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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 T T L T T k h c k h c = = F H G I K J L N M M O Q P P = F H G I K J α α α α θ µ α α θ µ 2 1 2 1 2 1 2 2 1 b ga f b g b g 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) α α θ µ 2 1 6 6 24 10 1 15 10 1 1 20 18 5 0 42 32 0 275 F H G I K J = × ° × ° F H G I K J ° ° = b g b g L T T k h c tan . tan . . . C C m C mm (e) If α α 2 1 < , 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 3 96 10 2 . × mol P19.6 T T C = ° ° + ° 1 33 20 0 . . C S C S b g P19.38 3 67 . cm 3 P19.8 0.313 m P19.40 between 10 kg 1 and 10 kg 2 P19.10 1.20 cm P19.42 2 41 10 11 . × molecules P19.12 15 8 . µ m P19.44 (a) 2.24 m; (b) 9 28 10 5 . × Pa P19.14
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