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552_Physics ProblemsTechnical Physics

# 552_Physics ProblemsTechnical Physics - 554 P19.15...

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554 Temperature P19.15 (a) L T L T Al Al Brass Brass 1 1 + = + α α b g b g T L L L L T T T = = × × = − ° = ° Al Brass Brass Brass Al Al C so C. This is attainable. α α 10 01 10 00 10 00 19 0 10 10 01 24 0 10 199 179 6 6 . . . . . . a f a f e j a f e j (b) T = × × 10 02 10 00 10 00 19 0 10 10 02 24 0 10 6 6 . . . . . . a f a f e j a f e j T = − ° 396 C so T = ° 376 C which is below 0 K so it cannot be reached. P19.16 (a) A A T i = 2 α : A = × ° ° 2 17 0 10 0 080 0 50 0 6 1 2 . . . C m C e j b g a f A = × = 1 09 10 0 109 5 . . m cm 2 2 (b) The length of each side of the hole has increased. Thus, this represents an increase in the area of the hole. P19.17 V V T i = = × × = β α 3 5 81 10 3 11 0 10 50 0 20 0 0 548 4 6 b g e j e j b ga f . . . . . gal gal P19.18 (a) L L T i = + 1 α a f : 5 050 5 000 1 24 0 10 20 0 6 1 . . . . cm cm C C = + × ° ° T a f T = ° 437 C (b) We must get L L Al Brass = for some T , or L T L T T T i i , , . . . . Al Al Brass Brass cm C cm C 1 1 5 000 1 24 0 10 5 050 1 19 0 10 6 1 6 1 + = + + × ° = + × ° α α
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