A container of liquid of density ρ has a surface area A.?
A container of liquid of density ρ has a surface area A exposed to normal atmospheric pressure. A distance h below the surface of the liquid is a tiny hole of area A0. The depth of the liquid in the container is H.
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If A0 ≪ A, which expression correctly gives the velocity v0 (in terms of only g, h, H, A0 and A) at which the fluid escapes through the hole of area A0?
I thought that it was sqrt(2gh) but that turned out to be wrong.
Update:that is the same answer that I got but it was wrong. This is part of the problem that I didn't see before i'm guessing that it helps come up with the right answer but I don't know how to incorporate it into what we did.
1/((1 − x)^n) ≈ 1 + n x if x ≪ 1.
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Bernoulli principle :
P1 + 1/2 ρ * v1^2 + ρ * g * h1 = P2 + 1/2 ρ * v2^2 + ρ * g * h2
---> P1 = P2
A0 ≪ A -----> v1 ≈ 0
g * H = 1/2 * v2^2 + g * (H - h)
2 * g * H = v2^2 + 2 * g * H - 2 * g * h
---> v2^2 = 2 * g * h
---> v2 = √(2gh)