A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate Q through the hole. Do not leave in terms of velocities, only leave in terms of r, R and d.

Explanation:

v₁ , v₂ be velocity of flow of water at the surface and hole , h₁ and h₂ be height of surface and hole  , P₁ and P₂ be pressure at surface and hole .

Using Bernoulli's formula for flow of liquid in tank

1/2 ρ v₁² + ρgh₁ + P₁ = 1/2 ρ v₂² + ρgh₂ + P₂

P₁ = P₂

1/2 ρ v₁² + ρgh₁  = 1/2 ρ v₂² + ρgh₂

1/2  ( v₁² -  v₂² ) =  g(h₂ -h₁ )

v₁² -  v₂² =  - 2gd

a₁ v₁ = a₂ v₂

π R² v₁ = π r² v₂

v₁ =  r² v₂ / R²

v₁² -  v₂² =  -2gd

r⁴ v₂² / R⁴ - v₂² = -2gd

v₂² ( R⁴ -  r⁴ ) = 2gd R⁴

v₂² = 2gd R⁴ / ( R⁴ -  r⁴ )

v₂ = √ [ 2gd R⁴ / ( R⁴ -  r⁴ ) ]

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

whichever one it is, it's definitely not one that you included on the list you provided.