HW3Sol - Physics 112 - HW #3 Solutions 3600 s T = (24 h) h...

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Physics 112 - HW #3 Solutions 3-29 (a) Rotation period: T = (24 h) 3600 s h = 8.64 × 10 4 s v = 2 π R T = 2 π (6.38 × 10 6 m) 8.64 × 10 4 s = 4.01 × 10 7 m 8.64 × 10 4 s = 464 m/s a rad = v 2 R = (464 m/s) 2 6.38 × 10 6 m = 0.0337 m/s 2 = 0.0034 g = 3.4 × 10 -3 g (b) a rad = v 2 R > g v > gR = (9.8 m/s 2 )(6.38 × 10 6 m) = 7910 m/s T = 2 π R v < 2 π (6.38 × 10 6 m) 7910 m/s = 5070 s = 1.41 h 3-34 (a) a rad = v 2 r = (3.00 m/s) 2 14.0 m = 0.643 m/s 2 (b) a tan = 0.500 m/s 2 a = a rad 2 + a tan 2 = (0.643 m/s 2 ) 2 + (0.500 m/s 2 ) 2 = 0.814 m/s 2 tan θ = a tan a rad = 0.500 m/s 2 0.643 m/s 2 = 0.778 θ = 38° a rad a tan a v θ 3-37 Let "w" = woman, "s" = sidewalk, and "g" = ground. Using the textbook's notation: v w/g = v w/s + v s/g (a) v w/g = 1.5 m/s + 1.0 m/s = 2.5 m/s t = x v w/g = 35.0 m 2.5 m/s = 14 s (b) v w/g = - 1.5 m/s + 1.0 m/s = - 0.5 m/s t = x v w/g = -35.0 m -0.5 m/s = 70 s (Now why would anyone want to do this?)
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3-81 Let "p" = plane, "a" = air, and "g" = ground. Using the textbook's notation: (a) In x-y (E-N) components: v p/a = (-220 km/h , 0) v p/g = -120 km 0.50 h , -20 km 0.50 h = (-240 km/h , -40 km/h) v p/a v p/g v
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This note was uploaded on 04/02/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW3Sol - Physics 112 - HW #3 Solutions 3600 s T = (24 h) h...

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