dane: t = 1 s V = 1000 m³ = 10³ m³ d = 1 g/cm³ = 1000 kg/m³ = 10³ kg/m³ h = 50 m = 5*10 m g = 10 m/s² = 10 N/kg szukane: P = ? Rozwiązanie P = W/t W = Ep = m·g·h ale m = V·d = 10³m³·10³kg/m³ = 10⁶ kg W = 10⁶kg·10N/kg·5*10m = 5*10⁸ J = 500*10⁶ J 1 MJ = 10⁶ J W = 500 MJ P = 500MJ/1s P = 500 MW ========= Odp. Szukana moc wodospadu wynosi 500 MW (megawatów).
[latex]P=frac{Delta W}{Delta t}=frac{F cdot s}{t}=frac{m cdot g cdot s}{t}=frac{ ho cdot v cdot g cdot s}{t} \ ho=frac{1g}{cm^3}=frac{1g}{10^{-6}m^3}=frac{10^6 g}{m^3}=frac{10^3 kg}{m^3} \ P=frac{ ho cdot v cdot g cdot s}{t}=frac{10^3 frac{kg} {m^3} cdot 1000 m^3 cdot 50m cdot 10 frac{m}{s^2}}{1s}=5 cdot 10^8 W[/latex]
