[tex]dane:\\V = 5 \ cm\times 15 \ cm\times 3 \ cm = 225 \ cm^{3} = 0,000225 \ m^{3}\\d = 1300\frac{kg}{m^{3}} \ - \ gestosc \ gliceryny\\g = 10\frac{N}{kg}\\szukane:\\F_{w} = ?\\\\Rozwiazanie\\\\F_{w} = d\cdot g\cdot V\\\\F_{w} = 1300\frac{kg}{m^{3}}\cdot10\frac{N}{kg}\cdot0,000225 \ m^{3}\\\\F_{w} = 2,925 \ N[/tex]
