[tex]dane:\\V = 1 \ m^{3}\\d = 920\frac{kg}{m^{3}}\\g = 10\frac{N}{kg}\\szukane:\\F_{w} = ?\\\\Rozwiazanie\\\\F_{w} = d\cdot g\cdot V\\\\F_{w} = 920\frac{kg}{m^{3}}\cdot10\frac{N}{kg}\cdot1 \ m^{3}}\\\\F_{w} = 9 \ 200 \ N =9,2 \ kN\\\\(1 \ kN = 1000 \ N)[/tex]
Odp. Siła wyporu ma wartość 9,2 kN.
