An open tank holds water 1.25m 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= 1000kg/m^3)

Answer 1

Wouldn't Bernoulli's equation reduced to

rho*g(h)=1/2 rho v^2
h=1.25, rho divides out, but it is given, g is given.

V^2=gh=1.25*10
V= 3.53m/s

massrate=V*rho*area=3.53m/s*1000kg/m^2*.03^2
= 1.06kg/sec