sol to assignment7

# sol to assignment7 - 7.4 The earth moves through 2 mdin...

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Unformatted text preview: 7.4- The earth moves through 2:: mdin ongyear (3.156 x 107 3). Thus, 2" “‘1 -1.99x10r7-“;—. ‘3.156 x 1073 .- Alternaﬁvely, the earth moves through 360 m one year (365.242 days). 360' ﬁg The ”mass... ‘ . . W,” n . . ..,..-—-.-~_—.-—- —-""l'”.. ~-- wr—w‘wrvr”"“” " ' ' i a) “‘ifg'" """33’-"100‘Eiﬁm - 10.47 rad/s cu - mg 0 - 10.47 rad/s a " - 2.00 rad/s2 (b) sacs-LtZ-ﬂr-Q-lLf—emg—s1524e—zmm (a) t: I 5.24 S. (b) Vt am .. (1.27:: 101 m)(8.17 rad/s) - 1.04 m/s. (c) At t— 1.005, ata 3.45 XIO'I m/sz. From an =mo+ at, we find 00 -2.72 rad/sat t=- 1.005. Thus, a, -1153 - mZ - (0.127 m}(2.72 rad/SF a0.940~m/32. ammuwatﬂ + (ax-)7- = LOOm/sz, and tenth at/ar -= .367, or 9- 20" 7.23 (a) a1- - mZ - (2.00 m)(3.00 rad/s)2 - 18.0 m/sZ . (b) F-mar - (50.0 kg)(18.0 111/32) - 900 N (c) We‘know the centripetal acceleration is "caused by the force of friction. Therefore, I" a 900 N. Also, the normal force, N, is equal to 490 N. Thus, p.= f/ N = 900 Nx’490 N = 1.84. A coefﬁcient of friction greater thanlone is unreasonable. Thus, she is not going to be able to stay on the ride. . 7.29 (a) At A the forcesou the car are'the normal force, N, and its weight. mvz ' ' 2 'Wehave FC- 1, xN—mg,or N=mg+grv— ,whichgives 2 N-(sookg)(9.30m/s2)+ 500 1030:: “”5 =2.49x104N. (b) Athe have, Fc-T nmg-M orNaMg-Vz 2 NzOWhichmeans ngI—or Vs Ig Thus, vmx- Vs Quin-5.0 m)(9.30~ m/sz) = 12.1 m/s "‘ " 'I ftheforceexertedbythe. 7.34 At the uﬂlbnum posmon, the magmmde o earth ofthe object is equal to the magmtude of the force exerted by the sunon the object. Thus, . ehm Q52 g. . 2 *. 2 ' “mg (1.50 x 1011 - r) r r2 M: 1.991 x 1030 (150 x 101‘1 - r)2 " MB 5.98 :5 1024 Taking the square root of both sides gives W " 577’ WM gives ‘ "r"; 1.51).}; 1011 m . x -1 l- - =- 3.33 x 105 ...
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