[tex]dane:\\m = 1 \ 200 \ kg\\r = 300 \ m\\v = 30\frac{m}{s}\\szukane:\\F_{d} = ?\\\\Rozwiazanie\\\\F_{d} = \frac{mv^{2}}{r}\\\\F_{d} = \frac{1200 \ kg\cdot(30\frac{m}{s})^{2}}{300 \ m}\\\\\boxed{F_{d} = 3 \ 600 \ N = 3,6 \ kN}\\\\(1 \ kN = 1000 \ N)[/tex]
