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Hw 9 Solutions

Hw 9 Solutions - Bell Evan Homework 9 Due 11:00 pm Inst...

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Bell, Evan – Homework 9 – Due: Mar 25 2005, 11:00 pm – Inst: Furlan 1 This print-out should have 26 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 1) 10 points The four tires of an automobile are inflated to a gauge pressure of 239000 Pa. Each tire has an area of 0 . 0245 m 2 in contact with the ground. Determine the weight of the automobile. Correct answer: 23422 N. Explanation: Let W be its weight. Then each tire sup- ports W 4 , so that P = F A = W 4 A , yielding: W = 4 A P = 4 (0 . 0245 m 2 )(239000 Pa) = 23422 N . 002 (part 1 of 1) 10 points Blood of density 1260 kg / m 3 that is to be ad- ministered to a patient is raised about 0 . 944 m higher than the level of the patient’s arm. The acceleration of gravity is 9 . 8 m / s 2 . How much greater is the pressure of the blood than it would be if the container were at the same level as the arm? Correct answer: 11656 . 5 Pa. Explanation: We use Δ P = ρ g h, where h is the height above the arm. Thus, Δ P = (1260 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 944 m) = 11656 . 5 Pa . 003 (part 1 of 2) 10 points The cross-sectional area of the output piston in a hydraulic device is 11 times the input piston’s area. By how much will the device multiply the input force? Correct answer: 11 . Explanation: Basic concepts Pressure is force per unit area. F in A in = F out A out = F out n A in F out = n F in . 004 (part 2 of 2) 10 points By what factor will the output piston move compared to the distance the input piston is moved? Correct answer: 0 . 0909091 . Explanation: F in d in = F out d out = n F in d out d out = 1 n d in . 005 (part 1 of 1) 10 points A U-tube of constant cross-sectional area, open to the atmosphere, is partially filled with a heavy liquid with density 10 . 3 g / cm 3 . A light liquid with density 1 . 79 g / cm 3 is then poured into both arms. h 0 . 567 cm light liquid 1 . 79 g / cm 3 heavy liquid 10 . 3 g / cm 3 If the equilibrium configuration of the tube is as shown in figure, with a difference in the height of the heavy liquid of 0 . 567 cm, determine the value of the difference in height of the light liquid h . Correct answer: 2 . 69563 cm. Explanation: Let : ρ = 1 . 79 g / cm 3 , ρ h = 10 . 3 g / cm 3 , and h h = 0 . 567 cm , Also, let h be the height of the light liquid column added to the right side of the U-tube.

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Bell, Evan – Homework 9 – Due: Mar 25 2005, 11:00 pm – Inst: Furlan 2 Consider the pressure at the elevation of the light-heavy liquid interface in the left column and at point at the same elevation in the right column. By Pascal’s Principle, the absolute pressure is the same as that at elevation in both columns. But, P right = P atm + ρ h g h h + ρ g h and P left = P atm + ρ g ( h h + h + h ) . Thus, from Pascal’s Principle, P left = P right ρ ( h h + h ) = ρ h ( h h ) , or h = ρ h ρ - 1 h h = 10 . 3 g / cm 3 1 . 79 g / cm 3 - 1 h h = 2 . 69563 cm .
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