[tex]dane:\\F_1 = 300 \ kN = 300 \ 000 \ N\\S_2 = 4 \ cm^{2}\\F_2 = 400 \ N\\szukane:\\S_1 = ?\\\\Rozwiazanie\\\\Z \ prawa \ Pascala\\\\\frac{F_1}{S_1} = \frac{F_2}{S_2}\\\\F_2\cdot S_1=F_1\cdot S_2 \ \ /:F_2\\\\S_1 = \frac{F_1}{F_2}\cdot S_2\\\\S_1 = \frac{300 \ 000 \ N}{400 \ N}\cdot4 \ cm^{2}}\\\\S_1 = 3 \ 000 \ cm^{2}[/tex]
Odp. Powierzchnia dużego tłoka S₁ = 3 000 cm².