569_Physics ProblemsTechnical Physics

569_Physics ProblemsTechnical Physics - Chapter 19 e j...

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Chapter 19 571 P19.71 (a) µπ ρπ == × × −− r 24 2 33 5 00 10 7 86 10 6 17 10 .. . m kg m kg m 3 ej e j (b) f v L 1 2 = and v T = µ so f L T 1 1 2 = Therefore, TL f × × × = 2 6 17 10 2 0 800 200 632 1 2 3 2 bg a f N (c) First find the unstressed length of the string at 0°C: LL T AY L L TAY AY =+ F H G I K J = + = × = × natural natural 2 so m m a n d P a 1 1 500 10 7854 10 200 10 4 2 71 0 π . Therefore, T AY = ×× 632 7 854 10 20 0 10 402 10 0 3 . e j , and L natural m m = = 0800 1 4 02 10 07968 3 . . . af . The unstressed length at 30.0°C is 30 13 0 0 0 0 ° ° ° C natural CC α a f , or L 30 6 1 110 10 300 079706 ° × = C m . . . Since T AY L N M O Q P ° 30 1 C , where T is the tension in the string at 30.0°C, ′= L N M O Q P × L N M O Q P = ° TA Y L L 30 0 1 7 854 10 20 0 10 15 8 0 C N . . e j . To find the frequency at 30.0°C, realize that = f f T T 1 1 so = f 1 200 580 192 Hz N 632 N Hz a f . *P19.72 Some gas will pass through the porous plug from the reaction chamber 1 to the reservoir 2 as the
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