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# hw9solutions - Niu (qn269) HW09 Tsoi (58020) 1 This...

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Unformatted text preview: Niu (qn269) HW09 Tsoi (58020) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 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 420 kg / m 3 is then poured into one arm of the tube, forming a column 14 cm in height, as shown. h 14 cm light liquid 420 kg / m 3 heavy liquid 1000 kg / m 3 What is the difference in the heights of the two liquid surfaces? Correct answer: 8 . 12 cm. Explanation: Let : = 14 cm , = 420 kg / m 3 , and 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 the liquid of height h poured into the right branch, so P + ( h ) h g = P + g . h = parenleftbigg 1 h parenrightbigg = (14 cm) parenleftbigg 1 420 kg / m 3 1000 kg / m 3 parenrightbigg = 8 . 12 cm . 002 (part 1 of 2) 10.0 points Calculate the pressure at an ocean depth of 1490 m. The acceleration of grav- ity is 9 . 8 m / s 2 , atmospheric pressure is 1 . 01 10 5 Pa , and the density of the sea wa- ter is 1024 kg / m 3 . Correct answer: 1 . 50534 10 7 Pa. Explanation: Let : g = 9 . 8 m / s 2 , P = 1 . 01 10 5 Pa , = 1024 kg / m 3 , and h = 1490 m . P = P + g h = 1 . 01 10 5 Pa + (1024 kg / m 3 ) (9 . 8 m / s 2 ) (1490 m) = 1 . 50534 10 7 Pa . 003 (part 2 of 2) 10.0 points Calculate the total force exerted on the out- side of a circular submarine window of diam- eter 28 . 9 cm at this depth. Correct answer: 9 . 87465 10 5 N. Explanation: Let : r = 14 . 45 cm = 0 . 1445 m . F = P A = P r 2 = (1 . 50534 10 7 Pa) (0 . 1445 m) 2 = 9 . 87465 10 5 N . 004 10.0 points One method of measuring the density of a liq- uid is illustrated in the figure. One side of the Niu (qn269) HW09 Tsoi (58020) 2 U-tube is in the liquid being tested, and the other side is in water of density 1000 kg / m 3 . The air is partially removed at the upper part of the tube and the valve is closed. The height of the water above its pool surface is . 74 m . The height of the liquid above its pool surface is 0 . 38 m . The difference in the heights of the pool surfaces is 0 . 21 m . Valve test liquid water . 38 m . 74 m . 21 m Find the density of the liquid on the left. Correct answer: 1947 . 37 kg / m 3 . Explanation: Let : w = 1000 kg / m 3 , h w = 0 . 74 m , h = 0 . 38 m , and h = 0 . 21 m . The pressure at the upper surface of each liquid is given by P = P atm w g h w = P atm g h....
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## This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas at Austin.

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hw9solutions - Niu (qn269) HW09 Tsoi (58020) 1 This...

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