[tex]dane:\\V = 1000 \ dm^{3} = 1 \ m^{3}\\d = 1000\frac{kg}{m^{3}}\\h = 30 \ m\\t = 1 \ min = 60 \ s\\g = 10\frac{m}{s^{2}}\\szukane:\\P = ?[/tex]
Rozwiązanie
Masa wody m:
[tex]d = \frac{m}{V} \ \ /\cdot V\\\\m = V\cdot d\\\\m = 1 \ m^{3}\cdot1000\frac{kg}{m^{3}}\\\\m = 1000 \ kg[/tex]
Moc silnika P:
[tex]P = \frac{W}{t}\\\\W = E_{p} = mgh\\\\P = \frac{mgh}{t}\\\\P = \frac{1000 \ kg\cdot10\frac{m}{s^{2}}\cdot30 \ m}{60 \ s}\\\\P = 5 \ 000 \ W = 5 \ kW\\\\(1 \ kW = 1000 \ W)[/tex]
Odp. Moc silnika pompy strażackiej ma wartość 5 kW.
