Odpowiedź :
[tex]F_{wyp} = 45 \ N\\m = 15 \ kg\\a = ?\\\\Z \ II \ zasady \ dynamiki\\\\a = \frac{F_{wyp}}{m}\\\\a = \frac{45 \ N}{15 \ kg} = 3\frac{N}{kg} = 3\frac{m}{s^{2}}[/tex]
[tex]F_{wyp} = 45 \ N\\m = 15 \ kg\\a = ?\\\\Z \ II \ zasady \ dynamiki\\\\a = \frac{F_{wyp}}{m}\\\\a = \frac{45 \ N}{15 \ kg} = 3\frac{N}{kg} = 3\frac{m}{s^{2}}[/tex]