Odpowiedź :
[tex]Dane:\\m = 1 \ kg\\r = 20 \ cm = 0,2 \ m\\T = 2 \ s\\Szukane:\\F_{d} = ?\\\\Rozwiazanie\\\\F_{d} = \frac{4\pi ^{2}mr}{T^{2}}\\\\F_{d} = \frac{4\cdot3,14^{2}\cdot1 \ kg\cdot0,2 \ m}{(0,2 \ s)^{2}}\\\\\boxed{F_{d}\approx1,97 \ N}[/tex]