Question #87982
A hose ejects water at 80cl/s through a hole 2mm in diameter. The water impinged on a wall and drops off without rebounding. What is the force on the wall?
1
Expert's answer
2019-04-15T09:55:20-0400
F=d(mv)dt=vd(m)dt=vd(ρV)dt=vρdVdtF= \frac{d(mv)}{dt}= v\frac{d(m)}{dt}=v\frac{d(\rho V)}{dt}=v\rho \frac{dV}{dt}

dVdt=d(As)dt=Adsdt=Av=(0.25πd2)v\frac{dV}{dt}=\frac{d(As)}{dt}= A \frac{ds}{dt}=Av=(0.25\pi d^2)v

F=4ρ(dVdt)2πd2=4(1000)(0.0008)2π(0.002)2=200NF=\frac{4\rho (\frac{dV}{dt})^2}{\pi d^2}=\frac{4(1000)(0.0008)^2}{\pi (0.002 )^2}=200N


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