[tex]Dane:\\V = 500 \ cm^{3} = 0,0005 \ m^{3}\\d_{w} = 1000\frac{kg}{m^{3}}\\g = 10\frac{m}{s^{2}}+10\frac{N}{kg}\\Szukane:\\F_{w} = ?\\\\\\F_{w} = d_{cieczy}\cdot g\cdot V_{zanurzonego \ ciala}\\\\F_{w} = 1000\frac{kg}{m^{3}}\cdot 10\frac{N}{kg}\cdot0,0005 \ m^{3} \\\\\boxed{F_{w} = 5 \ N}[/tex]
Odp. Wartość siły wyporu Fw = 5 N.
