Solution anual for Physics for the Life Sciences 2nd Edition by Allang Link download full: http://testbankair.co/download/solution-anual-forphysics-for-the-life-sciences-2nd-edition-by-allang/ CHAPTER ONE Physics and the Life Sciences ULTIPLE CHOICE QUESTIONS precise nuber since the sallest power of ten ultiple Choice 1.1 is 10 11 for the last digit. (a) follows with a Correct Answer (c). directly proportional precision of 10 6, (b) with 10-2, (c) with 10-6, to brain and (e) is the ost precise with 10-18 being the body eans that b = +1 in: a b sallest power of ten in the nuber. (1) ultiple Choice 1.4 brain body Note that the coefficient b is the slope of the curve after the natural logarith is taken on both sides of Eq. [1]. ultiple Choice 1.2 Correct Answer (c). We can argue in two ways: physically, we note that the slope represents an actual physical relation. Replotting data by using another unit syste cannot change the physical facts. The original relationship is given by: a b (1) brain body atheatically, plotting brain in unit g eans that we use values that are larger by a factor of 1000 on the left side in Eq. [1]. Thus, for Eq. [1] to reain correct, the prefactor ust also be larger by a factor of 1000. In Eq. [2] we take the natural logarith on both sides of Eq. [1], with the brain ass in unit kg on the right hand side: ln brain (kg) ln a b ln body (kg) (2) In Eq. [3] we rewrite Eq. [1] once ore with natural logariths, but use the brain ass in unit g: ln brain ( g) ln(1000a) b ln body (kg) (3) in which ln (1000 a) = ln 1000 + ln a. Eqs. [2] and [3] differ only in that a constant ter, ln1000 = 6.908, is added in the second Correct Answer (a). The nuber of significant figures in the nuber represents its accuracy. The ore significant figures a nuber has, the ore accurate the nuber is. In order of increasing accuracy, (e) has only one significant figure, (d) has two, both (b) and (c) have four, and (a) is the ost accurate with five significant figures. CONCEPTUAL QUESTIONS Conceptual Question 1.1 case. This represents a vertical shift of the curve, but not a change in its slope. ultiple Choice 1.3 Correct Answer (e). The precision of each nuber is represented by the saller power of ten in the nuber. Saller powers of ten indicate ore precise nubers. (d) is the least Copyright 2013 Nelson Education Liited 1
No. We build physical odels to describe observations of the world around us. These observations face liitations that are then unavoidably transferred to the physical odel. Upon iproving on our observations, the odel ight still be valid, it ight need corrections, or ight be wrong in a fundaental way. Siilarly, to build physical odels we are required to ake soe assuptions as a starting point, these assuptions for the basis for the physical odel. However, further observations ight confir or deny the initial assuptions and thus validate or invalidate the physical odel. Physical odels are therefore under constant revision and continued testing. Conceptual Question 1.2 (a) The nuber 11 is represented by the digit. Note that 10 in base-12 actually represents the nuber 12 in base-10. This is because the sequence of digits one-zero in base-12 eans 0 12 0 + 1 12 1 when expanded in base-10. (b) Repeated integer division of 3498572 by 12 will yield the various digits in base-12: 3498572 / 12 = 291547 with 8 as residue, so that 8 is the digit for 12 0 ; 291547 / 12 = 24298 with 7 as residue, so that 7 is the digit for 12 1. 2 Copyright 2013 Nelson Education Liited
Chapter 1 Continuing this division we find the sequence of residues: 8 12 0 + 7 12 1 + 7 12 2 + 8 12 3 + 0 12 4 + 2 12 5 + 1 12 6 = 3498572. Thus, 3498572 in base 12 is represented a 1208778. (c) The nuber 6 3 expands as 10 12 3 + 6 12 2 + 11 12 1 + 3 12 0, where we have converted fro to 10, to 11, and ultiplied each digit by the respective power of 12 for the place in the nuber. The result is 18279 in base-10. Conceptual Question 1.3 This office is roughly 4 wide by 5 long by 3 high. Therefore, in 3 the volue is 60 3. Since 1 = 10 2 c = 10 3 = 10-3 k we can express the volue of the office as 60 3 = 6 10 7 c 3 = 6 10 10 3 = 6 10-8 k 3. Although all these values are correct expressing volues of this order of agnitude would be sipler in 3. Conceptual Question 1.4 To convert a teperature T F in Fahrenheit to T C in Celsius we use the expression: T C T F 32 5 9 Therefore, in ters of the chirps per inute, the teperature in degrees Celsius is: TC 18 N 40 5 9 4 Although the forula is diensionally incorrect as N is expressed in chirps per inute and all other nubers are diensionless, it can still be used to correctly predict the teperature. Siplifications have been to ake it easier to write the forula by oitting all the units. If we wanted to include the units, the correct values would be 18 F, 40 chirps/in, 4 in/(chirp F), and 5/9 C/ F. Conceptual Question 1.5 We can expect a saller slope of the graph for the birds as copared to the aals. A saller slope on this log-log graph represents a saller ratio of brain to body and thus saller brain size relative to the size of the subjects in the class. Fro the data discussed in the chapter, on average and insofar as we can infer the intelligence of non-huan subjects, larger brain ass to body ass ratio sees to be correlated to intelligence. Conceptual Question 1.6 In y household we average around 1400 kwh per onth of electricity consuption. Converting kwh/onth we find an average power consuption of 1900 W and thus an average of 1900 joules of electrical energy used each second. There are seven individuals living in y household so each person uses 270 W or about 300 joules every second. If we are considering a city of population 10 6 individuals, then the power consued will be: 10 6 300 J s 3 10 8 J s Since the bob released 10 6 Joules of energy, it will power the city for: Tie 10 14 J 5 3 10 8 J s 3 10 s 4 days Note that this result is of the sae order of agnitude as the result found in Exaple 1.11. Averaging first the electric bill fro a nuber of households we could do a better estiate. We could also add the electric bills fro a nuber of businesses to add it to the total. However, we would expect the result to be still within the sae order of agnitude. ANALYTICAL PROBLES Proble 1.1 (a) 1.23 10 2 (b) 1.230 10 3 (c) 1.23000 104 since the last zero is significant it ust be expressed (d) 1.23 10-1 (e) 1.23 10-3 the zeros between the decial point and the 1 are not significant (f) 1.23000 10-6 the zeros between the decial point and the 1 are not significant, however the three zeros to the right of the 3 are significant Copyright 2013 Nelson Education Liited 3
Instructor s Solution anual to accopany Physics for the Life Sciences, Second Edition Proble 1.2 (a) 5 significant figures (b) 4 significant figures (c) 4 significant figures; the zero to the right of the decial point and before the first non-zero digit is not a significant figure (d) 4 significant figures; when a nuber is written in scientific notation, all digits expressed are significant figures Proble 1.3 The product ust be given to the nuber of significant figures of the least accurate nuber, that is the nuber with the sallest nuber if significant figures: (a) 5.61 10-1 both nubers have three significant figures since the zeros in 0.00456 after the decial point and before the 4 are not significant (b) 5.61 10 2 note that the last zero in 1230 is not significant (c) 5.6088 10 0 = 5.6088 both nubers have five significant figures (d) 5.609 10 0 = 5.609 note that the zero in 0.01230 after the decial point and before the 1 is not significant while the last zero is significant Proble 1.4 The quotient follows the sae rules as the ultiplication; the result ust be given to the nuber of significant figures of the least accurate nuber: (a) 2.70 10 4 both nubers have three significant figures since the zeros in 0.00456 after the decial point and before the 4 are not significant (b) 2.70 10 3 note that the last zero in 1230 is not significant (c) 2.6974 10-7 both nubers have five significant figures (d) 2.697 10-5 note that the zero in 0.01230 after the decial point and before the 1 is not significant while the last zero is significant Proble 1.5 Sus and differences ust be given with the precision of the least precise nuber, where the precision is found by the sallest power of ten present in each nuber: (a) 5.79 10 2 both nubers are precise to 10 0 so the result ust be quoted to that precision as 579 and then expressed in scientific notation (b) 1.23 10 3 while 0.456 is precise to 10-3, 1230 is only precise to 10 1 as the last zero is not significant and the result includes significant figures up to 10 1 (c) 3.33 10-1 both nubers are precise to 10-3 so the result is 0.333 which is then written in scientific notation (d) 3.34 10-1 the nuber 123.123 is the least precise of the two to only 10-3 so the result is 0.334 which is then written in scientific notation Proble 1.6 The standard precedence of operations eans quotients and ultiplications ust be perfored first coputing values with the result quoted to the significant figures of the least accurate nuber involved. Then differences and sus are perfored with the result quoted to the significant figures of the least precise nuber involved: (a) 1.27 10 2 the ultiplication requires three significant figures and the su ust be precise to 10 0 which is the precision of both nubers (b) 5.62 10 4 the ultiplication requires three significant figures so the partial result is precise only to 10 1 and thus the su will be quoted to a precision of 10 1 (c) -3.71 10 5 the quotient requires three significant figures and the partial result is precise only to 10 3 and thus the difference will be quoted to a precision of 10 3 (d) 6.85 10-6 the ultiplication requires three significant figures and the partial result is precise to 10-8, while the nuber to be added is precise to 10-10. Therefore the final result ust be quoted to a precision of 10-8 4 Copyright 2013 Nelson Education Liited
Chapter 1 Proble 1.7 y height is 165 c or 1.65, written in scientific notation it is: (a) 1.65 (10 9 n / 1 ) = 1.65 10 9 n (b) 1.65 (10 3 / 1 ) = 1.65 10 3 (c) 1.65 (10 2 c / 1 ) = 1.65 10 2 c (d) 1.65 100 = 1.65 (e) 1.65 (1 k / 10 3 ) = 1.65 10-3 k Representing the length in (d) is best suited for lengths of the order of y height. Proble 1.8 y ass is approxiately 75 kg so y weight would be: (a) 75 kg 10 /s 2 (10 6 N / 1 N) = 7.5 10 8 N (b) 75 kg 10 /s 2 (10 3 N / 1 N) = 7.5 10 5 N (c) 75 kg 10 /s 2 = 7.5 10 2 N (d) 75 kg 10 /s 2 (1 kn / 10 3 N) = 7.5 10-1 kn (e) 75 kg 10 /s 2 (1 GN / 10 9 N) = 7.5 10-7 GN Representing the weight as 750 N as done in part (c) or 0.75 kn as done in part (d) would be well suited for forces of the order of y weight. Proble 1.9 6 We start with a total tie of 7 10 s so that we can calculate: (a) 7 10 6 s (1 in / 60 s) = 10 5 in (b) 10 5 in (1 h / 60 in) = 2 10 3 h (c) 2 10 3 h (1 day / 24 h) = 8 10 1 day (d) 8 10 1 day (1 year / 365 day) = 2 10-1 year Representing the total tie as 80 days (c) or 0.2 years (d) would be well suited for ties of the order of the tie spent brushing your teeth. Proble 1.10 We have a distance of 42.195 k, and a tie of 2 h 2 in 11 s. A suitable cobination of distance and tie that leads to a quantity with the diensions of speed [L]/[T] would be the ratio of the distance to the tie. For the calculations, we can convert the tie into seconds as 7331 s. The average speed is thus: (a) (42.195 k / 7331 s) (3600 s / 1 h) = 2.072 10 1 k/h (b) (42.195 k / 7331 s) (10 3 / 1 k) = 5.756 10 0 /s = 5.756 /s (c) (42.195 k / 7331 s) = 5.756 10-3 k/s (d) (42.195 k / 7331 s) (3600 s / 1 h) (10 3 / 1 k) = 2.072 104 /h (e) (42195 / 7331 s) (1 s / 10 9 ns) (10 6 / 1 ) = 5.756 10-3 /ns Proble 1.11 You ust be careful when converting units to any power other than one. For exaple, 1 s 2 = 1 s 2 (10 3 s / 1 s) 2 = 10 6 s 2. With this in ind: (a) 10 /s 2 (10 3 / 1 ) = 10 4 /s 2 (b) 10 /s 2 (1 s / 10 3 s) 2 = 10-5 /s 2 (c) 10 /s 2 (1 k / 10 3 ) (3600 s / 1 h) 2 = 105 k/h 2 (d) 10 /s 2 (1 / 10 6 ) [(3600 s / 1 h) (24 365 h / 1 yr)] 2 = 10 10 /yr 2 (e) 10 /s 2 (10 6 / 1 ) (1 s / 10 3 s) 2 = 10 1 /s 2 = 10 /s 2 Proble 1.12 The area will be height width. Since 1 c = 10 = 10 4 = 10-2 = 10-5 k, the area in various units is: (a) 30 c 20 c = 6 10 2 c 2 ; note the nuber of significant figures (b) 300 200 = 6 10 4 2 (c) (30 10 4 ) (20 10 4 ) = 6 10 10 2 (d) (30 10-2 ) (20 10-2 ) = 6 10-2 2 (c) (30 10-5 k) (20 10-5 k) = 6 10-8 k 2 Proble 1.13 For conversion purposes, first note that 1 k = 10 3 = 10 4 d = 10 5 c, and for litres 1 L = 10 3 c 3. Furtherore, although the radius given has four significant figures, we the ultiplicative factor for the volue has been approxiated to 4 with only one significant figure. With these in ind: (a) 4 (6378 k) 3 (10 5 c / 1 k) 3 = 1.038 10 27 c 3 = 10 27 c 3 (b) 4 (6378 k) 3 (10 3 / 1 k) 3 = 1.038 10 21 3 = 10 21 3 (c) 4 (6378 k) 3 = 1.038 10 11 k 3 = 10 11 k 3 (d) 4 (6378 k) 3 (10 4 d / 1 k) 3 = 1.038 10 24 d 3 = 10 24 d 3 (e) Using (a) 1.038 10 27 c 3 (1 L / 10 3 c 3 ) = 1.038 10 24 L = 10 24 L Copyright 2013 Nelson Education Liited 5
2 2 1 Instructor s Solution anual to accopany Physics for the Life Sciences, Second Edition Proble 1.14 The density of an object easures the ass per unit volue. Since in Proble 1.13 we are told that the volue of a sphere is approxiately V = 4 R 3, the density should be: Density V 4R 3 We can now use this forula with R = 6378 k and = 5.9742 10 24 kg to find the required densities. Note that although the radius has four significant figures and the ass has five significant figures, we were told to approxiate the volue by ultiplying by 4, a nuber with only one significant figure. Note that 1 k = 10 3 = 10 4 d = 10 5 c = 10 9 = 10 15 p and 1 kg = 10 3 g = 10 9 g = 10 12 ng = 10-3 g. The densities are: (a) (5.9742 10 24 10 3 g) / [4 (6378 10 5 c) 3 ] = 5.757 10 0 g/c 3 = 6 g/c 3 (b) (5.9742 10 24 kg) / [4 (6378 10 3 ) 3 ] = 5.757 10 3 kg/ 3 = 6 10 3 kg/ 3 (c) (5.9742 10 24 10-3 g) / [4 (6378 k) 3 ] = 5.757 10 9 g/k 3 = 6 10 9 g/k 3 (d) (5.9742 10 24 10 9 g) / [4 (6378 10 15 p) 3 ] = 5.757 10-24 g/p 3 = 6 10-24 3 g/p (e) (5.9742 10 24 10 12 ng) / [4 (6378 10 9 ) 3 ] = 5.757 10-3 ng/ 3 = 6 10-24 ng/ 3 Proble 1.15 An equation cannot be both diensionally correct and wrong. It is iportant to note that the su or difference of two ters is diensionally correct only if both ters have the sae diensions. (a) Wrong. On the left-hand side we have [A] = [L 2 ], while on the right-hand side we have [4 ] [R] = 1 [L] since 4 is a diensionless quantity. Since [L 2 ] [L] the equation is diensionally wrong. (b) Wrong. On the right-hand side we have [x 1 ] = [L] being added to [v 1 t 2 ] = ([L]/[T]) [T 2 ] = [L] [T]. Since [L] [L] [T], the two quantities cannot be added and the forula is diensionally wrong. (c) Correct. On the left-hand side we have [V] = [L 3 ], and on the right-hand side we have [xyz] = [L] [L] [L] = [L 3 ]. Since both sides atch the equation is diensionally correct. (d) Wrong. On the right-hand side we have [/2] = [T]/[2] = [T] being added to [7] = 1. Since [T] 1, and a quantity with diensions cannot be added to a diensionless quantity the forula is diensionally wrong. Proble 1.16 We are told that v t 2, where v is the speed of the object, and t is the tie it has travelled. Let t 1 = 4 s and t 2 = 8s so that v 1 = 10 /s and v 2 is the speed we want to find. Fro the proportionality relationship: 1 1 v 1 t and v 2 t 2 1 2 Taking the ratio v 2 /v 1 we find: v v1 With the given values: 1 2t t 2 1 t 2 t 2 2 1 2 4s 2 v v t 10 s 2.5 2 1 s t 2 8s 2 The proportionality between v and t described in this proble is often called an Inverse- Square Law. In our case, doubling the tie reduces the speed by a factor of four; halving the tie would increase the speed by a factor of four. Proble 1.17 Fro Figure 1.9 we derived the relationship: brain body 0.68 We can use one of the data points listed in Table 1.5 alongside the proportionality relationship to find out the ass of the brain for the pygy sloth. I will use line 12 in the table that lists values for the ainland threetoed sloth with brain ass s = 15.1 g and body ass s = 3.121 kg. We know that the pygy sloth has a body ass of p = 3 kg and we want to find its brain ass p. Fro the proportionality relationship: and 0.68 0.68 s s p p 1 6 Copyright 2013 Nelson Education Liited
Chapter 1 We can then solve for : p p 0.68 s s 3 0.68 15.1g 14.7 g 3.121 Thus, a ass of approxiately 15 g is what we would expect for the brain of the pigy three-toed sloth. A brain ass of 200 g would put in question the easureent for the body ass, highlight an experiental error, an anoalous saple of the species (aybe with a severe brain defect), or siply an error in the reported data or calculations (possibly by one order of agnitude). Proble 1.18 (a) The resulting double logarithic plot is shown in Figure 1. An organized approach to plotting these data is based on extending Table 1.6 to include the logarith values of wingspan and ass, as shown here in Table 1. Figure 1 Using these logarithic data, the given power law W = a b is rewritten in the for ln W = b ln + ln a. Table 1 Bird W ln ln (c) W (g) Huingbird 7 1.95 10 2.30 Sparrow 15 2.71 50 3.91 Dove 50 3.91 400 5.99 Andean 320 5.77 11500 9.35 condor California 290 5.67 12000 9.39 condor The constants a and b are deterined fro this equation in the anner described in the Appendix Graph analysis ethods on p. 9 12 in Chapter 1. For the analysis we do not choose data pairs fro Table 1. As the graph in Fig. 1 illustrates, actual data points deviate fro the line that best fits the data (represented by the solid line). To avoid that the deviation of actual data affects our results, the two data pairs used in the analysis are obtained directly fro the solid line in Fig. 1. We choose ln W 1 = 2 with ln 1 = 2.6 and ln W 2 = 6 with ln 2 = 9.6. This leads to: ( I ) 2.0 ln(a) 2.6 b (II ) 6.0 ln(a) 9.6 b ( II ) (I ) 4.0 (9.6 2.6)b (1) Thus, b = 0.57. Due to the fluctuations of the original data and the systeatic errors you coit when reading data off a given plot, values in the interval 0.5 b 0.6 ay have been obtained. Substituting the value we found for b in forula (I) of Eq. [1] yields: 2 = 1.48 + ln a, i.e., ln a = 0.52 which corresponds to a value of a = 1.7. (b) We use the given value for the pterosaurs wingspan: W = 11 = 1100 c. The value has been converted to unit c since that is the unit used when we developed our forula in part (a). We first rewrite the forula for the wingspan with the ass as the dependent variable: W 1/b W a b a Entering the given value for the wingspan then leads to: 1100 1/0.57 1.7 85400 g 85.4 kg The ass of pterosaurs did not exceed 85 kg. Copyright 2013 Nelson Education Liited 7
Instructor s Solution anual to accopany Physics for the Life Sciences, Second Edition (c) We use again the power law relation we found in part (a) and insert the given value for the person s ass: W 1.7 70000 0.57 580 c A person of ass 70 kg would need a 5.8 wing span. Notice that a pterosaur has a only 20 % larger ass than a huan, but requires a 90 % increase in wingspan in order to achieve flight. Whereas a sparrow has a 500 % larger ass than a huingbird, yet it requires only a 100 % increase in wingspan. This shows the exponential nature of this relationship. Proble 1.19 Both the car and the cow produce the sae ass, the cow producing ethane, the car producing carbon dioxide. The olar ass of ethane (CH 4 ) is 16 g/ol while the olar ass of carbon dioxide (CO 2 ) is 44 g/ol. This eans in one gra of CH 4 you will find 1/16 ol, while in one gra of CO 2 you will find 1/44 ol. Therefore, if you have the sae ass of CH 4 and CO 2, there will be 44/16 ore oles of CH 4 than of CO 2. Since per ole CH 4 has 3.7 ties the global waring potential of CO 2, the cow will have 3.7 (44/16) = 10 ties the global waring potential of the car. Proble 1.20 According to the Fuel Consuption Guide (2011) published by Natural Resources Canada, the copact car with the best fuel efficiency in the city consues on average 4.5 L of regular gas to travel a distance of 100 k. In downtown Toronto, I have noticed gas stations being soewhere between 10 to 30 blocks apart. In contrast, gas stations see to be closer to each other in the suburban areas roughly 5 to 20 blocks apart. If a block is about 100 it sees a good average distance between gas stations to be 2 k. This eans that on average you will spend 90 L of gas to go fro one station to the next. Last week, the price of regular gas in Toronto oscillated between 120 /L and 130 /L. If the starting station had a price at 125 /L then it would cost (125 /L) 0.09 L = 11.3 to drive to the next station. Therefore, to break even as you go searching for another gas station, the cost at the next station should be about 11 /L cheaper. Proble 1.21 To copare and rank the relative agnitudes we need to establish the point of coparison. Regardless of the coparison standard, the relative agnitudes and thus the ranking should coe up the sae. I will use the force exerted by y other as the point of coparison and will call that force F. We will use the proportionality relationship: F, R 2 where is the ass of the object exerting the gravitational force, and R is the distance between that object and the baby. y other s ass was about 70 kg and the oent I was born I was really close to y other. The distance between our centers was probably around 10 c. The ass of the oon is 7 10 22 kg and it is at a distance of 4 10 8 fro Earth. So the force exerted by the oon F will be related to the force exerted by y other F according to: F F R 2 R 2 2 R R 2 1021 6 10 20 Thus, F = 60 F. The ass of the Sun is 2 10 30 kg and it is at a distance of 2 10 11 fro Earth. So the force exerted by the Sun F S will be related to the force exerted by y other F according to: S 2 R 2 F R S S S 2 F R S R 2 Thus, F S = 10 4 F. 3 1028 3 10 25 8 Copyright 2013 Nelson Education Liited
J Chapter 1 The ass of Jupiter is 2 10 27 kg and it is on average at a distance of 8 10 11 fro Earth. So the force exerted by Jupiter F J will be related to the force exerted by y other F according to: F J 2 R2 R J J 2 F R 2 R J 3 1025 2 10 26 Thus, F J = 0.6 F. The ass of ars is 6 10 23 kg and it is on average at a distance of 2 10 11 fro Earth. So the force exerted by ars F ars will be related to the force exerted by y other F according to: F ars ars R ars ars 2 F R 2. 2 R 2 R 1023 3 10 25 ars Thus, F ars = 0.03 F. The correct ranking fro sallest to largest is: F ars F J F F F S Proble 1.22 Assue you do not need any breaks, for resting or eating. Furtherore, assue that you are oving the dirt just to your back so that you do not take tie taking the dirt to another place. Furtherore, assue that you have an unliited air supply, which is not realistic if you are taking the dirt fro the front of the tunnel and placing it at your back. For your body to fit through the tunnel ust have a diaeter of at least your shoulder width, but you will need extra roo to aneuver. Let s say the tunnel needs to be 1 in diaeter. The total volue of the tunnel is then the volue of a cylinder of radius R = 0.5 and total length L = 100 : V R 2 L 78.5 3 80 3 Since each spoonful is about 5 c 3, it will take 80 3 / 5 c 3 ~ 10 9 spoonfuls of dirt to dig the tunnel. If it takes you about one second per spoonful, then you will need 10 9 s ~ 40 years to dig the full volue of the tunnel. This result quite clearly negates all the previous assuptions about breaks for rest and food, and you will definitely need ore tie to find a place for the reoved dirt. Proble 1.23 The current life expectancy of a huan is about 70 years ~ 2 10 9 s. Although the heart rate changes with age and with the level of activity, on average we could use the rest heart rate of an adult, which is around 70 bp, or 70 beats per inute. Since 70 bp = 1.2 beats/s, throughout your life your heart will beat: 2 10 9 s 1.2 beats s 2 10 9 beats In other words, your heart will beat about 2 billion ties. Proble 1.24 I tri y fingernails about once a week and the piece cut is about 1 wide, so the speed of growth is about 1 /week: (a) 1 /week = (10-3 )/(7 24 3600 s) = 2 10-9 /s (b) 1 /week = (10 3 )/(7 day) = 1 10 2 /day = 100 /day (c) 1 /week = (10-1 c)/(7/365 year) = 5 c/year Proble 1.25 The nuber of grains of sand will be V b /V g, where V b is the volue of the beach, and V g is the volue of a grain of sand. Since the beach is box-shaped, V b = l w d, where l is the length, w is the width, and d is the depth. (a) Since 1 3 = 1 3 (1 / 10 3 ) 3 = 10-9 3, with h = 4 we have: V b 100 10 4 4 10 12 grains V g 10 9 3 (b) Since the grains have the sae volue of 10-9 3 but h = 2 we have: V b 100 10 2 2 10 12 grains V g 10 9 3 Copyright 2013 Nelson Education Liited 9
Instructor s Solution anual to accopany Physics for the Life Sciences, Second Edition 10 Copyright 2013 Nelson Education Liited
Chapter 1 Proble 1.28 If two aals A and B have the sae densities then: V A V B, A B where is the ass and V the volue of the aal. Therefore we can reorganize for the ratio of the volues: V B B V A A That is, if all aals have approxiately the sae density, the ratio of the volues of two aals is the sae as the ratio of their asses. We can use Table 1.5 to find the required ratios: V Hippopotaus Hippopotaus 1351 26 V Sheep 52.1 and: Sheep V Lion V Chipunk Lion 190.8 2500 Chipunk 0.075 The volue of the ark is V Ark = (300 50 30) cubit 3 (0.45 / cubit) 3 ~ 4 10 4 3. Since the density of water is 1000 kg/ 3 we can use Table 1.5 to find the volue of each and all the anials according to: V Anial Anial Anial D Water 1000 kg 3 Adding the asses of all land aals in Table 1.5 (all except 34, 40, 43 and 49), we get 1.41 10 4 kg but since we need two of each anial Total = 2.82 10 4 kg. Therefore: 2.82 10 4 kg 3 V Total 30 Total D Water 1000 kg 3 Since V Total < V Ark, two of every land aal fro Table 1.5 will fit in the ark. For the last part of the question, we can calculate the volue of the blue whale: V BlueWhale 58059kg 60 3 D BlueWhale water 1000 kg 3 Since V < V, two of every land Total Blue Whale aal fro Table 1.5 could be housed in the volue equivalent to that of a blue whale. Copyright 2013 Nelson Education Liited 1