Odpowiedź :
[tex]m = 5 \ kg\\F_{w} = 20 \ N\\\\a = \frac{F_{w}}{m}\\\\a = \frac{20 \ N}{5 \ kg} = \frac{20 \ kg\cdot\frac{m}{s^{2}}}{5 \ kg} = 4\frac{m}{s^{2}}[/tex]
[tex]m = 5 \ kg\\F_{w} = 20 \ N\\\\a = \frac{F_{w}}{m}\\\\a = \frac{20 \ N}{5 \ kg} = \frac{20 \ kg\cdot\frac{m}{s^{2}}}{5 \ kg} = 4\frac{m}{s^{2}}[/tex]