[tex]dane:\\m = 24 \ kg\\r = 4 \ m\\T = 5 \ s\\szukane:\\a) \ F_{d} = ?\\b) \ a = ?\\\\Rozwiazanie\\\\a)\\v = \frac{2\pi r}{T} = \frac{2\cdot3,14\cdot4 \ m}{5 \ s}= 5,024\frac{m}{s}\\\\F_{d} = \frac{mv^{2}}{2}\\\\F_{d} = \frac{24 \ kg\cdot(5,024\frac{m}{s})^{2}}{5 \ s} = 151,4 \ N[/tex]
[tex]b)\\a = \frac{v^{2}}{r}\\\\a=\frac{(5,024\frac{m}{s})^{2}}{4 \ m} = 6,31\frac{m}{s^{2}}[/tex]
