571_Physics ProblemsTechnical Physics

571_Physics ProblemsTechnical Physics - Chapter 19 f ga L L...

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Chapter 19 573 ∆∆ L L xL xL T L L TT L k hc k =− F H G I K J L N M M O Q P P F H G I K J αα θ µ 21 2 2 1 bg a f b g b g tan tan . At dawn the next day the point P is farther down the roof by the distance L . It represents the displacement of every other point on the plate. (d) 66 24 10 1 15 10 1 1 20 18 5 042 32 0 275 F H G I K J −= × ° −× ° F H G I K J ° °= −− b g L k tan . tan . . . CC m C m m (e) If < , the diagram in part (a) applies to temperature falling and the diagram in part (b) applies to temperature rising. The weight of the plate still pulls it step by step down the roof. The same expression describes how far it moves each day. ANSWERS TO EVEN PROBLEMS P19.2 (a) 1 06 . atm; (b) −° 124 C P19.32 (a) 900 K; (b) 1 200 K P19.4 (a) 37 0 310 . C K ; (b) = 20 6 253 .C K P19.34 see the solution P19.36 396 10 2 . × mol P19.6 C ° + ° 133 200 .. CS C S P19.38 367 . c m 3 P19.8 0.313 m P19.40 between 10 kg 1 and 10 kg 2 P19.10 1.20 cm P19.42 241 10 11 . × molecules P19.12 15 8 . m P19.44 (a) 2.24 m; (b) 9 28 10 5 . × Pa P19.14 0.663 mm to the right at 78.2 ° below the
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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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