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Unformatted text preview: Overton, Mays Homework 30 Due: Apr 18 2005, 4:00 am Inst: Turner 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The small piston of a hydraulic lift has a crosssectional area of 9 . 3 cm 2 and the large piston has an area of 93 cm 2 , as in the figure below. F 9 . 3 cm 2 area 93 cm 2 What force F must be applied to the small piston to raise a load of 75 kN? Correct answer: 7500 N. Explanation: Let : A 1 = 9 . 3 cm 2 , A 2 = 93 cm 2 , W = 75 kN , and F = F . According to Pascals law, the pressure ex erted on A 1 must be equal to the one exerted on A 2 . The pressure P 1 = F A 1 must be equal to the pressure P 2 = W A 2 due to the load. F A 1 = W A 2 , so F = A 1 A 2 W = (9 . 3 cm 2 ) (93 cm 2 ) (75000 N) = 7500 N . 002 (part 1 of 2) 10 points A heavy liquid with a density 11 g / cm 3 is poured into a Utube as shown in the first figure below. The lefthand arm of the tube has a crosssectional area of 9 . 03 cm 2 , and the righthand arm has a crosssectional area of 2 . 55 cm 2 . A quantitly of 103 g of a light liquid with a density 1 . 6 g / cm 3 are then poured into the righthand arm as shown in the second figure below. h 1 h 2 9 . 03 cm 2 2 . 55 cm 2 heavy liquid 11 g / cm 3 L 9 . 03 cm 2 2 . 55 cm 2 light liquid 1 . 6 g / cm 3 Determine the height L of the light liquid in the column in the right arm of the Utube, as shown in the second figure above. Correct answer: 25 . 2451 cm. Explanation: Let : m = 103 g , A 1 = 9 . 03 cm 2 , A 2 = 2 . 55 cm 2 , = 1 . 6 g / cm 3 , and h = 11 g / cm 3 . Using the definition of density = m V = m A 2 L L = m A 2 = 25 . 2451 cm . 003 (part 2 of 2) 10 points If the density of the heavy liquid is 11 g / cm 3 , Overton, Mays Homework 30 Due: Apr 18 2005, 4:00 am Inst: Turner 2 by what height h 1 does the heavy liquid rise in the left arm? Correct answer: 0 . 808604 cm. Explanation: After the light liquid has been added to the right side of the tube, a volume A 2 h 2 of heavy liquid is displaced to the left side, raising the heavy liquid on the left side by a height of h 1 with a displaced volume of A 1 h 1 . Since the volume of heavy liquid is not changed, we have A 1 h 1 = A 2 h 2 . At the level of the heavylight liquid interface in the right side, the absolute pressure is P = P atm + g L , and at the same level in the left tube, P = P atm + h g [ h 1 + h 2 ] . Equating these two values, we obtain g L = h g [ h 1 + h 2 ] = h g h 1 + A 1 A 2 h 1 = h g h 1 1 + A 1 A 2 ....
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 Spring '08
 Turner
 Physics, Work

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