2.
[tex]\rho = 0,5\frac{g}{cm^{3}} = 0,5\cdot\frac{0,001 \ kg}{0,000001 \ m^{3}} = 0,5\cdot1000\frac{kg}{m^{3}} = 500\frac{kg}{m^{3}}[/tex]
4.
[tex]dane:\\m = 60 \ kg\\S = 200 \ cm^{2}} = 0,02 \ m^{2}\\g = 10\frac{N}{kg}\\szukane:\\p = ?\\\\Rozwiazanie\\\\p = \frac{F}{S}\\\\F = m\cdot g = 60 \ kg\cdot10\frac{N}{kg} = 600 \ N\\\\p = \frac{600 \ N}{0,02 \ m^{2}}\\\\p = 30 \ 000 \ Pa = 30 \ kPa\\\\(1 \ kPa = 1000 \ Pa)[/tex]
Odp. Ten człowiek wywiera ciśnienie p = 30 kPa.
