P28_057 - (0 . 0125 s < < 12 . 5 s). (d) The lower...

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57. (a) The four tires act as resistors in parallel, with an equivalent value given by 1 R eq = 4 X n =1 1 R tire = 4 R tire = R eq = R tire 4 . Using the stated values ( C =5 . 0 × 10 10 Fand10 8 <R tire < 10 11 Ω) we ±nd the capacitive time constant τ = R eq C in the range 0 . 012 s <τ< 13 s. (b) Eq. 26-22 leads to U 0 = 1 2 CV 2 = 1 2 ( 5 . 00 × 10 10 F )( 30 . 0 × 10 3 V ) 2 =0 . 225 J . (c) As demonstrated in Sample Problem 28-5, the energy “decays” exponentially according to U = U 0 e 2 t/τ . Solving for the time which gives U =0 . 050 J, we ±nd t = τ 2 ln µ U 0 U = τ 2 ln µ 0 . 225 0 . 050 which yields, for the range of time constants found in part (a), values of t in the range 0 . 094 s < t< 9 . 4 s. To obtain these particular values, we used 3-±gure versions of the part (a) results
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Unformatted text preview: (0 . 0125 s < < 12 . 5 s). (d) The lower range of resistance leads to the smaller times to discharge, which is the more desirable situation. Based on this criterion, low resistance tires are favored. (e) There are a variety of ways to safely and quickly ground a large charged object. A large metal cable connected to, say, the (metal) building frame and held at the end of, say, a long lucite rod might be used (to touch a part of the car that does not have much paint or grease on it) to make the car safe to handle....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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