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Unformatted text preview: . 01 sec to make the trip. At what speed was the train traveling? Answer: The passenger sees a length contracted distance of L = L/γ = 10 6 m/γ . Combine with his observation of trip time to conclude that . 01 sec = L v = L γv = 10 6 γv (5) J. Gunion 9D, Spring Quarter 2 or 10 6 m = 0 . 01 sec × γv = 0 . 01 sec × v q 1v 2 c 2 . (6) Dividing out the . 01 and squaring, we get 10 16 m 2 /s 2 = v 2 1v 2 c 2 (7) or 10 1610 16 v 2 9 × 10 16 = v 2 (8) or 10 16 = v 2 (1 + 1 9 ) = v 2 10 9 (9) or v = 10 8 r 9 10 m/s = c √ 10 . (10) J. Gunion 9D, Spring Quarter 3...
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This note was uploaded on 04/11/2009 for the course ENG 100 taught by Professor Delwiche during the Spring '08 term at UC Davis.
 Spring '08
 Delwiche

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