# physics

An open tank holds water 1.25 m deep. If a small hole of cross section area 3cm^2 is made at the bottom of the tank, calculate the mass of water per second initially flowing out of the hole. (g=10m/s^2, density of water= 1000 kgm^-3)

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1. Sice you are still showing no work, my advice will be brief.

Use the Bernoulli equation (with altitude change)to get the velocity and the continuity equation to get the mass flow rate.

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2. Don't know how to solve it!pleas help me out

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3. Mass of water flowing out per second = mv

v is the velocity of the water coming out of the hole.

Since 'm' = V¶
Where 'V' is the volume of hole
'¶' is the density of water flowing out.

Therefore,mass of water flowing out per second= 0.03*1000*√2*10*1.25

=150kg/sec

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