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quiz22_solutions - P1 rho v1^2/2=P2 rho v2^2/2 so P2=P1...

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05/02/08 Physics 201 Quiz 22 Water flows along a pipe shown below (like demo in class). The cross-sectional area is A1 at the skinny part and A2 at the fat part. The water flows in at a rate G (liters/sec) and a pressure P1 . a) find the pressure and velocity in both parts. ANS: v1=G/A1; v2=G/A2;
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Unformatted text preview: P1+rho v1^2/2=P2+rho v2^2/2 so P2=P1+rho(v1^2-v2^2)/ 2=P1+rho G^2(1/A1^2-1/A2^2)/2. b) find the height of the water in the vertical tubes. The tubes are open at the top where the atmospheric pressure is Pa ANS: Pa=P1-rho g h1, so h1=(P1-Pa)/(rho g). Likewise h2=(P2-Pa)/(rho g)...
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