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What is the percent length contraction of an aircraft traveling at Mach 2?

So, we know that Mach 2= 680.58 m/s

and that L'=L[tex]\sqrt{1-(v/c)^2}[/tex]

If you divide over the L to get:

L'/L=[tex]\sqrt{1-(v/c)^2}[/tex]=% length contraction

Plug-n-chug from here to get:

L'/L=[tex]\sqrt{1-(680.58/c)^2}[/tex]

=[tex]\sqrt{1-(5.15X10^-12)}[/tex]

=[tex]\sqrt{1}[/tex]

=1

Is this correct? It only contracts 1%?

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# Homework Help: Percent Length Contraction (check solution)

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