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Water is flowing a speed of $\;1.5 ms^{-1} \;$ through a horizontal tube cross - sectional area $\;10^{-2}m^{2}\;$ and you are trying to stop the flow by your palm . Assuming that the water stops immediately after hitting the palm , the minimum force that you must exert should be (density of water =$\; 10^{3}\;kgm^{-3}\;$).

$(a)\;15\;N\qquad(b)\;22.5\;N\qquad(c)\;33.7\;N\qquad(d)\;45\;N$

1 Answer

Since the water attains zero velocity immediately after impact with palm , the force acting on it will be (Density)(Area) [(velocity)^2] because this is actually the momentum of the water flowing per second, which is actually force.

I was able to write this because AreaxVelocity is Quantity of Water flowing per second. Multiply this by Density , you get Mass of water flowing per second. Multiply this by velocity and you get momentum flowing per second. You can check that the dimensions of this formula have the same dimensional formula as that of force.

Now putting the values, we obtain the answer b.) 22.5 N
answered Dec 28, 2016 by riteshkartik
 

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