hw21a - aldawsari (ma28683) HW21 Betancourt (16856) 1 This...

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Unformatted text preview: aldawsari (ma28683) HW21 Betancourt (16856) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 17.1-17.6 001 (part 1 of 2) 10.0 points At 11 C, an aluminum ring has an inner di- ameter of 8 cm and a brass rod has a diameter of 8 . 06 cm. If the temperature coefficient of expansion for brass is b = 1 . 9 10 5 ( C) 1 and the temperature coefficient of expansion for alu- minum is a = 2 . 4 10 5 ( C) 1 , to what temperature must the ring be heated so that it will just slip over the rod? Correct answer: 323 . 5 C. Explanation: Given : T = 11 C , d a = 8 cm , d b = 8 . 06 cm , b = 1 . 9 10 5 ( C) 1 , and a = 2 . 4 10 5 ( C) 1 . The new length will be d b = d a (1 + T ) d b = d a + d a T T 1 = T + T = T + d b- d a d a a = 11 C + 8 . 06 cm- 8 cm (8 cm)(2 . 4 10 5 ( C) 1 ) = 323 . 5 C . 002 (part 2 of 2) 10.0 points To what temperature must both be heated so that the ring just slips over the rod? Correct answer: 1555 C. Explanation: We need L Al = L brass for some T . Using the same law, L Al (1 + a T ) = L brass (1 + b T ) d a + d a a T = d b + d b b T T = T + d b- d a a d a- b d b a d a- b d b = bracketleftbig 2 . 4 10 5 ( C) 1 bracketrightbig (8 cm)- bracketleftbig 1 . 9 10 5 ( C) 1 bracketrightbig (8 . 06 cm) = 3 . 886 10 5 m / C so T = T 2- T T 2 = T + T = T + d b- d a...
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hw21a - aldawsari (ma28683) HW21 Betancourt (16856) 1 This...

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