chap7 solutions

chap7 solutions - 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11...

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Unformatted text preview: 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 CHAPTER SEVEN SOLUTIONS (a) w = (33 rev/min)(2n rad/rev)(1 min/60 s) : 3.5 rad/s. (b) B= wt= (3.5 rad/s)(1.5 s) = 5.2 rad. The earth moves through 21: rad in one year (3.156 x 107 3). Thus, 2" rad =199x10-79—q 3.156 x 1075 5 Alternatively, the earth moves through 360° in one year (365.242 days). 360' _ mg Thus, m = 3652 days — 0.986 day . Use m = (90+ at, with to = 0.20 rev/s = 1.256 rad/s. 1.256 rad/s=0 + a(30 s), or a =42 x 10-2 rad/52. a): 9 = 4.7 rev = 29.5 radians, there is a constant angular acceleration. Thus, 9=wot+%at2 yields 29.51311=0+%a(1.23)2,0r a=41rad/s2. In this problem, (02 = (1203 + 20:6 becomes, (2.2 I‘adz’s)2 = (0.60 rad/s)2 + 2(0.70 rad/32w. so 9== 3.2 rad. (Hf = 2.51 x 104 rev/min = 2.63 x 103 rad/s wf'mo 163x103 rad/s—O 2 2 t =—-—-—--"3_20 S =8.22x10 rad/s. l (b) 9=w0t+ —ar2 =0+ l (8.22 x102 rad/s2)(3.20 s)2 =4.21 x 103 rad 2 2 wo= 100 rev/min= 10.47 rad/s gm - 009:0 - 10.47 rad/s (a) a= (b) 9=6t=m Ewr=mggw(s24s)=2z4ra¢ (of = 5 rev/s = 103: rad/s. We will break the motion into two stages: (1) an. acceleration period and (2) a deceleration period. While speeding up, 91 asr=w§flg~4§ (8.0 s) = 125.7 rad. While slowing down, 32 = swig-JEEle (12.0 s) = 188.5 m Btmal=31+92=314lrad=50reu Q (37.0 revMZn: rad/lrevg t= 3.00 S =77.S rad/s. But also, 5):“) :00 =W=775 rad/s. Thus, (.00 = 57.0 rad/s. Then, an -t :90 = 98.0 rad/33; S57.0 rad/s 29 2(40 rev!(2nrad/rev2 —= = .14rad/ 2. 12 (60 s)2 0 S (b) a) = 010+ at= 0 + (0.14)(60) = 8.4 rad/s. 80, w: = 13.7 rad/52. a: (a) Using 6=%at2, we have a= 89 CHAPTER SEVEN SOLUTIONS s=vt2 - 0= (14.14 m/sz2 Eat 2(0500 m/sz) ’ (c) Vt = yot + at: gives 14.14 m/s = 0 + (0.500 We?) r, or := 28.3 s. or s=200m. 7.2 1 Let m be the mass of a red corpuscle and let 1' be the radius of the centrifuge. We are given 2 FC 240x 10‘111&1=T=mrw2 , so 1 [Fe 4.00 x 10-11 = _= M: 4 _ g _ (0 1111‘ (3.00 x 10‘16)(0_15) 9 28md/s 150reV/s 7.22 The friction force must supply the force producing the centripetal acceleration. mvz Thus, we must have f =ch where f= ping, and Fc =7. 2&2 1" v2 = arg = 070(20 rn)(9.80 m/sz), from which, v = 12 m/s. Therefore, umg = . The mass cancels, and we have 7.23 (a) a1- = m2 = (2.00 m)(3.00 rad/s)2 = 18.0ni/s2. (b) F= mar = (50.0 kg)(18.0 m/sZ) = 900 N (c) We know the centripetal acceleration is caused by the force of friction. Therefore, I” = 900 N. Also, the normal force, N, is equal to 490 N. Thus, y. = f/ N = 900 N/ 490 N = 1.84. A coefficient of friction greater than one is unreasonable. Thus, she is not going to be able to stay on the ride. 7.24 (a) The radial acceleration must not exceed 73 = 7(9.80 111/82): 68.6 m/sZ. 2 Thus, from a- =VTt , we have r =(100 m/s}2/68.6 m/s2 = 1.462; 102 m. (b) P: mar z m(7g) = 7(mg) = 7(pilot's weight) = 7(80.0 kg)(9.80 3%) = 5488 N. 7.25 (a) From Ef‘y = 0, we have \\ I N cost): 111 g (1) where N is the normal force exerted on N the car by the ramp. Now, use Fradial inward = mar: _Radial_ _ _ _ _ Nsin8= mVZ/r (2) Diwali” Divide (2) by (1) cans: VZ/rg (b) tane = (13.4 n1/s)2/(50.0 m)(9.80 111/52) tan9= 0366,50 9: 20.1“. 7.26 (a) Since the 1 kg mass is in equilibrium, we have for the tension in the string, T= mg = (1.0)(9.80) = 9.8 N. (b) Since the centripetal acceration of the puck is produced by the tension in the string, we have Fc= T= 9.8 N. 91 CHAPTER SEVEN SOLUTIONS II. _ (6.67 x 1011 NmZ/kg2)(3.00 x 104- kg)§5.98 x 1024 kg) _325N 1‘ (1.92 x 108 m)2 " ‘ Gmlmz The force exerted on the'ship by the moon is F2 = r2 , or =(6.67 x 10-11 NmZ/kg2)(3.00 x 104 kg}(7.36 x 1022 kg) _ N F2 (1.92 x 103 m)2 4'00 Thus, the resultant force is (325 N — 4.00 N) = 321 N directed toward the earth. 7.32 (a) The Sun-Earth distance is 1.496 2th11 m, and the Earth-Moon distance is 3.84 x103, so the distance from the Sun to the Moon during a solar eclipse is 1.496 x1011 m - 3.8421 108 = 1.49216 x1011 m. The mass of the Sun, Earth, and Moon are mg = 1.991 x1030 kg, m1; = 5.98 x1024 kg, and m1 = 7.36 x1022 kg. We have Gmlmz (6.67 x 1011 NmZ/kg2){1.991x 1030x736 x1022) PSNF 2 = 11 2 1' (1.49216 x10 ) 133.71 = 4.39 x1020 N. =(6.67 x 10'11Nm2/kg2)(5.98 1110242936 x1022) (h) FE“ (3.84 x108)2 = 1.99 x1020 N. (C) FSE=(6.67 x 101leZ/kg21(1.991x1030)(5.98 x 10241 (1.496 x1011)2 = 3.55 x1022 N. 7.33 The force exerted on the 2 kg mass by the 3 kg mass is in the positive y direction and given by: Gmimz (6.67 x 10-1 1 Nmz/kg22g30 kgMZfl kg) F = = =1.00 10-10N. 1 r2 (2.0 m)2 X The force exerted on the 2.0 kg mass by the 4.0 kg mass is in the positive X direction, and is G I —11 2 2 I I F25 m1m2fi§667 x 10 Nm /kg )(40 kgnzo 1<g1=334x10_11 N_ r2 (4.0 m)2 The resultant F is found from the Pythagorean theorem to be P FR=1.1x10'10 N. Also, e=tan-1(-F%)=72". 7.34 At the equilibrium position, the magnitude of the force exerted by the earth on the object is equal to the magnitude of the force exerted by the sun on the object. Thus, GMEm cgm . . (1.50 x 1011 - 1-)2~ r2 ’gmng r2 Ms 1.991 x 1-030 —=—=——=3.33 105 (1.50 x 1011 - 192 MB 5.98 x 1024 x Taking the square root of both sides gives r = 577, which gives r = 1.50 x 1011 In. (1.50 x 1011 - r} 7.35 Assume that the Moon moves around the center of the Earth, and 93 ...
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This homework help was uploaded on 02/05/2008 for the course PHYS 111 taught by Professor Shapiro during the Fall '01 term at Wesleyan.

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chap7 solutions - 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11...

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