You could use Torricelli, but you would have to add the 3 atm as additional water at the top.
Better is Bernoulli's law:
v^2/2+gz+p/rho = constant
This is true everywhere. At the top, v=0 (height drops very slowly) z is zero, so
4atm/density= constant
At the bottom
v^2/2+g*11+1atm/rho= constant
They are the same constant, so set them equal
v^2/2+ 11g+1atm/density=4atm/density
change the atm to standard units, Pascals, and solve for v in m/s . Use for g 9N/kg, and density of water in kg/m^2
A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom though a small hole. Calculate the speed with which the water comes out of the tank.
I believe this has something to do with Torricelli's theorem, but I am unsure what to do because the tank is closed with pressure?
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