HW 6 solutions (Ch 15,16,17)

HW 6 solutions (Ch 15,16,17) - Physics 6C HW#6 solution May...

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Unformatted text preview: Physics 6C HW#6 solution May 22, 2008 15.30 Volume flow rate for a steady current is, A 1 v 1 = A 2 v 2 a) 20 . l 60 . s ( 1000 cm 3 1 l ) = (1 . 00 cm ) 2 v hose v hose = 333 cm 3 /s 3 . 14 cm 2 = 106 cm/s b) 20 . l 60 . s ( 1000 cm 3 1 l ) = (0 . 500 cm ) 2 v nozzle v nozzle = 333 cm 3 /s . 785 cm 2 = 424 cm/s 15.33 We can use the Bernoullis equation, P 1 + 1 2 v 2 1 + gh 1 = P 2 + 1 2 v 2 2 + gh 2 We need two points in order to apply the Bernoullis equation. One point is where the steam exits the tank(the point of interest) and the other point can be any other place in the tank. In this solution, I choose the top surface as the second point. Then, the equation becomes, At the first point P air + 1 2 v 2 x + gh At the second point P air + 1 2 (0) 2 + gh two equation are equal P air + 1 2 v 2 x + gh = P air + 1 2 (0) 2 + gh v 2 x = 2 g ( h- h ) v 2 x = p 2 g ( h- h ) Then the horizontal distance, x ,is, x = v x t y = h = 1 2 gt 2 t = s 2 h g plug t into x...
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This note was uploaded on 07/02/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.

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HW 6 solutions (Ch 15,16,17) - Physics 6C HW#6 solution May...

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