435_Physics ProblemsTechnical Physics

435_Physics ProblemsTechnical Physics - Chapter 14 *P14.74...

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Chapter 14 437 *P14.74 (a) Take point 1 at the free water surface in the tank and point 2 at the bottom end of the tube: Pg y v y v d P v vg d 11 1 2 22 2 2 00 2 2 2 1 2 1 2 1 2 2 ++ = = + + = ρρ The volume flow rate is V t Ah t vA == 2 . Then t Ah Ah Ag d = = 2 2 . (b) t = × = 05 298 10 44 6 2 4 .. . . m m 21 0 m m s m s af ej *P14.75 (a) For diverging stream lines that pass just above and just below the hydrofoil we have y vPg y v tt t bb b = 1 2 1 2 . Ignoring the buoyant force means taking yy tb Pn v P v PP vn b b bt b += + −= 1 2 1 2 1 2 1 2 2 ρ bg The lift force is PPA A b 1 2 1 . (b) For liftoff, 1 2 1 2 1 2 12 AM g v Mg nA b b = F H
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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