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If you have an open tank filled with water, and you poke a hole half way down the tank, you can solve for things like the speed of the water coming out of the hole with the Bernoulli's equation, $P_1 + \rho gh + \frac{1}{2}\rho v_1^2=P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2$

Why do these pressures cancel out?

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If the tank is open, the surface of the liquid inside is exposed to atmospheric pressure. Same for the liquid flowing out the hole in the side.

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