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Unformatted text preview: homework 27 FIERRO, JEFFREY Due: May 4 2008, 11:00 pm 1 Question 1, chap 18, sect 5. part 1 of 1 10 points A small plastic ball (similar to a pingpong ball) with mass 3 . 5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is totally sub merged, the tension in the thread is 0 . 033 N . The density of water is 1000 kg / m 3 and the acceleration of gravity is 9 . 81 m / s 2 . 3 . 5 g Determine the diameter of the ball. Correct answer: 2 . 35789 cm (tolerance 1 %). Explanation: V sphere = 4 3 r 3 = 4 3 parenleftbigg d 2 parenrightbigg 3 = 1 6 d 3 . Let : w = 1000 kg / m 3 , g = 9 . 81 m / s 2 , T = 0 . 033 N , and m = 3 . 5 g = 0 . 0035 kg . Consider the forces acting on the PingPong ball B mg T Applying Archimedes principle, we have B = w f = m f g = w V ball g = 1 6 w g d 3 . Applying summationdisplay F y = 0 to the ball, B mg T = 0 1 6 w g d 3 mg T = 0 . Then we have d = 3 radicalBigg 6 ( T + mg ) w g = 3 radicalBigg 6 [0 . 033 N + (0 . 0035 kg) (9 . 81 m / s 2 )] (1000 kg / m 3 ) (9 . 81 m / s 2 ) 100 cm m = 2 . 35789 cm . Question 2, chap 18, sect 2. part 1 of 1 10 points A fountain sends a stream of water 25 . 4 m up into the air. The acceleration of gravity is 9 . 8 m / s 2 . If the base of the stream is 6 . 61 cm in diameter, what power is required to send the water to this height? Correct answer: 19058 . 9 W (tolerance 1 %). Explanation: Let : h 1 = 25 . 4 m , d = 6 . 61 cm , r = 6 . 61 cm 2 = 0 . 03305 m , and g = 9 . 8 m / s 2 . Using conservation of energy, we may write K max = U max = E total 1 2 mv 2 = mg h, and v = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (25 . 4 m) = 22 . 3123 m / s . homework 27 FIERRO, JEFFREY Due: May 4 2008, 11:00 pm 2 is the velocity of the stream at its base. The flow rate at the base is r 2 v = (0 . 03305 m) 2 (22 . 3123 m / s) = 0 . 0765663 m 3 / s . Using the density of water as 1000 kg / m 3 , we see that 76 . 5663 kg of water emerges from the fountain each second. m t = r 2 v = (1000 kg / m 3 ) (0 . 03305 m) 2 (22 . 3123 m / s) = 76 . 5663 kg . Thus, each second, sufficient work is done to raise 76 . 5663 kg to a height of 25 . 4 m or P = W t = mg h t = r 2 v g h t = (1000 kg / m 3 ) (0 . 03305 m) 2 (22 . 3123 m / s) (9 . 8 m / s 2 ) (25 . 4 m) (1 s) = 19058 . 9 W . Question 3, chap 18, sect 5. part 1 of 1 10 points An empty rubber balloon has a mass of . 0151 kg. The balloon is filled with helium at a density of 0 . 0925 kg / m 3 . At this density the balloon has a radius of 0 . 318 m. The acceleration of gravity is 9 . 8 m / s 2 . If the filled balloon is fastened to a vertical line, what is the tension in the line?...
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This note was uploaded on 05/11/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Mass, Work

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