Physics

A beam of particles have a measured life time of 1.5 x 10-8 s when traveling 2.0 x 108 m/s. What would their lifetime be if they were at rest?

My Answer:

tm= tp/√(1-(v2/c2))
(1.5*10-8) = tp/√(1-((2*108)2/(3*108)2))
(1.5*10-8) = tp/√(1-((4*1016)/(9*1016))
(1.5*10-8) = tp/√(1-(4/9))
(1.5*10-8) = tp/√(5/9)
tp =(√(5)/3)(1.5*10-8)
tp =1.118033989*10-8

The lifetime of the particles if they were at rest would be 1.1*10-8s.

Is this correct? Thanks.

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asked by Ann
  1. yes; good work!

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    posted by drwls
  2. tm= tp/√(1-(v2/c2))
    (1.5*10-8) = tp/√(1-((2*108)2/(3*108)2))
    (1.5*10-8) = tp/√(1-((4*1016)/(9*1016))
    (1.5*10-8) = tp/√(1-(4/9))
    (1.5*10-8) = tp/√(5/9)
    tp =(√(5)/3)(1.5*10-8)
    tp =1.118033989*10-8

    on the 6th line, why does it become √(5)/3)? Is it not supposed to be √(5/9) from the line before?

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    posted by mike
  3. yes this is right
    in anwser to mike, they square rooted 3 but not 5
    so it because (√(5)/3) = (√(5/9))

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    posted by pro

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