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Suppose the tube in the previous problem is kept verticla with B upward. Water enters through B at the rate of `1cm^3s^-1`. Repeat parts a, b, and c. Note that the speed decreases as the water falls down. |
Answer» Correct Answer - `187.5 N//m^(2)` `A_(1)v_(1) = A_(2)v_(2)` `P_(y) + (1)/(2)rhov_(y)^(2) + rhogh = P_(x) + (1)/(2) rhov_(x)^(2) + 0`. `P_(x) - P_(y) = (1)/(2)rho(v_(y)^(2) - v_(x)^(2)) + rhogh` `P_(x) - P_(y) = (1)/(2) 1000[((1)/(2))^(2) - ((1)/(4))^(2)] + 1000 xx 10 xx (15)/(1600)` `P_(x) - P_(y) = 187.5 N//m^(2)` |
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