[tex]Dane:\\m = 0,5 \ kg\\r = 25 \ cm = 0,25 \ m\\v = 5\frac{m}{s}\\Szukane:\\F_{d} = ?\\\\Rozwiazanie\\\\F_{d} = \frac{mv^{2}}{r}\\\\F_{d} = \frac{0,5 \ kg\cdot(5\frac{m}{s})^{2}}{0,25 \ m}\\\\F_{d} = \frac{12,5 \ kg\cdot\frac{m^{2}}{s^{2}}}{0,25 \ m}\\\\\boxed{F_{d} = 50 \ N}[/tex]