569_Physics ProblemsTechnical Physics

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

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Chapter 19 571 P19.71 (a) µ π ρ π = = × × = × r 2 4 2 3 3 5 00 10 7 86 10 6 17 10 . . . m kg m kg m 3 e j e j (b) f v L 1 2 = and v T = µ so f L T 1 1 2 = µ Therefore, T Lf = = × × × = µ 2 6 17 10 2 0 800 200 632 1 2 3 2 b g e j a f . . N (c) First find the unstressed length of the string at 0°C: L L T AY L L T AY A Y = + F H G I K J = + = × = × = × natural natural 2 so m m and Pa 1 1 5 00 10 7 854 10 20 0 10 4 2 7 10 π . . . e j Therefore, T AY = × × = × 632 7 854 10 20 0 10 4 02 10 7 10 3 . . . e je j , and L natural m m = + × = 0 800 1 4 02 10 0 796 8 3 . . . a f e j . The unstressed length at 30.0°C is L L 30 1 30 0 0 0 ° = + ° ° C natural C C α . . a f , or L 30 6 0 796 8 1 11 0 10 30 0 0 797 06 ° = + × = C m m . . . . b g e j a f . Since L L 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 = ° T AY L L 30 7 10 1 7 854 10 20 0 10 0 800 0 797 06 1 580 C N . . . . e je j . To find the frequency at 30.0°C, realize that = f f T T 1 1 so ′ = = f
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