[tex]dane:\\F_1=500 \ N\\S_1 = 0,4 \ m^{2}\\F_2 = 2000 \ N\\szukane:\\S_2 = ?\\\\Rozwiazanie\\\\\frac{F_1}{S_1} = \frac{F_2}{S_2}\\\\F_1\cdot S_2 = F_2\cdot S_1 \ \ /:F_1\\\\S_2 = \frac{F_2}{F_1}\cdot S_1\\\\S_2 = \frac{2000 \ N}{500 \ N}\cdot0,4 \ m^{2} = 4\cdot0,4 \ m^{2} = 1,6 \ m^{2}[/tex]
Odp. Powierzchnia tłoka S₂ = 1,6 m².
