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hw17 - karna(pk4534 HW 17 coker(58245 This print-out should...

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karna (pk4534) – HW 17 – coker – (58245) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is the last chapter covered on Quiz 4. 001 10.0 points The four tires of an automobile are inflated to an absolute pressure of 2 . 5 × 10 5 Pa. Each tire has an area of 0 . 022 m 2 in contact with the ground. Determine the weight of the automobile. Correct answer: 22000 N. Explanation: Let : P = 2 . 5 × 10 5 Pa and A = 0 . 022 m 2 . F g = 4 P A = 4 (2 . 5 × 10 5 Pa) (0 . 022 m 2 ) = 22000 N . 002 10.0 points The air pressure above the liquid in figure is 1 . 52 atm . The depth of the air bubble in the liquid is 45 . 3 cm and the liquid’s density is 795 kg / m 3 . 45 . 3 cm air air Determine the air pressure in the bubble suspended in the liquid. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 57505 × 10 5 Pa. Explanation: Let : ρ = 795 kg / m 3 , P 0 = 1 . 52 atm , g = 9 . 8 m / s 2 , and h = 45 . 3 cm = 0 . 453 m . P = P 0 + ρ g h = (1 . 52 atm) · 1 . 013 × 10 5 Pa atm + (795 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 453 m) = 1 . 57505 × 10 5 Pa . 003 10.0 points A simple U-tube that is open at both ends is partially filled with a heavy liquid of density 1000 kg / m 3 . A liquid of density 286 kg / m 3 is then poured into one arm of the tube, forming a column 13 cm in height, as shown. h 13 cm light liquid 286 kg / m 3 heavy liquid 1000 kg / m 3 What is the difference in the heights of the two liquid surfaces? Correct answer: 9 . 282 cm. Explanation: Let : = 13 cm , ρ = 286 kg / m 3 , and ρ h = 1000 kg / m 3 . Because the liquid in the U-tube is static, the pressure exerted by the heavy liquid col- umn of height - h in the left branch of the tube must balance the pressure exerted by
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karna (pk4534) – HW 17 – coker – (58245) 2 the liquid of height h poured into the right branch, so P 0 + ( - h ) ρ h g = P 0 + ℓ ρ g . h = parenleftbigg 1 - ρ ρ h parenrightbigg = (13 cm) parenleftbigg 1 - 286 kg / m 3 1000 kg / m 3 parenrightbigg = 9 . 282 cm . 004 10.0 points The small piston of a hydraulic lift has a cross-sectional area of 8 . 1 cm 2 and the large piston has an area of 63 cm 2 , as in the figure below. F 8 . 1 cm 2 area 63 cm 2 What force F must be applied to the small piston to maintain the load of 89 kN at a constant elevation? Correct answer: 11442 . 9 N. Explanation: Let : A 1 = 8 . 1 cm 2 , A 2 = 63 cm 2 , and W = 89 kN . According to Pascal’s 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 = (8 . 1 cm 2 ) (63 cm 2 ) (89000 N) = 11442 . 9 N . 005 10.0 points Compared to an empty ship, would a ship loaded with a cargo of Styrofoam sink deeper or rise in the water? 1. The ship will float at the same level be- cause its density hasn’t changed. 2. Since Styrofoam is less dense than water, the ship rises. 3. Since Styrofoam pushes down on the ship with its weight, the ship sinks deeper. cor- rect 4. It depends on the weight of the Styro- foam.
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