PHY2048_Fall09_Homework_Set13

# PHY2048_Fall09_Homework_Set13 - PHY2048 Fall 2009 Homework...

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Unformatted text preview: PHY2048 Fall 2009 Homework Set 13 Solutions University of Florida Page 1 of 4 Department of Physics PHY 2048 Homework Set 13 Solutions (Chapters 14 & 15) Problem 14-62**: The volume of water flowing out of the end of the tube in time t is (a) 3 2 1 2 1 362 . 6 min 1 sec 60 min) 10 )( / 15 ( 2 03 . m s m m t v r Avt t R V V = = = = = π π . The continuity equation tells us that 1 1 2 2 A v A v = and hence (b) s m cm cm s m d d v A A v v / 4 . 5 5 3 ) / 15 ( 2 2 2 1 1 2 1 1 2 = = = = . Bernoulli’s equation tells us that 2 1 2 1 1 2 2 2 1 2 v P v P ρ ρ + = + , since there is no change in height ( i.e. h 1 = h 2 ). Hence, (c) ( 29 Pa s m s m m kg v v P P P atm gauge 4 2 2 3 3 2 1 2 2 2 1 2 1 2 10 792 . 9 ) / 4 . 5 ( ) / 15 ( ) / 10 ( ) ( × =- =- =- = ρ where I used the fact that P 1 = P atm . Problem 14-65**: Bernoulli’s equation tells us that ) ( 2 2 1 2 2 1 h H g v P gH v P B B A A- + + = + + ρ ρ ρ ρ . The continuity equation tells us that B B A A A v A v = , where A B is the area of the hole and A A is the area of the top of the tank. Hence, ) ( 2 2 1 2 2 2 1 h H g v P gH v A A P B B B A B A- + + = + + ρ ρ ρ ρ , Using the fact that P A = P B = P atm and solving for v B gives gh A A gh v A B B 2 ) / ( 1 2 2 ≈- = , where I have used the fact that A B /A A << 1. Now using the kinematic equations of motion on a small piece of the water gives t v t x B = ) ( and 2 2 1 ) ( ) ( gt h H t y-- = , where the piece of water is at the hole at t = 0. the piece of water is at the hole at t = 0....
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PHY2048_Fall09_Homework_Set13 - PHY2048 Fall 2009 Homework...

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