What is the rate at which water flows out the tank in m^3/s?

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SUMMARY

The rate at which water flows out of a tank with a 6m² cross-section hole at the bottom, filled to a depth of 0.6m, can be calculated using Bernoulli's principle. The pressure at the hole is determined by the hydrostatic pressure formula, P = ρgh, resulting in a pressure of 5880 N/m². The flow rate can be derived from the area of the hole and the velocity of the water exiting, assuming laminar flow and ignoring viscosity for simplification.

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A tank of large area is filled with water to a depth of 0.6m. A hole of 6m^2 cross section in the bottom allows water to drain out in a continuous stream. What is the rate at which water flows out the tank in m^3/s?
This is how I figure the problem should be solved:
Q= pi*r^4(p1-p2)/8L*viscosity

Area= 6e-4m^2
h= 0.6m
P= rho*g*h= 1.00e3kg/m^3*9.8m/s^2*0.6m= 5880N/m^2
what I can't figure out is how do I get viscosity from the area
 
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This kind of question probably belongs in the Homework Help section. Anyways...

Forget viscosity. (Unless they want you to include it; in which case: Lot's of luck!) Assume laminar flow. Apply Bernoulli's principle. Hint: Since the tank has a "large area", ignore the 1/2ρV2 term at the surface.
 

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