Physics For The Life Sciences , 2nd UK Ed. Edition Solution Manual

Preview (16 of 533 Pages)

100%

Purchase to unlock

Loading page image...

1CHAPTER ONEPhysics and the Life SciencesMULTIPLE CHOICE QUESTIONSMultiple Choice 1.1Correct Answer (c).mbraindirectly proportionaltoMbodymeans that b = +1 in:bbrainbodyma M(1)Note that the coefficientbis the slope of thecurve after the natural logarithm is taken onboth sides of Eq. [1].Multiple Choice 1.2CorrectAnswer(c).Wecanargueintwoways:physically,wenotethatthesloperepresentsanactualphysicalrelation.Replotting data by using another unit systemcannot change the physical facts. The originalrelationship is given by:bbrainbodyma M(1)Mathematically, plottingmbrainin unit g meansthat we use values that are larger by a factor of1000 on the left side in Eq. [1]. Thus, for Eq.[1] to remain correct, the prefactor must alsobe larger by a factor of 1000. In Eq. [2] wetake the natural logarithm on both sides of Eq.[1], with the brain mass in unitkgon the right–hand side:ln()lnln()brainbodymkgabMkg(2)In Eq. [3] we rewrite Eq. [1] once more withnatural logarithms, but use the brain mass inunitg:ln()ln(1000 )ln()brainbodymgabMkg(3)in which ln (1000a) = ln 1000 + lna.Eqs. [2] and [3] differ only in that a constantterm, ln1000 = 6.908, is added in the secondcase. This represents a vertical shift of thecurve, but not a change in its slope.Multiple Choice 1.3CorrectAnswer(e).Theprecisionofeachnumber is represented by the smaller power ofteninthenumber.Smallerpowersoftenindicate more precise numbers. (d) is the leastprecise number since the smallest power of tenis 1011for the last digit. (a) follows with aprecision of 106, (b) with 10-2, (c) with 10-6,and (e) is the most precise with 10-18being thesmallest power of ten in the number.Multiple Choice 1.4Correct Answer (a). The number of significantfigures in the number represents its accuracy.The more significant figures a number has, themoreaccuratethenumberis.Inorderofincreasingaccuracy,(e)hasonlyonesignificant figure, (d) has two, both (b) and (c)have four, and (a) is the most accurate withfive significant figures.CONCEPTUAL QUESTIONSConceptual Question 1.1No.Webuildphysicalmodelstodescribeobservations of the world around us. Theseobservationsfacelimitationsthatarethenunavoidably transferred to the physical model.Uponimprovingonourobservations,themodelmightstillbevalid,itmightneedcorrections,ormightbewronginafundamental way. Similarly, to build physicalmodelswearerequiredtomakesomeassumptionsasastartingpoint,theseassumptions form the basis for the physicalmodel. However, further observations mightconfirm or deny the initial assumptions andthus validate or invalidate the physical model.Physical models are therefore under constantrevision and continued testing.Conceptual Question 1.2(a)The number 11 is represented by the digit. Note that 10 in base-12 actually representsthe number 12 in base-10. This is because thesequence of digits one-zero in base-12 means 0120+ 1121when expanded in base-10.(b)Repeated integer division of 3498572 by12 will yield the various digits in base-12:3498572 / 12 = 291547 with 8 as residue, sothat 8 is the digit for 120; 291547 / 12 = 24298with 7 as residue, so that 7 is the digit for 121.

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition2Continuing this division we find the sequenceof residues: 8120+ 7121+ 7122+ 8123+ 0124+ 2125+ 1126= 3498572.Thus, 3498572 in base 12 is represented a1208778.(c)The number63 expands as 10123+ 6122+ 11121+ 3120, where we haveconvertedfromto10,to11,andmultiplied each digit by the respective powerof 12 for the place in the number. The result is18279 in base-10.Conceptual Question 1.3This office is roughly 4 m wide by 5 m long by3 m high. Therefore, in m3the volume is 60m3. Since 1 m = 102cm = 103mm = 10-3kmwe can express the volume of the office as 60m3= 6107cm3= 61010mm3= 610-8km3.Althoughallthesevaluesarecorrectexpressing volumes of this order of magnitudewould be simpler in m3.Conceptual Question 1.4To convert a temperatureTFin Fahrenheit toTCin Celsius we use the expression:5 932CFTTTherefore, in terms of the chirps per minute,the temperature in degrees Celsius is:4051894CNTAlthoughtheformulaisdimensionallyincorrect asNis expressed in chirps per minuteand all other numbers are dimensionless, it canstillbeusedtocorrectlypredictthetemperature.Simplificationshavebeentomake it easier to write the formula by omittingall the units. If we wanted to include the units,thecorrectvalueswouldbe18F,40chirps/min, 4 min/(chirpF), and 5/9C/F.Conceptual Question 1.5We can expect a smaller slope of the graph forthe birds as compared to the mammals. Asmaller slope on this log-log graph represents asmaller ratio ofmbraintoMbodyand thus smallerbrain size relative to the size of the subjects intheclass.Fromthedatadiscussedinthechapter, on average and insofar as we can inferthe intelligence of non-human subjects, largerbrain mass to body mass ratio seems to becorrelated to intelligence.Conceptual Question 1.6In my household we average around 1400 kWhpermonthofelectricityconsumption.Converting kWh/month we find an averagepower consumption of 1900 W and thus anaverage of 1900 joules of electrical energyused each second. There are seven individualsliving in my household so each person uses270 W or about 300 joules every second. If weareconsideringacityofpopulation106individuals, then the power consumed will be:68JJss10300310Since the bomb released 106Joules of energy,it will power the city for:1458 J s10JTime310 s4 days310Note that this result is of the same order ofmagnitude as the result found in Example 1.11.Averaging first the electric bill from a numberof households we could do a better estimate.We could also add the electric bills from anumber of businesses to add it to the total.However, we would expect the result to be stillwithin the same order of magnitude.ANALYTICAL PROBLEMSProblem 1.1(a)1.23102(b)1.230103(c)1.23000104sincethelastzeroissignificant it must be expressed(d)1.2310-1(e)1.2310-3the zeros between the decimalpoint and the 1 are not significant(f)1.2300010-6thezerosbetweenthedecimal point and the 1 are not significant,however the three zeros to the right of the 3 aresignificant

Loading page image...

Chapter 13Problem 1.2(a)5 significant figures(b)4 significant figures(c)4 significant figures; the zero to the right ofthe decimal point and before the first non-zerodigit is not a significant figure(d)4 significant figures; when a number iswritteninscientificnotation,alldigitsexpressed are significant figuresProblem 1.3The product must be given to the number ofsignificantfiguresoftheleastaccuratenumber, that is the number with the smallestnumber if significant figures:(a)5.6110-1bothnumbershavethreesignificant figures since the zeros in 0.00456after the decimal point and before the 4 are notsignificant(b)5.61102note that the last zero in 1230 isnot significant(c)5.6088100= 5.6088 both numbers havefive significant figures(d)5.609100= 5.609 note that the zero in0.01230 after the decimal point and before the1isnotsignificantwhilethelastzeroissignificantProblem 1.4The quotient follows the same rules as themultiplication; the result must be given to thenumberofsignificantfiguresoftheleastaccurate number:(a)2.70104bothnumbershavethreesignificant figures since the zeros in 0.00456after the decimal point and before the 4 are notsignificant(b)2.70103note that the last zero in 1230 isnot significant(c)2.697410-7bothnumbershavefivesignificant figures(d)2.69710-5note that the zero in 0.01230after the decimal point and before the 1 is notsignificant while the last zero is significantProblem 1.5Sums and differences must be given with theprecision of the least precise number, wherethe precision is found by the smallest power often present in each number:(a)5.79102both numbers are precise to 100so the result must be quoted to that precision as579 and then expressed in scientific notation(b)1.23103while 0.456 is precise to 10-3,1230 is only precise to 101as the last zero isnotsignificantandtheresultincludessignificant figures up to 101(c)3.3310-1both numbers are precise to 10-3so the result is 0.333 which is then written inscientific notation(d)3.3410-1the number 123.123 is the leastprecise of the two to only 10-3 so the result is0.334whichisthenwritteninscientificnotationProblem 1.6The standard precedence of operations meansquotientsandmultiplicationsmustbeperformedfirstcomputingvalueswiththeresult quoted to the significant figures of theleastaccuratenumberinvolved.Thendifferences and sums are performed with theresult quoted to the significant figures of theleast precise number involved:(a)1.27102the multiplication requires threesignificant figures and the sum must be preciseto 100which is the precision of both numbers(b)5.62104the multiplication requires threesignificantfiguressothepartialresultisprecise only to 101and thus the sum will bequoted to a precision of 101(c)-3.71105thequotientrequiresthreesignificantfiguresandthepartialresultisprecise only to 103and thus the difference willbe quoted to a precision of 103(d)6.8510-6the multiplication requires threesignificantfiguresandthepartialresultisprecise to 10-8, while the number to be added isprecise to 10-10. Therefore the final result mustbe quoted to a precision of 10-8

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition4Problem 1.7My height is 165 cm or 1.65 m, written inscientific notation it is:(a)1.65 m(109nm / 1 m) = 1.65109nm(b)1.65 m(103mm / 1 m) = 1.65103mm(c)1.65 m(102cm / 1 m) = 1.65102cm(d)1.65100 m = 1.65 m(e)1.65 m(1 km / 103m) = 1.6510-3kmRepresenting the length in m (d) is best suitedfor lengths of the order of my height.Problem 1.8My mass is approximately 75 kg so my weightwould be:(a)75 kg10 m/s2(106N / 1 N) = 7.5108N(b)75 kg10 m/s2(103mN / 1 N) = 7.5105mN(c)75 kg10 m/s2= 7.5102N(d)75 kg10 m/s2(1 kN / 103N) = 7.510-1kN(e)75 kg10 m/s2(1 GN / 109N) = 7.510-7GNRepresenting the weight as 750 N as done inpart (c) or 0.75 kN as done in part (d) would bewellsuitedforforcesoftheorderofmyweight.Problem 1.9We start with a total time of 7106s so thatwe can calculate:(a)7106s(1 min / 60 s) = 105min(b)105min(1 h / 60 min) = 2103h(c)2103h(1 day / 24 h) = 8101day(d)8101day(1 year / 365 day) = 210-1yearRepresenting the total time as 80 days (c) or0.2 years (d) would be well suited for times ofthe order of the time spent brushing your teeth.Problem 1.10We have a distance of 42.195 km, and a timeof 2 h 2 min 11 s. A suitable combination ofdistance and time that leads to a quantity withthe dimensions of speed [L]/[T] would be theratioofthedistancetothetime.Forthecalculations,wecanconvertthetimeintoseconds as 7331 s. The average speed is thus:(a)(42.195 km / 7331 s)(3600 s / 1 h) =2.072101km/h(b)(42.195 km / 7331 s)(103m / 1 km) =5.756100m/s = 5.756 m/s(c)(42.195 km / 7331 s) = 5.75610-3km/s(d)(42.195 km / 7331 s)(3600 s / 1 h)(103m / 1 km) = 2.072104 m/h(e)(42195 m / 7331 s)(1 s / 109ns)(106m / 1 m) = 5.75610-3m/nsProblem 1.11You must be careful when converting units toany power other than one. For example, 1 s2=1 s2(103ms / 1 s)2= 106ms2. With this inmind:(a)10 m/s2(103mm / 1 m) = 104mm/s2(b)10 m/s2(1 s / 103ms)2= 10-5m/ms2(c)10 m/s2(1 km / 103m)(3600 s / 1 h)2=105km/h2(d)10 m/s2(1 Mm / 106m)[(3600 s / 1 h)(24365 h / 1 yr)]2= 1010Mm/yr2(e)10 m/s2(106m / 1 m)(1 s / 103ms)2=101m/ms2= 10m/ms2Problem 1.12The area will be heightwidth. Since 1 cm =10 mm = 104m = 10-2m = 10-5km, the areain various units is:(a)30 cm20 cm = 6102cm2; note thenumber of significant figures(b)300 mm200 mm = 6104mm2(c)(30104m)(20104m) = 61010m2(d)(3010-2m)(2010-2m) = 610-2m2(c)(3010-5km)(2010-5km) = 610-8km2Problem 1.13For conversion purposes, first note that 1 km =103m = 104dm = 105cm, and for litres 1 L =103cm3.Furthermore,althoughtheradiusgivenhasfoursignificantfigures,wethemultiplicative factor for the volume has beenapproximated to 4 with only one significantfigure. With these in mind:(a)4(6378 km)3(105cm / 1 km)3= 1.0381027cm3= 1027cm3(b)4(6378 km)3(103m / 1 km)3= 1.0381021m3= 1021m3(c)4(6378 km)3= 1.0381011km3= 1011km3(d)4(6378 km)3(104dm / 1 km)3= 1.0381024dm3= 1024dm3(e)Using (a) 1.0381027cm3(1 L / 103cm3) = 1.0381024L = 1024L

Loading page image...

Chapter 15Problem 1.14The density of an object measures the mass perunit volume. Since in Problem 1.13 we are toldthat the volume of a sphere is approximatelyV= 4R3, the density should be:3Density4mmVRWe can now use this formula withR= 6378km andm= 5.97421024kg to find therequireddensities.Notethatalthoughtheradius has four significant figures and the masshas five significant figures, we were told toapproximate the volume by multiplying by 4, anumber with only one significant figure. Notethat 1 km = 103m = 104dm = 105cm = 109m= 1015pm and 1 kg = 103g = 109g = 1012ng= 10-3Mg. The densities are:(a)(5.97421024103g) / [4(6378105cm)3] = 5.757100g/cm3= 6 g/cm3(b)(5.97421024kg) / [4(6378103m)3]= 5.757103kg/m3= 6103kg/m3(c)(5.9742102410-3Mg) / [4(6378km)3] = 5.757109Mg/km3= 6109Mg/km3(d)(5.97421024109g) / [4(63781015pm)3] = 5.75710-24g/pm3= 610-24g/pm3(e)(5.974210241012ng) / [4(6378109m)3] = 5.75710-3ng/m3= 610-24ng/m3Problem 1.15Anequationcannotbebothdimensionallycorrect and wrong. It is important to note thatthesumordifferenceoftwotermsisdimensionally correct only if both terms havethe same dimensions.(a)Wrong. On the left-hand side we have [A]= [L2], while on the right-hand side we have[4][R]=1[L]since4isadimensionless quantity. Since [L2][L] theequation is dimensionally wrong.(b)Wrong. On the right-hand side we have [x1]= [L] being added to [v1t2] = ([L]/[T])[T2] =[L][T]. Since [L][L][T], the twoquantities cannot be added and the formula isdimensionally wrong.(c)Correct. On the left-hand side we have [V]= [L3], and on the right-hand side we have[xyz] = [L][L][L] = [L3]. Since both sidesmatch the equation is dimensionally correct.(d)Wrong. On the right-hand side we have[m/2] = [T]/[2] = [T] being added to [7] = 1.Since [T]1, and a quantity with dimensionscannot be added to a dimensionless quantitythe formula is dimensionally wrong.Problem 1.16We are told that2vt, where v is the speedof the object, and t is the time it has travelled.Lett1= 4 s andt2= 8s so thatv1= 10 m/s andv2is the speed we want to find.From the proportionality relationship:12212211andvvttTaking the ratiov2/v1we find:222122112121ttvtvtWith the given values:221m1m2ss224s102.58stvv tThe proportionality betweenvandtdescribedin this problem is often called an Inverse-Square Law. In our case, doubling the timereduces the speed by a factor of four; halvingthe time would increase the speed by a factorof four.Problem 1.17From Figure 1.9 we derived the relationship:0.68brainbodymMWe can use one of the data points listed inTable1.5alongsidetheproportionalityrelationship to find out the mass of the brainfor the pygmy sloth. I will use line 12 in thetable that lists values for the mainland three-toed sloth with brain massms= 15.1 g andbody massMs= 3.121 kg. We know that thepygmy sloth has a body mass ofMp= 3 kg andwe want to find its brain massmp. From theproportionality relationship:0.680.68andssppMMmm

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition6We can then solve formm:0.680.68315.1g14.7 g3.121ppssMmmM Thus, a mass of approximately 15 g is what wewould expect for the brain of the pigmy three-toed sloth. A brain mass of 200 g would put inquestion the measurement for the body mass,highlight an experimental error, an anomaloussample of the species (maybe with a severebrain defect), or simply an error in the reporteddata or calculations (possibly by one order ofmagnitude).Problem 1.18(a)The resulting double–logarithmic plot isshown in Figure 1. An organized approach toplotting these data is based on extending Table1.6toincludethelogarithmvaluesofwingspan and mass, as shown here in Table 1.Figure 1Using these logarithmic data, the given powerlawW=aMbis rewritten in the form lnW=blnM+ lna.Table 1BirdW(cm)lnWM(g)lnMHummingbird71.95102.30Sparrow152.71503.91Dove503.914005.99Andeancondor3205.77115009.35Californiacondor2905.67120009.39The constants a and b are determined from thisequationinthemannerdescribedintheAppendixGraph analysis methodson p. 9 – 12inChapter1.Fortheanalysiswedonotchoose data pairs from Table 1. As the graphin Fig. 1 illustrates, actual data points deviatefrom the line that best fits the data (representedby the solid line). To avoid that the deviationof actual data affects our results, the two datapairs used in the analysis are obtained directlyfrom the solid line in Fig. 1. We choose lnW1= 2 with lnM1= 2.6 and lnW2= 6 with lnM2= 9.6. This leads to:( )2.0ln( )2.6()6.0ln( )9.6()( ) 4.0(9.62.6)IabIIabIIIb(1)Thus,b= 0.57. Due to the fluctuations of theoriginal data and the systematic errors youcommit when reading data off a given plot,values in the interval 0.5b0.6 may havebeen obtained.Substitutingthevaluewefoundforbinformula (I) of Eq. [1] yields: 2 = 1.48 + lna,i.e., lna= 0.52 which corresponds to a valueofa= 1.7.(b)We use the given value for the pterosaurs’wingspan:W= 11 m = 1100 cm. The value hasbeen converted to unit cm since that is the unitused when we developed our formula in part(a).Wefirstrewritetheformulaforthewingspanwiththemassasthedependentvariable:1/bbWWa MMa Entering the given value for the wingspan thenleads to:1/0.5711008540085.41.7MgkgThe mass of pterosaurs did not exceed 85 kg.

Loading page image...

Chapter 17(c)We use again the power law relation wefound in part (a) and insert the given value forthe person’s mass:0.571.7 70000580WcmA person of mass 70 kg would need a 5.8 mwing span. Notice that a pterosaur has a only20 % larger mass than a human, but requires a90 % increase in wingspan in order to achieveflight. Whereas a sparrow has a 500 % largermass than a hummingbird, yet it requires onlya 100 % increase in wingspan. This shows theexponential nature of this relationship.Problem 1.19Both the car and the cow produce the samemass,thecowproducingmethane,thecarproducing carbon dioxide. The molar mass ofmethane (CH4) is 16 g/mol while the molarmass of carbon dioxide (CO2) is 44 g/mol. Thismeans in one gram of CH4you will find 1/16mol, while in one gram of CO2you will find1/44 mol. Therefore, if you have the samemass of CH4and CO2, there will be 44/16more moles of CH4than of CO2. Since permole CH4has 3.7 times the global warmingpotential of CO2, the cow will have 3.7(44/16)=10timestheglobalwarmingpotential of the car.Problem 1.20AccordingtotheFuelConsumptionGuide(2011)publishedbyNaturalResourcesCanada, the compact car with the best fuelefficiency in the city consumes on average 4.5L of regular gas to travel a distance of 100 km.IndowntownToronto,Ihavenoticedgasstations being somewhere between 10 to 30blocks apart. In contrast, gas stations seem tobe closer to each other in the suburban areasroughly 5 to 20 blocks apart. If a block is about100mitseemsagoodaveragedistancebetween gas stations to be 2 km. This meansthat on average you will spend 90 mL of gas togo from one station to the next. Last week, thepriceofregulargasinTorontooscillatedbetween 120 ¢/L and 130 ¢/L. If the startingstation had a price at 125 ¢/L then it wouldcost (125 ¢/L)0.09 L = 11.3 ¢ to drive to thenext station. Therefore, to break even as yougo searching for another gas station, the cost atthenextstationshouldbeabout11¢/Lcheaper.Problem 1.21To compare and rank the relative magnitudeswe need to establish the point of comparison.Regardlessofthecomparisonstandard,therelativemagnitudesandthustherankingshould come up the same. I will use the forceexertedbymymotherasthepointofcomparison and will call that forceFm. We willuse the proportionality relationship:2MFR,whereMis the mass of the object exerting thegravitationalforce,andRisthedistancebetween that object and the baby. My mother’smass was about 70 kg and the moment I wasborn I was really close to my mother. Thedistancebetweenourcenterswasprobablyaround 10 cm.The mass of the Moon is 71022kg and it isat a distance of 4108m from Earth. So theforce exerted by the MoonFMwill be relatedtotheforceexertedbymymotherFmaccording to:2222212010610MMMmMmmmMmmFmRRmFmRRThus,FM= 60Fm.The mass of the Sun is 21030kg and it is at adistance of 21011m from Earth. So the forceexerted by the SunFSwill be related to theforce exerted by my motherFmaccording to:22282225310310SSSmSmmmSmmFmRRmFmRRThus,FS= 104Fm.

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition8The mass of Jupiter is 21027kg and it is onaverage at a distance of 81011m from Earth.So the force exerted by JupiterFJwill berelated to the force exerted by my motherFmaccording to:22252226310210JJJmJmmmJmmFmRRmFmRR.Thus,FJ= 0.6Fm.The mass of Mars is 61023kg and it is onaverage at a distance of 21011m from Earth.So the force exerted by MarsFMarswill berelated to the force exerted by my motherFmaccording to:2222232510310MarsMarsMarsmMarsmmmMarsmmFmRRmFmRRThus,FMars= 0.03Fm.The correct ranking from smallest to largest is:MarsJmMSFFFFFProblem 1.22Assumeyoudonotneedanybreaks,forresting or eating. Furthermore, assume that youare moving the dirt just to your back so thatyou do not take time taking the dirt to anotherplace. Furthermore, assume that you have anunlimited air supply, which is not realistic ifyou are taking the dirt from the front of thetunnel and placing it at your back. For yourbody to fit through the tunnel must have adiameter of at least your shoulder width, butyou will need extra room to maneuver. Let’ssay the tunnel needs to be 1 m in diameter. Thetotal volume of the tunnel is then the volumeof a cylinder of radiusR= 0.5 m and totallengthL= 100 m:23378.5m80 mLRVSince each spoonful is about 5 cm3, it will take80 m3/ 5 cm3~ 109spoonfuls of dirt to dig thetunnel. If it takes you about one second perspoonful, then you will need 109s ~ 40 yearsto dig the full volume of the tunnel. This resultquiteclearlynegatesallthepreviousassumptions about breaks for rest and food,and you will definitely need more time to finda place for the removed dirt.Problem 1.23The current life expectancy of a human isabout 70 years ~ 2109s. Although the heartrate changes with age and with the level ofactivity, on average we could use the rest heartrate of an adult, which is around 70 bpm, or 70beats per minute. Since 70 bpm = 1.2 beats/s,throughout your life your heart will beat:99beats s210 s1.2210 beatsIn other words, your heart will beat about 2billion times.Problem 1.24I trim my fingernails about once a week andthe piece cut is about 1 mm wide, so the speedof growth is about 1 mm/week:(a)1 mm/week = (10-3m)/(7243600 s) =210-9m/s(b)1 mm/week = (103m)/(7 day) = 1102m/day = 100m/day(c)1 mm/week = (10-1cm)/(7/365 year) = 5cm/yearProblem 1.25The number of grains of sand will beVb/Vg,whereVbis the volume of the beach, andVgisthe volume of a grain of sand. Since the beachis box-shaped,Vb=lwd, wherelis thelength,wis the width, anddis the depth.(a)Since 1 mm3= 1 mm3(1 m / 103mm)3=10-9m3, withh= 4 m we have:1293100 m10 m4 m410grains10mbgVV(b)Since the grains have the same volume of10-9m3buth= 2 m we have:VbVg100 m10 m2 m109m321012grains

Loading page image...

Chapter 19(c)Withh= 4 m and the grains with averagevolume 2 mm3= 210-9m3we have:1293100 m10 m4 m210grains210mbgVVFrom (b) to (c) we have doubled the depth ofthe box. Doubling one of the dimensions of thebox will only double the total volume of thebox. If we had instead doubled each one of thedimensions of the box, the volume would haveincreasedbyafactorof23=8.Wealsodoubled the total volume of a grain of sand andthisthereforecompensatesexactlythedoublingofthedepthoftheboxandthedoubling of the volume of the box. This isdifferent from our conversions of cubed unitsbecauseintheconversionsallthreedimensions are being converted. Therefore, thefinal effect is the cube of the conversion in oneof the linear dimensions.Problem 1.26Earth is roughly a sphere of radius 6400 km.The surface area of a sphere of radiusRisgivenby4R2.However,about¾ofthesurface of the Earth is covered by water.Therefore the dry surface of the Earth is anarea of:22142146400 km10m4dryARIfapersonstretcheshis/herarmsout,thedistance between the fingertips will be aboutthe same as the height of the person. Thisdistance isknown as the arm span. If weassume and average height of 1.70 m then theperson can reach around in a circle of diameter1.70 m. However, if the person leans, the reachmight be extended. We assume the reach of theperson extends to a circle 2 m in diameter. Thearea for the person not to touch another personis then:223m2persondAThe number of people able to stand withouttouching each other is then:141310310trillion330drypersonAAThis is about four thousand times the currentworld population.Problem 1.27Earth is roughly a sphere of radius 6400 km.The surface area of a sphere of radiusRisgivenby4R2.However,about¾ofthesurfaceoftheEarthiscoveredbywater.Therefore the dry surface of the Earth is anarea of:22214146400 km10m4dryARAn average person is about 1.70 m tall andabout 60 cm wide at the shoulders. To simplifythe estimate, a coffin would then be 2 m longby 1 m wide so each dead person will take upAperson= 2 m2. There is enough room to bury:141310510people2drypersonAAAccordingtoanestimatedonebythePopulationReferenceBureau,about100billion people (1011) have lived and died onEarth.Tothesignificantfiguresweareestimating, this number is not relevant. Thecurrent population on Earth is close to 7 billion(7109), and according to theUnited Nationsthe current death rate is about 8.4 deaths per1000 people per year. If the population onEarth remains stable at 7 billion, each yearabout (8.4/1000)(7109) = 6107peopledie. Therefore to cover the dry surface of Earthwith graves we will have to wait:peopl1e57yr3510people810 yr1Myr610However, this estimate ignores the changes inbirth rate as well as death rate. Both thesequantities are quite hard to predict.

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition10Problem 1.28IftwomammalsAandBhavethesamedensities then:MAVAMBVB,whereMis the mass andVthe volume of themammal. Therefore we can reorganize for theratio of the volumes:BABAVMVMThat is, if all mammals have approximately thesame density, the ratio of the volumes of twomammals is the same as the ratio of theirmasses. We can use Table 1.5 to find therequired ratios:13512652.1HippopotamusHippopotamusSheepSheepVMVMand:190.825000.075LionLionChipmunkChipmunkVMVMThe volume of the ark isVArk= (3005030)cubit3(0.45 m / cubit)3~ 4104m3. Sincethe density of water is 1000 kg/m3we can useTable 1.5 to find the volume of each and all theanimals according to:3kgm1000AnimalAnimalAnimalWaterMMVDAdding the masses of all land mammals inTable 1.5 (all except 34, 40, 43 and 49), we get1.41104kg but since we need two of eachanimalMTotal= 2.82104kg. Therefore:343kgm2.8210301000kgmTotalTotalWaterMVDSinceVTotal<VArk, two of every land mammalfrom Table 1.5 will fit in the ark. For the lastpartofthequestion,wecancalculatethevolume of the blue whale:33kgm5805960 m1000BlueWhaleBlueWhalewaterMkgDVSinceVTotal<VBlueWhale, two of every landmammal from Table 1.5 could be housed in thevolume equivalent to that of a blue whale.

Loading page image...

11CHAPTER TWOKinematicsMULTIPLE CHOICE QUESTIONSMultiple Choice 2.1CorrectAnswer(d).Theinitialandfinalposition vectors are equal so the difference willbe zero. That is, the displacement is zero ascalculatedfromEq.[2.1].However,thedistance travelled is not zero as this takes intoaccount the path used to travel. Using Eq. [2.2]we see that the average velocity will be zero.However, since the distance is not zero, theaverage speed will be non-zero and positive.Multiple Choice 2.2Correct Answer (d). The accelerations in thex-andy-directions are independent, so havingone the acceleration in one direction beingequal to zero does not imply anything specificfor the acceleration in the other direction.Multiple Choice 2.3Correct Answer (c). As the child moves in acircle, her acceleration is centripetal and thusdirected toward the centre of curvature. In thiscase, the centre of curvature is the centre of thecircle made by the child and is located wherethe father stands.CONCEPTUAL QUESTIONSConceptual Question 2.1Yes, when the object is slowing down.Conceptual Question 2.2If the average velocity in a given time intervaliszero,thedisplacementoverthattimeinterval must also be zero.Conceptual Question 2.3(a)Atthemomenttheobjectreachesitshighest altitude, its velocity is zero and itsacceleration is equal to the acceleration due togravity.(b)Same as part (a) since the acceleration ofthe object is constant.Conceptual Question 2.4The dog runs at twice the speed of its masterfor the same amount of time. That means thatthe dog runs twice as far as its master. Since itsmaster travels a distance of 1 km, the dog musttravel 2 km.Conceptual Question 2.5(a)Yes, because the object may change itsdirectionwithoutchangingitsspeed.Anexample of this is uniform circular motion in.(b)No.Conceptual Question 2.6(a)Yes.Iftheprojectile’svelocityhasahorizontal component, then at the top of thetrajectory the velocity is completely horizontaland the gravitational acceleration is vertical.(b)Along the entire path if the projectile isthrown vertically.Conceptual Question 2.7(a)The same as if thrown on steady ground.(b)The path is the projectile trajectory in thelast equation of Case Study 2.1, wherevinitial,xisthe constant horizontal velocity of the train.Conceptual Question 2.8Its component along the path of the objectaccelerates the object; i.e., its motion is nolonger uniform. The perpendicular componentcauses a curved path.Conceptual Question 2.9(a)Uniform circular motion(b)Accelerated motion along a straight linewith linear increase (or decrease) in speed.Conceptual Question 2.10(a)No;itsx–andy–componentsvarycontinuously, i.e., the velocity in a Cartesiancoordinate system is not constant.(b)Yes; its speed along the path is constant asno acceleration acts in that direction for auniform motion.(c)Yes if we refer to its magnitude, no if werefer to acceleration as a vector in a Cartesiancoordinate system.

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition12Conceptual Question 2.11Thegravitationalaccelerationdownwardissmallerthanthecentripetalaccelerationrequired to hold it on a circular path; as a resultthe bottom of the bucket has to push the waterdownwhenitisatthetopoftheloop.Newton’s third law then states that the waterpushesthebottomofthebucketupward.Compare the water to a person on a fast rollercoaster, as opposed to a slow Ferris wheel.ANALYTICAL PROBLEMSProblem 2.1(a)We define the vector sum asr, so that wecan writer=a+b, or, in component notation,5( 14)95510xxxyyyrabrab  Thus,r= (–9, 10) as shown in Figure 1.Figure 1(b)The magnitude of vectorris calculatedwith the Pythagorean theorem:2222||(9)1013.5xyrrrTocharacterizethevectorrinpolarcoordinates, its angle with the positivex–axismust also be determined. This can be donegeometrically, as can be seen from Figure 1:tan1.11132yxrr   Your pocket calculator may show= – 48°.Take care to always draw the vector to checkthatyouranswermakessense.Toyourcalculator, –a/b cannot be distinguished froma/(–b). It is the job of the person using thecalculator to interpret the physical meaning ofthe number the calculator returns.Problem 2.2The problem implies two equations, that wewrite each in component form:( )6()2()5()8xxyyxxyyIabIIabIIIabIVab (1)Adding the first and third as well as the secondand fourth lines of Eq. [1], leads to vectora:21210xyaawecombinethesecomponentstofindthemagnitude ofa:220.555.02aSubtracting the third from the first and thefourth from the second lines in Eq. [1] in turngivesbin component form:21126xybb and the magnitude of vectorb:22=5.5 +( 3) =6.26bProblem 2.3Using thex1-coordinate axis the red dot islocatedatcoordinate3sothat13r.However, using thex2-coordinate axis the reddot is at coordinate 1 so that21r. Note that211rrx,where12xisthepositionvector for the origin of thex2-coordinate axisas measured on thex1-coordinate axis.

Loading page image...

Chapter 213Problem 2.4In the (x1,y1)-coordinate system the red dot islocated at coordinates (2,3) so that the positionvector is12,3r. In the (x2,y2)-coordinatesystem the red dot is located at coordinates(−2,2)sothatthepositionvectoris.Notethat211rra,where1(4,1)ais the position vector for the originof the (x2,y2)-coordinate system as measuredon the (x1,y1)-coordinate system.Problem 2.5Using Eq. [2.2],mkmshav10.136.4v.Problem 2.6(a)Figure 2 shows the relative position ofpointsAandBinthefractureplane.Forconvenience, this plane has been chosen as thexy–plane.Withthischoiceofcoordinatesystem, the horizontal slip is along thex–axis,where the pointAon the bone has moved topositionC. The vertical slip, along the verticaly–axis, moves the point on the bone from C topointB. To put it another way, pointAslips topointB; the horizontal slip is thex–componentof the vectorABand the vertical slip is they–componentofAB.Thus,thedisplacementfromAtoBcan be written in vector notation:224003(4 mm)(3 mm)5 mmABACCBABFigure 2Thetotaldisplacementwithinthefractureplane is 5 mm.(b)Of the two components of displacementconsideredinpart(a)thehorizontaldisplacementACdoesnotcontributetoadisplacement along the axis of the bone. Figure3 shows how the component of the verticaldisplacementfrompart(a),i.e.,CB(equivalent to displacementAD), relates to thedisplacement along the bone’s axis. In Figure 3the displacement from pointAto pointEis thedisplacement along the bone’s axis (and thedisplacement from pointEto pointDis thedisplacement perpendicular to the bone’s axis).Figure 3The angle of= 20° given in the problem textis the angle between the vectorsAEandAD.WecancalculatethelengthofAEusingtrigonometry:cos20(3 mm)cos 202.8 mmAEADAEThe displacement along the bone’s axis is 2.8mm. In an analogous fashion it can be shownthatthedisplacementperpendiculartothebone’s axis, i.e. the magnitude of the vectorgiven byED, is 1.0 mm.Problem 2.7(a)We choose the origin at the bottom of thefeet of the person, with thex–axis pointinghorizontal and toward the right and they–axispointing straight up. In this coordinate systemwe express the two vectors shown in Figure2.32 for the male person,d1andd2. Based onthe given lengths (magnitudes) and the givenangle we find:

Loading page image...

’s Solution Manual to accompany Physics for the Life Sciences, Second Edition141122200||150|| sin45.9|| cos65.5ddcmddcmdcm(1)Note the negative sign in they–component ofthe second vector! The vector from the bottomof the feet to the hand is the sum of the twovectors in Eq. [1]:1245.984.5maledddcmcm (b)Thecalculationsareanalogousforthefemale person and yield:37.376.8femaledcmcm Problem 2.8(a)The distance between the two intersectionsof the dashed line with the skull in the topsketch measures 54 mm. This provides us witha scale for Figure 2.33 because the problemstates that the dashed line measures 16 cm fora real skull. Before analyzing the figure acoordinatesystemmustbechosen.Thisisshown in Figure 4 with the direction up chosento be thez–direction for both sketches, thedirection to the ob-server’s right in the topsketch thex–direction and the direction towardthe back of the skull in the bottom sketch they–direction.For convenience we choose the pointAtocoincidewiththeoriginofthecoordinatesystem:A= (0, 0, 0). We use the skull in thetop sketch to obtain the length of thexz–projection of vectorABand then we use theskull in the bottom sketch to obtain the lengthof theyz– projection of the same vector. Thevector sum of both projections yields the actualvector fromAtoBin the last step. In the topsketchofFigure2.33wemeasureforthehorizontal component of the distance betweenAandB12.0mmandfortheverticalcomponent of the same distance 20.7 mm. Theprojection of the vector betweenAandBin thetop sketch is given byABxz= (3.56 cm, 0, –Figure 46.13 cm) where we used the scale conversionas found at the beginning of the solution. Theindex “xz” identifies that this is not the finalvectorABbut only its projection into thexz–plane. Thez–component is negative since thepointBlies at a negativez–value. Now werepeat the same analysis for the bottom sketchof Figure 2.33. We measure for the horizontalcomponent of the distance betweenAandB21.9 mm and for the vertical component of thesame distance again 20.7 mm. The projectionof the vectorABin the bottom sketch is givenbyAByz= (0, 6.49 cm, – 6.13 cm) with thescale conversion taken into account. Note thatit is not a coincidence that we find the samelength for the vertical direction in both plotssince it is physically the same distance. Aslong as your two measurements of that lengthvary within the margin of error for readingdata off your ruler, it is fine. Otherwise, this isa built–in self– control and you should go backandcheckyourmeasurementsoncemore.From the two projections we now find thevectorABand its length:2223.56 cm, 6.49 cm,6.13 cm3.566.49(6.13)9.6 cmABABAB 

Loading page image...

Chapter 215(b)For convenience we assume that the skullis left–right symmetric, i.e., we have only todetermine the position of one of the two lastmaxilla molars. That point coincides with pointBinFigure2.33.Letusdefinethepointbetween the two central maxilla incisor teethas pointC. Using the same approach as in part(a)andmeasuringonthetopsketchahorizontaldistanceof12.5mm,averticaldistance of 2.0 mm and, on the bottom sketch,a depth displacement of 20 mm we find for thevectorCB= (3.70 cm, 0.60 cm, 5.93 cm). Dueto the symmetry the vector to the other lastmaxilla molar (called pointB’) isCB’= (–3.70cm,0.60cm,5.93cm).Theanglebetween these two vectors follows from thescalar product:2222cos(3.7,0.6,5.93)(3.7,0.6,5.93)3.70.65.9313.690.3635.160.4413.690.3635.16CB CBCB CBCBCBCB Thus, we findφ= 64°.(c)There are two ways to confirm this angleusing Figure 2.34. Either you read the anglebetween the two straight-line segments off thefigure directly using a protractor, or you use aruler and Figure 5:Figure 5The figure shows how the distance between thetwo last maxilla molars, which is given as|BB’| = 2dx, and the distance from that line topointC,dy, are related to the anglewe seek.Measuring the distance between the centres ofthe two last molars asdx= 42.5 mm andmeasuring the distance perpendicular asdy=31.5 mm we find:12(42.5 mm)tan0.675231.5 mm)68xydd  Thisvalueagreeswithφ=64°withintheprecision of the analysis. The skull in Figure2.33 and the top view of the dentition in Figure2.34 are obtained from different sources, thus,normal anatomic variations may contribute tothe difference of the found values. Note thatwe did not convert the measurements in part(c) using a scale for Figure 2.33. This is due tothe fact that obtaining angles is based on thedivision of lengths, where such a scale factorcancels anyway.Problem 2.9Assumethatthebacteriummovesalongastraight line:42 8.4 cm4.810 s13.3 hr3.5m/sdtvActual bacteria move in random fashion; thistype of motion is discussed in Chapter 11.Problem 2.10Formotionwithconstantacceleration,theposition as a function of time can be foundusingEq.[2.9].Withthevaluesofinitialposition,initialvelocity,andconstantacceleration we can write:2555xtt(1)where we assume all quantities in SI units.Figure 6Graphically, Eq. [1] is an inverted parabolasuch that att= 0 s the position isx= 5 m, andastincreases,xdecreases passingx= 0 m attimet= 0.62 s. Att= 10 s the position reachesx=−545m.Qualitatively,thegraphisillustrated in Figure 6.

Preview Mode

This document has 533 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Physics For The Life Sciences , 2nd UK Ed. Edition Class Notes

Physics For The Life Sciences , 2nd UK Ed. Edition Test Bank

Exploring Light Intensity Through Polarizers

IPL Lab Report

Distance Time Velocity Gizmo Handout-2

Energy Skate Park Activity

Exploring Density Measuring Mass and Volume of Obj

Student Exploration Distance Time Graphs

Barron's Physics Practice Plus: 400+ Online Questions and Quick Study Review (2022)

College Physics: A Strategic Approach, 4th Edition Test Bank

Company

Explore

Study Tools