Svar 2.2:6

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Nuvarande version (20 december 2017 kl. 10.46) (redigera) (ogör)
 
(4 mellanliggande versioner visas inte.)
Rad 1: Rad 1:
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<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>
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<math>\displaystyle \mathbf{e}_F = \frac{\mathbf{F}}{F} = \frac{(6,10)\,\textrm{N}}{\sqrt{6^2+10^2}\,\textrm{N}} = \frac{(6,10)}{\sqrt{136}} = \frac{(6,10)}{11,66} </math>
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eller
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<math>e_F=-\frac{'''F'''}{F}</math>
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eller
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<math>\displaystyle \mathbf{e}_F = -\frac{\mathbf{F}}{F} </math>

Nuvarande version

\displaystyle \displaystyle \mathbf{e}_F = \frac{\mathbf{F}}{F} = \frac{(6,10)\,\textrm{N}}{\sqrt{6^2+10^2}\,\textrm{N}} = \frac{(6,10)}{\sqrt{136}} = \frac{(6,10)}{11,66}

eller

\displaystyle \displaystyle \mathbf{e}_F = -\frac{\mathbf{F}}{F}