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

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The discussion focuses on calculating the water flow rate from a tank with a 0.6m depth and a 6m² hole at the bottom. The initial approach suggests using the formula Q= pi*r^4(p1-p2)/8L*viscosity, but viscosity is deemed unnecessary for this scenario. Instead, applying Bernoulli's principle is recommended, simplifying the calculation by ignoring the velocity term at the water surface due to the tank's large area. The key takeaway is to focus on the hydrostatic pressure to determine the flow rate without complicating factors like viscosity. The conversation emphasizes the importance of using appropriate fluid dynamics principles for accurate calculations.
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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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