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Unformatted text preview: homework 36 – BAUTISTA, ALDO – Due: Apr 28 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Assume: h = 8 m, L = 1 m, and θ = 39 ◦ , and that the cross-sectional area at A is very large compared with that at B. Assume: y = 0 at B. Measure the height from the top of the angled tube. The figure below shows a water tank with a valve at the bottom. The acceleration of gravity is 9 . 8 m / s 2 . θ B L h A Valve If this valve is opened, what is the max- imum height attained by the water stream coming out of the right side of the tank? Correct answer: 2 . 91912 m (tolerance ± 1 %). Explanation: Let us first compute the speed with which the water leaves the tank at B by applying Bernoulli’s equation between A and B. Since the cross-sectional area at A is much larger than the one at B, we can neglect the speed of the water at A compared to that at B. The pressure at both points is going to be equal to the atmospheric pressure P atm . The equation is then P atm + ρ w g ( h- L sin θ ) = P atm + 1 2 ρ w v 2 B , and so v B = p 2 g ( h- L sin θ ) . Now the problem reduces to that of projec- tile motion, for which the maximum height is given by h M = v 2 B sin 2 θ 2 g = [ h- L sin θ ] sin 2 θ = [(8 m)- (1 m) sin(39 ◦ )] sin 2 (39 ◦ ) = 2 . 91912 m . Question 2 Part 1 of 1. 10 points. A large storage tank is filled to a height 88 cm. If the tank is punc- tured with a small hole at a height 35 cm from the bottom of the tank, h h How far from the tank will the stream land? Assume that the tank is very large so that the velocity of the liquid at the top of the tank is zero. Correct answer: 86 . 1394 cm (tolerance ± 1 %). Explanation: Basic Concepts: Bernoulli’s Equa- tion and Projectile Motion According to Bernoulli’s Equation, P + 1 2 ρ v 2 + ρ g y = constant Assume that the velocity of the liquid at the top of the tank is 0. And the pressure at the top of the bank and at the puncture are both P air .Hence, P air + ρ g h = P air + 1 2 ρ v 2 x + ρ g h From this, ρ g ( h- h ) = 1 2 ρ v x 2 homework 36 – BAUTISTA, ALDO – Due: Apr 28 2006, 4:00 am 2 Solving for v x , v x = p 2 g ( h- h ) Also, the time of projectile motion is t = s 2 h g Therefore, x = v x s 2 h g = p 2 g ( h- h ) s 2 h g = 2 p h ( h- h ) = 2 p 35 cm (88 cm- 35 cm) = 86 . 1394 cm Question 3 Part 1 of 1. 10 points. A vessel is filled with a liquid of density 1431 kg / m 3 . There are two holes (one above the other) in the side of the vessel. Liquid is streaming out of both of these holes in a horizontal direction. The acceleration of gravity is 9 . 8 m / s 2 . The upper hole is a distance 58 m down from the surface of the liquid. The lower hole is a distance 84 m down from the surface of the liquid....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08