Hwk4Solns - Assignment 4 Solutions Chpt 26 1. (a) The...

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Unformatted text preview: Assignment 4 Solutions Chpt 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 10 19 C) = 7.5 10 21 . 23. The resistance of conductor A is given by 2 A A r L R = , where r A is the radius of the conductor. If r o is the outside diameter of conductor B and r i is its inside diameter, then its cross-sectional area is ( r o 2 r i 2 ), and its resistance is ) ( 2 2 i B r r L R = The ratio is . 3 ) 50 . ( ) 50 . ( ) . 1 ( 2 2 2 2 2 2 = = = mm mm mm r r r R R A i B A . 26. Let be the radius of the kite string and 2.00 mm r = 0.50 mm t = be the thickness of the water layer. The cross-sectional area of the layer of water is ....
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Hwk4Solns - Assignment 4 Solutions Chpt 26 1. (a) The...

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