Physics 115 Final With Answers

81x1501470000joules 27 a 392n b 176n c 108n d 687n

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Unformatted text preview: o it’s always faster in the last second than in the first so it covers more distance in the last second than in the first. PHYS115 – Spring 2012 10 Final Test - 1 May 2012 24. 1 Bernoulli again: p 2 v 2 gh must be constant, so the same everywhere. Call the height at sea level, h = 0. 1 At the pump station: p 2 v 2 gh 4,000 , 000 0 0 4,000,000 Pa At the highest possible point v=0, and there’s zero pressu re (that’s why it is the highest point), 1 so there p 2 mv 2 gh 1,000 9.81 h 4 ,000,000 408 m 1, 000 9.81 At sea level we have h=0, and we also have atmospheric pressure 100,000 Pa, so then c o We get 4 ,000,000 1,000 9.81 h so h 1 1 p 2 v 2 gh 100 , 000 2 v 2 gh . 1 Then 100 ,000 2 1, 000 v 2 4, 000,000 and we so v 2 v 2 (4,000,000 100,000) and 1, 000 2 ( 4,000,000 100,000) 80 m/s 1,000 25. It rotates 3x slower, so its angular speed is 1/3rd of what it was before. We know that (rotational kinetic energy) = 1/2 x (rotatio nal mass) x (angular speed)2, as in 1 I 2 . So if the angular speed is 1/3rd of what it was, the (rotational kinetic energy) 2 is 9x smaller 26. PHYS115 – Spring 2012 11 Final Test - 1 May 2012 a. b. c. d. 9,810 J 1,470,000 J 150,000 J 1470 J (change in potential energy) = mass x g x (change in h eight) = 1000 x 9.81 x 150 = 1,470,000 joules 27. a. 39.2 N b. 176 N c. 108 N d. 68.7 N Two forces act on the log: (1) the buoyant force; and (2) i ts weight. (1) buoyant force = weight of the 11 kg of diplaced water. This is weight = (mass of 11 kg) x g = 11 x 9.81 = 107.9 N (upward) (2) weight of the log = (mass of log) x g = 7 x 9.81 = 68.67 N (downward) Net force is then 107.9 (up) + 68.67 (down) = 39.2 N (u p) 28. A solid block of metal has temperature 80oC. What is its a bsolute t...
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This document was uploaded on 02/23/2014 for the course PHYSICS 115 at UNL.

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