1.
[tex]dane:\\F_2 = 1 \ 200 \ N\\S_2 = 4 \ m^{2}\\F_1 = 400 \ N\\szukane:\\S_1 = ?\\\\Rozwiazanie\\\\\frac{F_1}{S_1} = \frac{F_2}{S_2}\\\\S_1\cdot F_2 = S_2\cdot F_1 \ \ /:F_2\\\\S_1 = \frac{F_1}{F_2}\cdot S_2\\\\S_1 = \frac{400 \ N}{1200 \ N}\cdot4 \ m^{2}=\frac{1}{3}\cdot4 \ m^{2}\\\\S_1 \approx 1,33 \ m^{2}[/tex]
Odp. Powierzchnia małego tłoka S₁ = 4/3 m² ≈ 1,33 m².
2.
[tex]dane:\\F = 4 \ kN = 4 \ 000 \ N\\S = 5 \ m^{2}''szukane:\\p = ?\\\\Rozwiazanie\\\\p = \frac{F}{S}\\\\p = \frac{4000 \ N}{5 \ m^{2}} = 800 \ Pa[/tex]
Odp. Szukane ciśnienie p = 800 Pa.
