HW 06-solutions - cooper(mlc4285 HW 06 florin(55245 1 This...

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cooper (mlc4285) – HW 06 – florin – (55245) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points An object of volume V is floating at the in- terface between water and a denser liquid. 4 5 of the object’s volume is displacing water, while 1 5 of the object’s volume is displacing the heavier liquid. ρ w ρ If the density of the heavier liquid is ρ and the density of water is ρ w , what is the total buoyant force on the object? 1. 4 ρ w V g 5 2. ( ρ - ρ w ) V g 3. ( ρ + 4 ρ w ) V g 5 correct 4. ( ρ + ρ w ) V g 5. ρ V g 5 6. ρ w V g 5 7. (4 ρ + ρ w ) V g 5 Explanation: The buoyant force is the density of the fluid displaced times the volume displaced, so F B = 4 5 ρ w V g + 1 5 ρ V g = ( ρ + 4 ρ w ) V g 5 . 002(part1of2)10.0points Assuming that the ocean has a constant den- sity like that near the surface, Calculate the pressure at an ocean depth of 1060 m. The ac- celeration of gravity is 9 . 8 m / s 2 , atmospheric pressure is 1 . 01 × 10 5 Pa , and the density of the sea water is 1024 kg / m 3 . Correct answer: 1 . 07383 × 10 7 Pa. Explanation: Let : g = 9 . 8 m / s 2 , P 0 = 1 . 01 × 10 5 Pa , ρ = 1024 kg / m 3 , and h = 1060 m . P = P 0 + ρ g h = 1 . 01 × 10 5 Pa + (1024 kg / m 3 ) (9 . 8 m / s 2 ) (1060 m) = 1 . 07383 × 10 7 Pa . 003(part2of2)10.0points Calculate the total force exerted on the out- side of a circular submarine window of diam- eter 24 . 3 cm at this depth. Correct answer: 4 . 9801 × 10 5 N. Explanation: Let : r = 12 . 15 cm = 0 . 1215 m . F = P A = P π r 2 = (1 . 07383 × 10 7 Pa) π (0 . 1215 m) 2 = 4 . 9801 × 10 5 N . 004(part1of2)10.0points A loaded flatbottom barge floats in fresh wa- ter. The bottom of the barge is 4 . 9 m below the water line. When the barge is empty the barge’s bottom is only 1 . 33 m below the water line. What is the difference between the pressure on the bottom of the loaded barge and the pressure at the water line?
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cooper (mlc4285) – HW 06 – florin – (55245) 2 Correct answer: 48020 Pa. Explanation: Let : d 1 = 4 . 9 m and ρ = 1000 kg / m 3 . P = d 1 ρ g = 4 . 9 m (1000 kg / m 3 ) (9 . 8 m / s 2 ) = 48020 Pa . 005(part2of2)10.0points If the surface area of the bottom of the barge is 456 m 2 what is the weight of the load in the barge? Correct answer: 1 . 59536 × 10 7 N. Explanation: Let : d 2 = 1 . 33 m and a 1 = 456 m 2 . Weight in a bouyant, stable system is equal to the mass of the fluid displaced times grav- ity. This is known as Archimedes’ Principle. The mass of water displaced, and thus the mass of the load, is W = m g = ρ ( d 1 - d 2 ) a 1 g = (1000 kg / m 3 ) (4 . 9 m - 1 . 33 m) · (456 m 2 ) (9 . 8 m / s 2 ) = 1 . 59536 × 10 7 N . 006 10.0points A block of wood weighs 50 . 6 N in air. A sinker is hanging from the block, and the weight of the wood-sinker combination is 234 . 8 N when the sinker alone is immersed in water. When the wood-sinker combination is com- pletely immersed, the weight is 139 . 3 N.
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