1.
[tex]dane:\\\Delta I = 2 \ A\\L = 0,08 \ H\\\epsilon = 4 \ V\\szukane:\\t = ?\\\\Rozwiazanie\\\\\epsilon = \frac{L\cdot \Delta I}{t} \ \ \rightarrow \ \ t = \frac{L\cdot\Delta I}{\epsilon}\\\\t = \frac{0,08\frac{V\cdot s}{A}\cdot2 \ A}{4 \ V}\\\\\underline{t = 0,04 \ s}[/tex]
2.
[tex]dane:\\x_1 = 1 \ cm = 0,01 \ m\\W_1 = E_{ps_1} = 2 \ J\\x_2 = 3 \ cm\\szukane:\\W_2 = ?[/tex]
Rozwiązanie
Praca to zmiana energii (w tym przypadku energii sprężystości)
[tex]W = \Delta E = \frac{k\Delta x}{2}\\\\W_1 = \frac{k\Delta x_1^{2}}{2} = 2 \ J\\\\W_2 = \frac{k(3\Delta x_1)^{2}}{2} = 9W_1\\\\W_2 = 9\cdot2 \ J\\\\\underline{W_2 = 18 \ J}[/tex]
