Homework 22 - Gozick Brandon – Homework 22 – Due Dec 5...

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Unformatted text preview: Gozick, Brandon – Homework 22 – Due: Dec 5 2006, 7:00 pm – Inst: D Weathers 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points On a day when the temperature reaches 59 . 8424 ◦ F, what is the temperature in de- grees Celsius? Correct answer: 15 . 468 ◦ C. Explanation: Given : T F = 59 . 8424 ◦ F . From the relationship T F = 9 5 T C + 32 T C = 5 9 ( T F- 32) = 5 9 (59 . 8424 ◦ F- 32) = 15 . 468 ◦ C . 002 (part 2 of 2) 10 points What is the temperature in Kelvin? Correct answer: 288 . 468 K. Explanation: From the equation T C = T K- 273 T K = T C + 273 = 288 . 468 K . keywords: 003 (part 1 of 1) 10 points A square hole 4 . 2 cm along each side is cut in a sheet of copper. Calculate the change in the area of this hole if the temperature of the sheet is increased by 74 K. Correct answer: 0 . 0443822 cm 2 . Explanation: Given : L = 4 . 2 cm and Δ T = 74 K . Each side of the hole expands linearly by Δ L = L α Δ T . The area, therefore expands by Δ A = ( L + Δ L ) 2- A . Since Δ L is much smaller than L we should keep only the linear term in Δ L , so Δ A = 2 L Δ L = 2 α A Δ T = 2 £ 1 . 7 × 10- 5 ( ◦ C)- 1 /( 17 . 64 cm 2 ) (74 K) = 0 . 0443822 cm 2 keywords: 004 (part 1 of 1) 10 points A steel rod 3 . 7 cm in diameter is heated so that its temperature increases by 86 ◦ C and then is fastened between two rigid supports. The rod is allowed to cool to its original tem- perature. Assuming that Young’s modulus for the steel is 1 . 58 × 10 11 N / m 2 and that its av- erage coefficient of linear expansion is 1 . 3 × 10- 5 ( ◦ C)- 1 , calculate the tension in the rod. Correct answer: 189 . 929 kN. Explanation: Given : d = 3 . 7 cm = 0 . 037 m , Δ T = 86 ◦ C , Y = 1 . 58 × 10 11 N / m 2 , and α = 1 . 3 × 10- 5 ( ◦ C)- 1 . The elongation of the rod from the change of the temperature is Δ ‘ = α ‘ Δ T F A = Y Δ ‘ ‘ = Y α Δ T F = A Y α Δ T = π µ d 2 ¶ 2 Y Δ T = π µ . 037 m 2 ¶ 2 × ( 1 . 58 × 10 11 N / m 2 ) (86 ◦ C) = 189 . 929 kN . Gozick, Brandon – Homework 22 – Due: Dec 5 2006, 7:00 pm – Inst: D Weathers 2 keywords: 005 (part 1 of 2) 10 points...
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Homework 22 - Gozick Brandon – Homework 22 – Due Dec 5...

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