PROBLEM SOLUTIONS. g, recognizing that 2 is a dimensionless. 1.1 Substituting dimensions into the given equation T 2. constant, we have.

Transcription

1 PROBLEM SOLUTIONS 1.1 Substituting diensions into the given equation T g, recognizing that is a diensionless constant, we have T g L or T T T LT Thus, the diensions are consistent. 1. (a) Fro x = Bt, we find that B x t Thus, B has units of [ B] [ x] L [ t] T If x = A sin ( ft), then [A] = [x]/sin ( ft)] But the sine of an angle is a diensionless ratio. Therefore, [ A] [ x ] L 1. (a) The units of volue, area and height are: [V] = L, [A] = L, and [h] = L We then observe that or L = L L or [V] = [A][h] Thus, the equation is V = Ah is diensionally correct. Vcylinder R h R h Ah, where A R Vrectangular box wh w h Ah, where A w length width 1.4 In the equation 1 v 1 v 0 gh, L ML [ v ] [ v 0 ] M T T

2 while gh M L T L 1 M L. Thus, the equation is diensionally incorrect. T In L L v v 0 at,[ v] [ v 0 ] but [ at ] [ a][ t ] T L T T Hence, this equation is diensionally incorrect. L ML In the equation a v, we see that [ a] [ ][ a ] M T T while [ v ] L T L T Therefore, this equation is also diensionally incorrect. 1.5 Fro the universal gravitation law, the constant G is G = Fr /M. Its units are then G F r M s s 1.6 (a) Solving KE = p / for the oentu, p, gives p KE where the nueral is a diensionless constant. Diensional analysis gives the units of oentu as: p KE M M L T M L T M L T Therefore, in the SI syste, the units of oentu are s. Note that the units of force are /s or [F] = M L/T. Then, observe that F t M L T T M L T p Fro this, it follows that force ties tie is proportional to oentu: Ft p. (See the ipulse oentu theore in Chapter 6, F t = p, which says that a constant force F

3 ties a duration of tie t equals the change in oentu, p.) 1.7 Blindly adding the two lengths, we get 8.76 c. However, 15. c has only one decial place. Therefore, only one decial place accuracy is possible in the su, and the answer ust be rounded to 8.8 c. 1.8 A w c c Multiplying out this product of binoinals gives A c The first ter gives the best value of the area. The second and third ters add to give the uncertainty and the fourth ter is negligible in coparison to the other ters. The area and its uncertainty are found to be A 09 c 4 c 1.9 (a) has significant figures with the uncertainty in the tenths position has 4 significant figures (c).46 6 has significant figures (d) 0.00 =. has significant figures. The two zeros were originally included only to position the decial. 1. c = /s (a) Rounded to significant figures: c =.00 8 /s Rounded to 5 significant figures: c = /s (c) Rounded to 7 significant figures: c = /s 1.11 Observe that the length l = 5.6 c, the width w = 6.5 c, and the height h =.78 c all contain significant figures. Thus, any product of these quantities should contain significant figures. (a) w 5.6 c 6.5 c 5.7 c

4 V w h 5.7 c.78 c 99. c (c) wh 6.5 c.78 c 17.7 c V wh 17.7 c 5.6 c 99.5 c (d) In the rounding process, sall aounts are either added to or subtracted fro an answer to satisfy the rules of significant figures. For a given rounding, different sall adjustents are ade, introducing a certain aount of randoness in the last significant digit of the final answer. 1.1 (a) A r have Recognize that the last ter in the brackets is insignificant in coparison to the other two. Thus, we A C r (a) The su is rounded to 797 because 756 in the ters to be added has no positions beyond the decial = (. ) 56. = ust be rounded to 1.1 because. has only two significant figures. (c) 5.60 ust be rounded to because 5.60 has only four significant figures (a) Answer liited to three significant figures because of the accuracy of the denoinator Answer liited to the four significant figure accuracy of two of the operands ft 1 fatho d i fathos i 6 ft

5 The answer is liited to one significant figure because of the accuracy to which the conversion fro fathos to feet is given c.00 1 k s 1 i k 1 furlong 0.15 i 6 furlongs 1.49 s 6 furlongs and c 1.49 s s 1 day 14.0 days 1 egafurlong 1 fortnight 6 furlongs 6 yielding c 1.80 egafurlongs fortnight firkins 6.00 firkins 9 gal 1 firkin.786 L 1 gal c 1 L 1 c (a) k i 48 i 5.60 k c k 4 h 1 61 ft k c 5 80 ft k 5 (c) h 0 0 ft 6.19 k c 5 80 ft k 5 (d) d 8 00 ft.499 k c 5 80 ft In (a), the answer is liited to three significant figures because of the accuracy of the original data value, 48 iles. In, (c), and (d), the answers are liited to four significant figures because of the accuracy to which the kiloeters-to-feet conversion factor is given.

6 1.19 Note that 1 =.81 ft, so (1 ) = (.81 ft) or 1 = (.81) ft A w 1 0 ft 150 ft 1.50 ft ft yr 1 day 1 h Age of Earth 1 s yr 65.4 days 4 h 600 s ft Distance to Proxia Centauri ft 1 1. rate 1 in day 1 day 4 h 1 h.54 c 600 s 1.00 in 9 n c 9. n s This eans that the proteins are assebled at a rate of any layers of atos each second! 8 1. c.00 s 600 s 1 k 1 h 1 i k i h 1.4 Volue of house 50.0 ft 6 ft 8.0 ft.8 1 ft 0 c c Volue 5.0 acre ft 1.81 ft ft 1 acre 1.81 ft.08 4

7 1.6 1 Volue of pyraid area of base height acres ft acre 481 ft 9.08 ft ft.57 1 ft Volue of cube = L = 1 quart (Where L = length of one side of the cube.) Thus, L 1 quart 1 gallon 4 quarts.786 liter 1 gallon c 1 liter 946 c and L 946 c 9.8 c 1.8 nuber of pounds = (nuber of burgers)(weight/burger) = (5 burgers)(0.5 Ib/burgers) = 1.5 Ib, or Ib Nuber of head of cattle = (weight needed)/(weight per head) =(1.5 Ib)/(00 Ib/head) = head, or 7 head Assuptions are 0.5 lb of eat per burger and a net of 00 lb of eat per head of cattle. 1.9 Consider a roo that is 1 ft square with an 8.0 ft high ceiling. Recognizing that 1 =.81 ft, so (1 ) = (.81 ft) or 1 = (.81) ft, the volue of this roo is V 1 1 ft 1 ft 8.0 ft.81 ft roo A ping pong ball has a radius of about.0 c, so its volue is 4 4 Vball r The nuber of balls that would easily fit into the roo is therefore

8 n V roo ball.4 V or We assue that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each tie. Thus, on average, each person is sick for weeks out of each year (5 weeks). The probability that a particular person will be sick at any given tie equals the percentage of tie that person is sick, or probability of sickness weeks 1 5 weeks 6 The population of the Earth is approxiately 6 billion. The nuber of people expected to have a cold on any given day is then Nuber sick population probability of sickness (a) Assue that a typical intestinal tract has a length of about 7 and average diaeter of 4 c. The estiated total intestinal volue is then V A total d The approxiate volue occupied by a single bacteria is 6 18 Vbacteria typical length scale If it is assued that bacteria occupy one hundredth of the total intestinal volue, the estiate of the nuber of icroorganiss in the huan intestinal tract is n total or n 14 V 18 bacteria V The large value of the nuber of bacteria estiated to exist in the intestinal tract eans that they are probably not dangerous. Intestinal bacteria help digest food and provide iportant nutrients. Huans and bacteria enjoy a utually beneficial sybiotic relationship.

9 1. A blade of grass is 1/4 inch wide, so we ight expect each blade of grass to require at least 1/16 in = 4. 4 ft. Since, 1 acre = ft, the nuber of blades of grass to be expected on a quarter-acre plot of land is about n total area 0.5 acre ft acre area per blade 4. 4 ft blade =.5 7 blades, or 7 blades 1. A reasonable guess for the diaeter of a tire ight be ft, with a circuference (C = r = D = distance travels per revolution) of about 9 ft. Thus, the total nuber of revolutions the tire ight ake is n total distance traveled distance per revolution i 5 80 ft i 9 ft rev 7 rev, or ~ 7 rev 1.4 Answers to this proble will vary, dependent on the assuptions one akes. This solution assues that bacteria and other prokaryotes occupy approxiately one ten-illionth of ( 7 ) of the Earth s volue, and that the density of a prokaryote, like the density of the huan body, is approxiately equal to that of water ( / ). (a) estiated nuber n V V total Earth REarth 9 single single 6 perokaryote perokaryote V V length scale total water single perokaryote density total volue nv (c) The very large ass of prokaryotes iplies they are iportant to the biosphere. They are responsible for fixing carbon, producing oxygen, and breaking up pollutants, aong any other biological roles. Huans

10 depend on the! 1.5 The x coordinate is found as x r cos.5 cos 5.0 and the y coordinate y r sin.5 sin The x distance out to the fly is.0 and the y distance up to the fly is 1.0. Thus, we can use the Pythagorean theore to find the distance fro the origin to the fly as, = =. d x y 1.7 The distance fro the origin to the fly is r in polar coordinates, and this was found to be. in Proble 6. The angle is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as y 1.0 tan 0.50 x.0 1 and tan The polar coordinates are r =. and = The x distance between the two points is x x x 1.0 c 5.0 c 8.0 c and the y distance between the is y y y 1.0 c 4.0 c 1.0 c. The distance between the is found fro the Pythagorean theore: d x y 8.0 c 1.0 c 65 c = 8.1c 1.9 Refer to the figure given in Proble 1.40 below. The Cartesian coordinates for the two given points are x r cos.00 cos x r cos 5.00 cos y r sin.00 sin y r sin 5.00 sin The distance between the two points is then: s x y

11 1.40 Consider the figure shown at the right. The Cartesian coordinates for the two points are: x r cos x r cos y r sin y r sin The distance between the two points is the length of the hypotenuse of the shaded triangle and is given by 1 1 s x y x x y y or s r cos r cos r r cos cos r sin r sin r r sin sin r cos sin r cos sin r r cos cos sin sin Applying the identities cos sin 1 and cos 1 cos sin 1 sin cos 1, this reduces to s r r r r cos cos sin sin r r r r cos (a) With a = 6.00 and b being two sides of this right triangle having hypotenuse C = 9.00, the Pythagorean theore gives the unknown side as = 6.71 b c a a 6.00 tan b 6.71 (c) b 6.71 sin c Fro the diagra, cos (75.0 ) = d/l Thus,

12 d L cos cos The circuference of the fountain is, C r, so the radius is r C Thus, tan 55.0 h r h.9 which gives h = (.9 ) tan (55.0 ) = (a) sin side opposite hypotenuse so, side opposite (.00 ) (sin 0.0 ) = 1.50 cos adjacent side hypotenuse so, adjacent side (.00 ) (cos 0.0 ) = (a) The side opposite =.00 The side adjacent to =.00 (c) 4.00 cos (d) 4.00 sin (e) 4.00 tan Using the diagra at the right, the Pythagorean Theore yields c Fro the diagra given in Proble 1.46 above, it is seen that

13 5.00 tan and tan (a) and See the figure given below: (c) Applying the definition of the tangent function to the large right triangle containing the 1.0 angle gives: y/x = tan 1.0 [1] Also, applying the definition of the tangent function to the saller right triangle containing the 14.0 angle gives: x y 1.00 k tan 14.0 [] (d) Fro equation [1] above, observe that: y/x = tan 1.0 Substituting this result into equation [] gives: y y tan k tan 1.0 tan 14.0 Then, solving for the height of the ountain, y, yields y 1.00 k tan1.0 tan14.0 tan 14.0 tan k Using the sketch at the right: w 0 tan 5.0, or w 0 tan The figure at the right shows the situation described in the proble stateent:

14 Applying the definition of the tangent function to the large right triangle containing the angle in the figure, one obtains: y/x = tan [1] Also, applying the definition of the tangent function to the sall right triangle containing the angle gives: x y d tan [] Solving equation [1] for x and substituting the result into equation [] yields: y y tan tan or tan y tan d y d tan The last result siplifies to y tan y tan d tan tan Solving for y: y tan tan d tan tan or y d tan tan d tan tan tan tan tan tan 1.51 (a) Given that a F/M, we have F a Therefore, the units of force are those of a, [ F] [ a] [ ][ a ] M L T M L T L M L [ F] M so newton = T T s 1.5 (a) i i k k h h 1 i h v ax i i k h k h h 1 i h h

15 v (c) ax i i i k h k h h h 1 i h h 1.5 (a) Since 1 c, then c c c, giving 1.0 ass density volue c 6 c c 1 As rough calculation, treat as if 0% water. cell: 4 ass density volue = kidney: 4 4 ass density volue r fly: ass density volue density rh Assue an average of 1 can per person each week and a population of 00 illion. nuber cans person nuber cans year population weeks year week can person week 1 8 people 5 weeks yr cans yr, or ~ cans yr

16 nuber of tons weight can nuber cans year oz 1 lb 1 ton can 0.5 can 16 oz 000 lb yr 5 ton yr, or ~ 5 ton yr Assues an average weight of 0.5 oz of aluinu per can The ter s has diensions of L, a has diension of LT, and t has diensions of T. Therefore, the equaton, s = k a t n with k being diensionless, has diensions of L = (LT ) (T) n or L 1 T 0 = L T n The powers of L and T ust be the sae on each side of the equation. Therefore, L 1 = L and = 1 Likewise, equating powers of T, we see that n = 0, or n = =. Diensional analysis cannot deterine the value of k, a diensionless constant Assue the tub easures 1. by 0.5 by 0. (a) It then has a volue V = 0. and contains a ass of water V water water tub 0. 00, or ~ Pennies are now ostly zinc, but consider copper pennies filling 80% of the volue of the tub. Their ass is V pennies copper 0.8 tub 8.9 / , or ~ 1.57 The volue of oil equals V ass density If the slick is a circle of radius r and thickness equal to the diaeter, d, of a olecule, V d r thickness of slick area of oil slick, where r 0.418

17 Thus, d F r , or ~ (a) For a sphere, A = 4 R. In this case, the radius of the second sphere is twice that of the first, or R = R 1. Hence, A 4 R R R A R R R 4 4 For a sphere, the volue is V R Thus, V 4 R R R V R R R The nuber of tuners is found by dividing the nuber of residents of the city by the nuber of residents serviced by one tuner. We shall assue 1 tuner per,000 residents and a population of 8 illion. Thus, nuber of tuners population frequency of occurrence 6 1 tuner 8 residents 8 tuners or ~ tuners 4 residents 1.60 (a) The aount paid per year would be dollars s 65.5 days dollars annual aount s 1.00 day yr yr Therefore, it would take 9 1 dollars.16 dollars yr yr, or ~ yr The circuference of the Earth at the equator is C r The length of one dollar bill is so that the length of nine trillion bills is

18 encircle the Earth dollar dollars 1 1. Thus, the nine trillion dollars would n C , or 4ties 1.61 (a) 1 yr 1 yr 65. days 1 yr s 1 day.16 7 s Consider a segent of the surface of the oon which has an area of 1 and a depth of 1. When filled with eteorites, each having a diaeter 6, the nuber of eteorites along each edge of this box is n length of an edge 1 eteorite diaeter 6 6 The total nuber of eteorites in the filled box is then N n 6 18 At the rate of 1 eteorite per second, the tie to fill the box is t y s = s yr, or yr.16 7 s 1.6 We will assue that, on average, 1 ball will be lost per hitter, that there will be about hitters per inning, a gae has 9 innings, and the tea plays 81 hoe gaes per season. Our estiate of the nuber of gae balls needed per season is then nuber of balls needed nuber lost per hitter nuber hitters/gae hoe gaes/year hitters innings gaes 1 ball per hitter 9 81 inning gae year balls 7 00 or ~ year 4 balls year