chirp,whichever is most usefulforanswering the particular question that is posed.55.Picture the Problem: The snowy cricket chirps at a rate that is linearly dependent upon the temperature.Strategy:Use the given formula to determine the temperatureTthat corresponds to the given number of chirps perminute by your pet cricket.Solution: 1.Find the number of chirps per second:112 chirps1.87 chirps60.0 ssNt==2.Find the number of chirpsNper 13 s:1.87 chirps13.0 s24.3 chirps1 sN==3.Determine the temperature from the formula:40.040.024.340°F64.3°FNTTN=−=+=+=Insight:The number of significant figures might be limited by the precision of the numbers 13 and 40 that are given inthe description of the formula. In this case we interpreted them as exact and let the precision of the measurements 112and 60.0 s limit the significant digits of our answer.

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Chapter1:Introduction to PhysicsJames S. Walker,Physics,4thEdition1–1456.Picture the Problem: The cesium atom oscillates many cycles during the time it takes the cricket to chirp once.Strategy:Find the time in between chirps using the given formula and then find the number of cycles the cesium atomundergoes during that time.Solution: 1.Find the time in between chirps:40.065.040.0chirps1.9213.0 s13.0 ssNTt−−===2.Find the number of cesium atom cycles:99,192, 631, 770 cycles1 s4.7810cycles/chirps1.92 chirp=Insight:The number of significant figures might be limited by the precision of the numbers 13 and 40 that are given inthe description of the formula. In this case we interpreted them as exact and let the precision of the measurement 65.0°Flimit the significant digits of our answer.

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2–1Chapter2: One-Dimensional KinematicsAnswers to Even-Numbered Conceptual Questions2.An odometer measures the distance traveled by a car. You can tell this by the fact that an odometer has anonzero reading after a round trip.4.No. After one complete orbit the astronaut’s displacement is zero. The distance traveled, however, isroughly 25,000 miles.6.A speedometer measures speed, not velocity. For example, if you drive with constant speed in a circularpath,your speedometer maintains the same reading, even though your velocity is constantly changing.8.Yes. For example, your friends might have backed out of a parking place at some point in the trip, giving anegative velocity for a short time.10.No. If you throw a ball upward, for example, you might choose the release point to bey= 0. This doesn’tchange the fact that the initial upward speed is nonzero.12.(a)Yes. The object might simply be at rest.(b)Yes. An example would be a ball thrown straight upward; atthe top of its trajectory its velocity is zero, but it has a nonzero acceleration downward.14.Yes. A ball thrown straight upward and caught when it returns to its release point has zero average velocity,but it has been accelerating the entire time.16.When she returns to her original position, her speed is the same as it was initially; that is, 4.5 m/s.18.(a)No. Displacement is thechangein position, and therefore it is independent of the location chosen for theorigin.(b)Yes. In order to know whether an object’s displacement is positive or negative, we need to knowwhich direction has been chosen to be positive.Solutions to Problemsand Conceptual Exercises1.Picture the Problem:You walk in both the positive and negativedirections along a straight line.Strategy:The distance is the total length of travel, and thedisplacement is the net change in position.Solution:(a)Add thelengths:()()0.750.60 mi0.60 mi1.95 mi++=(b)Subtractxifromxfto find the displacement.0.750.00 mi0.75 mifixxx=−=−=Insight:The distance traveled is always positive, but the displacement can be negative.

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