[tex]dane:\\p = 20\frac{N}{m^{2}} = \frac{20 \ N}{0,0001 \ m^{2}} = 200 \ 000 \ \frac{N}{m^{2}}\\F = 600 \ N\\szukane:\\S = ?\\\\Rozwiazanie\\\\p = \frac{F}{S} \ \ /\cdot S\\\\p\cdot S = F \ \ /:p\\\\S = \frac{F}{p}\\\\S = \frac{600 \ N}{200 \ 000 \ \frac{N}{m^{2}}}\\\\S = 0,003 \ m^{2} = 30 \ cm^{2}[/tex]
Odp. Powierzchnia tłoka w silniku S = 0,003 m².
