Unformatted text preview: √ (0.36) =1/0.6 =5/3 = 1.67 So t = 1.67 × 9x1011 s = 1.5x1010 s b. In that time, traveling at 0.8c, the Kaon will cover 0.8 × (3.0x10 8 m/s) × (1.5x1010 s) = 3.6x102 m = 3.6 cm. c. Velocity addition with u ′ = 0.75c, the pion's velocity in the Kaon rest frame, and V=0.8c, the velocity of the Kaon rest frame relative to the lab. d. E=m c 2 = 8.85x1028 Kg × (3.0x10 8 m/s) 2 = 7.97x1011 J or 7.97x1011 J / (1.60x1019 J/eV) = 4.98x10 8 eV = 498 MeV. u = ′ u +V 1 + ′ u V c 2 = 0.75c +0.8c 1 +0.75 × 0.8 = 1.55c 1.60 = 0.97c...
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 Spring '06
 GaryR.Goldstein
 Physics, Acceleration, General Relativity, Special Relativity, Velocity, Kaon

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