Answer:
[tex]F = 1.114\,N[/tex]
Explanation:
A common stone may have densities between [tex]1600\,\frac{kg}{m^{3}}[/tex] and [tex]3300\,\frac{kg}{m^{3}}[/tex]. Let consider a rock with a density of [tex]2200\,\frac{kg}{m^{2}}[/tex]. The buoyant force acting on the stone has the following formula:
[tex]F = \frac{\rho_{w}}{\rho_{s}}\cdot m\cdot g[/tex]
[tex]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)[/tex]
[tex]F = 1.114\,N[/tex]
