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Suppose the tube in the previous problem...

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.

Text Solution

Verified by Experts

The correct Answer is:
`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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