[tex]dane:\\h = 80 \ cm = 0,8 \ m\\p_{a} = 98 \ kPa\\p_{c} = 109 \ kPa\\g = 9,8\frac{m}{s^{2}} = 9,8\frac{N}{kg}\\szukane:\\d_{c} = ?\\\\Rozwiazanie\\\\p_{c} = p_{a}+p_{h}\\\\p_{h} = p_{c} - p_{a} = 109 \ kPa -98 \ kPa = 11 \ kPa = 11 \ 000 \ Pa = 11 \ 000 \ \frac{N}{m^{2}}\\\\\\p_{h} = d_{c}\cdot g\cdot h \ \ /:(g\cdot h)\\\\d_{c} = \frac{p_{h}}{g\cdot h}\\\\d_{c} = \frac{11000\frac{N}{m^{2}}}{9,8\frac{N}{kg}\cdot0,8 \ m}\\\\d_{c} \approx 1400\frac{kg}{m^{3}}\approx1,4\frac{g}{cm^{3}}[/tex]
Odp. Gęstość cieczy wynosi 1,4 g/cm³.
