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Unformatted text preview: 998 10 8 m / s)(0 . 050) (0 . 0193 m / s ly)(1 + 0 . 050) = 7 . 4 10 8 ly . (g) rom the result of part (a), t = r c r = (7 . 4 10 8 ly)(9 . 46 10 15 m / ly) 2 . 998 10 8 m / s (0 . 0193 m / s ly)(7 . 4 10 8 ly) = 2 . 5 10 16 s , which is equivalent to 7 . 8 10 8 y. (h) Letting r = c t , we solve for t : t = r c = 7 . 4 10 8 ly c = 7 . 4 10 8 y . (i) The distance is given by r = c t = c (7 . 8 10 8 y) = 7 . 8 10 8 ly . (j) rom the result of part (f), r B = c ( / ) (1 + / ) = (2 . 998 10 8 m / s)(0 . 080) (0 . 0193 mm / s ly)(1 + 0 . 080) = 1 . 15 10 9 ly ....
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 Fall '08
 SPRUNGER
 Physics, Light

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