563_Physics ProblemsTechnical Physics

563_Physics ProblemsTechnical Physics - Chapter 19 P19.56 L...

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Chapter 19 565 P19.56 (a) T L g i i = 2 π so L Tg i i == = 2 2 2 2 4 1000 980 4 02482 ππ .. . s m s m 2 af ej ∆∆ LL T T g T i f i × ° ° = × = + −− α 19 0 10 0 284 2 10 0 4 72 10 22 02483 10000950 950 10 61 5 5 . . . . . C m C m m 9.80 m s s s 2 bg (b) In one week, the time lost is time lost 11 0 5 week 9.50 s lost per second time lost = F H G I K J × F H G I K J 700 86 400 5 ±dweek s 1.00 d s lost s time lost = 57 5 . s lost P19.57 Ir d m = z 2 and since rT T i =+ 1 for T << 1 we find IT T i 1 2 thus T i i af bg 2 (a) With =×° 17 0 10 .C and T 100 C we find for Cu: I I ° ° = 2 17 0 10 100 0 340% CC a f (b) With 24 0 10 and T 100 C we find for Al: I I ° ° = 2 24 0 10 100 0 480% P19.58 (a) Bg V =′ ρ ′= + PP g d 0 ′′ = PV i 0 B gP V P gP V Pg d ii = = + ρρ 00 0 (b) Since d is in the denominator, B must decrease as the depth increases.
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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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