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Svar 2.2:6

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(Ny sida: e_F=\frac{'''F'''}{F}=\frac{(6,10)N}{\sqrt{6^2+10^2}N}=p136(6;10)=11;66(6;10) eller eF=ÀFF)
Rad 1: Rad 1:
-
e_F=\frac{'''F'''}{F}=\frac{(6,10)N}{\sqrt{6^2+10^2}N}=p136(6;10)=11;66(6;10)
+
<math>e_F=\frac{'''F'''}{F}=\frac{(6,10)N}{\sqrt{6^2+10^2}N}=\frac{(6,10)}{\sqrt{136}}=\frac{(6,10)}{11,66}</math>
eller
eller
-
eF=ÀFF
+
<math>e_F=-\frac{'''F'''}{F}</math>

Versionen från 15 december 2009 kl. 15.43

eF=FF=(610)N62+102N=136(610)=1166(610) eller

eF=FF