Svar 2.2:7
FörberedandeFysik
(Skillnad mellan versioner)
(Ny sida: '''F'''<math>=Fe_F=120N\frac{(2,5)m-(0,3)m}{\sqrt{(2-0)^2+(5-3)^2}m}=120\frac{(2,2)}{\sqrt{8}}N=120\frac{(1,1)}{\sqrt{2}}N </math>) |
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| Rad 1: | Rad 1: | ||
| - | + | <math> \mathbf{F} = F\mathbf{e}_F = 120\, \textrm{N}\,\ \frac{(2,5)\,\textrm{m}-(0,3)\,\textrm{m}}{\sqrt{(2-0)^2+(5-3)^2}\,\textrm{m}} = 120\frac{(2,2)}{\sqrt{8}}\,\textrm{N} = 120\frac{(1,1)}{\sqrt{2}}\, \textrm{N} </math> | |
Nuvarande version
\displaystyle \mathbf{F} = F\mathbf{e}_F = 120\, \textrm{N}\,\ \frac{(2,5)\,\textrm{m}-(0,3)\,\textrm{m}}{\sqrt{(2-0)^2+(5-3)^2}\,\textrm{m}} = 120\frac{(2,2)}{\sqrt{8}}\,\textrm{N} = 120\frac{(1,1)}{\sqrt{2}}\, \textrm{N}
