Respuesta :
Answer:
h = 4.281 m
Explanation
Given Data,
Weight of disk = 2kg
Speed of water jet (V) = 10 m/s
diameter of water jet = 25 mm
Cal. the velocity of the water jet as function height by applying Bernoulli's eqtn of water surface to the jet
[tex]\frac{P}{\rho } +\frac{V^{2}}{2} + gz[/tex] = Constant
[tex]\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g[/tex]
[tex]V=\sqrt{V_{0}^{2}-2gh}[/tex]
Relation between [tex]V_{0}[/tex] & V
[tex]m=\rho V_{0}A_{0}[/tex]
[tex]\rho VA=\rho V_{0}A_{0}[/tex]
[tex]VA= V_{0}A_{0}[/tex]
Momentum
[tex]F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA[/tex]
[tex]-mg = w_{1}[-\rho VA] +w_{2}[\rho VA][/tex]
[tex]w_{1}[/tex] = V
[tex]w_{2}[/tex] = 0
[tex]mg = \rho V^{2}A[/tex]
[tex]mg = \rho VV_{0} A[/tex]
[tex]mg = \rho VV_{0} A_{0}[/tex]
[tex]mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }[/tex]
Solving for h
[tex]h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}][/tex]
g is gravitational acc.
[tex]= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ][/tex]
[tex]= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}][/tex]
[tex]= \frac{100-16.0078}{19.62}[/tex]
h = 4.281 m
h of disk on which it remains stationary.
