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HW 6 solutions (Ch 15,16,17)

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

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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 . 0 l 60 . 0 s ( 1000 cm 3 1 l ) = π (1 . 00 cm ) 2 v hose v hose = 333 cm 3 /s 3 . 14 cm 2 = 106 cm/s b) 20 . 0 l 60 . 0 s ( 1000 cm 3 1 l ) = π (0 . 500 cm ) 2 v nozzle v nozzle = 333 cm 3 /s 0 . 785 cm 2 = 424 cm/s 15.33 We can use the Bernoulli’s equation, P 1 + 1 2 ρv 2 1 + ρgh 1 = P 2 + 1 2 ρv 2 2 + ρgh 2 We need two points in order to apply the Bernoulli’s 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 ρv 2 x + ρgh At the second point P air + 1 2 ρ (0) 2 + ρgh 0 two equation are equal P air + 1 2 ρv 2 x + ρgh = P air + 1 2 ρ (0) 2 + ρgh 0 v 2 x = 2 g ( h 0 - h ) v 2 x = p 2 g ( h 0 - 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 x = v x s 2 h g = p 2 g ( h 0 - h ) s 2 h g = 2 p h ( h 0 - h ) 1
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16.12 Δ V = ( β CCl 4 - 3 α F e ) V i Δ T = (5 . 81 × 10 - 4 - 3(11 . 0 × 10 - 6 ))(50 . 0 gal )(20 . 0
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