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 ) ]

D. Cars have the same velocity at one instant of time between dots 4 and 5.

Explanation:

The two cars will eventually have the same velocity and it occurs between 4 and 5. Velocity is normally defined as displacement over time. This shows that a change in the value of either time or displacement will affect the value of the velocity. Thus, at a given time between the points 4 and 5, the two cars have the same value of velocity.