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Percent Length Contraction (check solution) 
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#1
Feb109, 11:59 PM

P: 50

Question:
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 Plugnchug 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%? 


#2
Feb209, 12:13 AM

HW Helper
P: 5,341

I'd recheck your calculation and be careful in taking your square root.
680/300,000 is what you were intending I trust? 


#3
Feb209, 12:18 AM

P: 50

Or is the percent contraction supposed to be very small since the aircraft, compared to the speed of light, is going extremely slow? 


#4
Feb209, 12:27 AM

HW Helper
P: 5,341

Percent Length Contraction (check solution)
Yes it is a small number.
And when you take the square root it gets closer to 1. Use a calculator, and don't approximate or round until you have an expression for the percentage. 


#5
Feb209, 12:40 AM

P: 181




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