[tex]dane:\\F = 0,45 \ N\\F_{z} = 0,25 \ N\\d = 1\frac{g}{cm^{3}} = 1000\frac{kg}{m^{3}}\\g = 10\frac{N}{kg}\\szukane:\\V = ?\\\\F_{w} = F - F_{z} = 0,45 \ N-0,25 \ N = 0,2 \ N\\\\F_{w} = d\cdot g\cdot V \ \ /:(d\cdot g)\\\\V = \frac{F_{w}}{d\cdot g}\\\\V = \frac{0,2 \ N}{1000\frac{kg}{m^{3}}\cdot10\frac{N}{kg}} = \frac{0,2 \ N}{10000\frac{N}{m^{3}}}\\\\V = 0,00002 \ m^{3} = 20 \ cm^{3}[/tex]
Odp. Objętość drutu wynosi 20 cm³.
