749_Physics ProblemsTechnical Physics

# 749_Physics ProblemsTechnical Physics - Chapter 26 P26.38...

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Chapter 26 89 P26.38 With switch closed, distance ′ = dd 0500 . and capacitance ′ = = = C A d A d C 00 2 2 . (a) QC V CV = == × = ∆∆ af ej 22 2 0 0 1 0 1 0 0 4 0 0 6 . F V C µ (b) The force stretching out one spring is F Q A CV A Add d = = = = 2 0 2 2 0 2 2 0 2 2 4 2 bg . One spring stretches by distance x d = 4 , so k F x d F H G I K J × × = 2 4 8 8 2 00 10 100 800 10 250 2 6 2 3 2 a f . . . F V m kN m . P26.39 The energy transferred is HQ V ET C V J ×= × 1 2 1 2 50 0 1 00 10 2 50 10 89 .. . a f and 1% of this (or E int J 250 10 7 . ) is absorbed by the tree. If m is the amount of water boiled away, then Em m int Jkg C C C Jkg J =⋅ ° ° ° + × = × 4186 100 30 0 2 26 10 2 50 10 67 . giving m = 979 k g . *P26.40 (a) UC V C V C V =+= 1 2 1 2 2 a f a f a f (b) The altered capacitor has capacitance ′ = C C 2 . The total charge is the same as before: CV CV CV C V ∆∆∆
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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