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Question-193345




Question Number 193345 by yaslm last updated on 11/Jun/23
Answered by aleks041103 last updated on 13/Jun/23
The trajectry of the water is a freefall parabolla  ⇒H=(1/2)gt^2 ⇒t=(√((2H)/g))  D=vt=v(√((2H)/g))  ⇒v=(√((gD^2 )/(2H)))  then the flow rate Φ is  Φ=vS=(1/4)πd^2 v  ⇒Φ=((πd^2 D)/4)(√(g/(2H)))
$${The}\:{trajectry}\:{of}\:{the}\:{water}\:{is}\:{a}\:{freefall}\:{parabolla} \\ $$$$\Rightarrow{H}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \Rightarrow{t}=\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$${D}={vt}={v}\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gD}^{\mathrm{2}} }{\mathrm{2}{H}}} \\ $$$${then}\:{the}\:{flow}\:{rate}\:\Phi\:{is} \\ $$$$\Phi={vS}=\frac{\mathrm{1}}{\mathrm{4}}\pi{d}^{\mathrm{2}} {v} \\ $$$$\Rightarrow\Phi=\frac{\pi{d}^{\mathrm{2}} {D}}{\mathrm{4}}\sqrt{\frac{{g}}{\mathrm{2}{H}}} \\ $$

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