Question

Calculate the buoyant force acting on it if a stone of mass 250g is thrown in water?​

Answer 1

`F = 1.114\,N`

Step-by-step explanation:

A common stone may have densities between
`1600\,(kg)/(m^(3))` and
`3300\,(kg)/(m^(3))`. Let consider a rock with a density of
`2200\,(kg)/(m^(2))`. The buoyant force acting on the stone has the following formula:

`F = (\rho_(w))/(\rho_(s))\cdot m\cdot g`

`F = (1000\,(kg)/(m^(3)) )/(2200\,(kg)/(m^(3)) ) \cdot (0.25\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)`

`F = 1.114\,N`