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Homework Statement
Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m^2 ; at point 3 it is 0.0160 m^2 . The area of the tank is very large compared with the cross-sectional area of the pipe.
[PLAIN]http://img132.imageshack.us/img132/4245/xcvbb.jpg
Homework Equations
[tex]A_1v_1=A_2v_2[/tex]
[tex]p_1+\rho gh_1+\frac{1}{2}\rho v_1^2=p_2+\rho gh_2+\frac{1}{2}\rho v_2^2[/tex]
The Attempt at a Solution
[tex]p_1+\rho gh_1+\frac{1}{2}\rho v_1^2 = 0+(\rho g*10.0 m)+\frac{1}{2}\rho(0 m/s)^2[/tex]
Is this correct for the left side of the equation? I figure p_1=0 because it's at the top of the tank and v_1=0 because it's not flowing.
I'm not sure what to put for the right side because I don't know p_2. I know that [itex]p=\rho gh[/itex] but that doesn't make sense in the equation.
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