Question

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

Answer 1

Answer:

$F = 1.114\,N$

Explanation:

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

$F = \frac{\rho_{w}}{\rho_{s}}\cdot m\cdot g$

$F = \frac{1000\,\frac{kg}{m^{3}} }{2200\,\frac{kg}{m^{3}} } \cdot (0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1.114\,N$