[tex]Dane:\\m = 30 \ kg\\T = 6 \ s\\r = 5 \ m\\Szukane:\\a_{d} = ?\\F_{d} = ?\\\\Rozwiazanie\\\\a_{d} = \frac{4\pi^{2} r}{T^{2}}\\\\a_{d} = \frac{4\cdot3,14^{2}\cdot5 \ m}{(6 \ s)^{2}}\\\\\boxed{a_{d} \approx5,5\frac{m}{s^{2}}}[/tex]
[tex]F_{d} = \frac{4\pi^{2} mr}{T^{2}}\\\\F_{d} = \frac{4\cdot3,14^{2}\cdot30 \ kg\cdot5 \ m}{(6 \ s)^{2}}\\\\\boxed{F_{d} \approx164,3 \ N}[/tex]
