This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Niu (qn269) HW09 Tsoi (58020) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A simple Utube 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 Utube 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 Utube 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....
View
Full
Document
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.
 Spring '09
 Staff
 Physics

Click to edit the document details