Question

A 4.0mm-diameter hold is 1.0m below the surface of a 2.0m diameter tank of water. How fast does water squirt out of the hole? (Show Derivation)

Answer #1

**let P1 is the pressure at the top of the tank and h1 is
the height of tank.**

**let P2 is the pressure at the hole and h2 is the height
of the hole from bottom of the tank.**

**let rho is the dencity of water.**

**let velocity of water at the top of the tank, v1 =
0**

**let v2 = v, is the velocity of the water through the
hole.**

**use Bernoulli's equation,**

**P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 +
rho*g*h2**

**PA + 0 + rho*g*h1 = PA + (1/2)*rho*v^2 + rho*g*h2 (here
PA is the atmospheric pressure)**

**(1/2)*rho*v^2 = rho*g*(h1 - h2)**

**v^2 = 2*g*(h1 - h2)**

**v = sqrt(2*g*(h1 - h2))**

**= sqrt(2*9.8*(2 - 1))
= 4.43 m/s
<<<<<<<<<<--------------Answer**

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