[tex]Dane:\\m = 10 \ t = 10 \ 000 \ kg\\v = 200\frac{m}{s}\\F_{d} = F_{g}\\g = 10\frac{m}{s^{2}} = 10\frac{N}{kg}\\Szukane:\\R = ?[/tex]
Rozwiązanie
Korzystamy ze wzoru na siłę dośrodkową:
[tex]F_{d} = \frac{mv^{2}}{R} \ \ /\cdot R\\\\F_{d} \cdot R = mv^{2} \ \ /:F_{d}\\\\R = \frac{mv^{2}}{F_{d}}\\\\F_{d} = F_{g} = m\cdot g = 10000 \ kg\cdot10\frac{N}{kg} = 100 \ 000 \ N\\\\R = \frac{10000 \ kg\cdot(20\frac{m}{s})^{2}}{100000 \ N}=\frac{10000 \ kg\cdot40000\frac{m^{2}}{s^{2}}}{100000 \ kg\cdot\frac{m}{s^{2}}}=0,1\cdot40000 \ m\\\\\boxed{R = 4 \ 000 \ m}[/tex]
Odp. Promień tego zakrętu R = 4 000 m.
