p41_056 - δd es = 15 cm Then since d em ∝ t δt/t = δd...

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56. (a) The distance from the Earth to the Moon is d em =3 . 82 × 10 8 m (see Appendix C). Thus, the time required is given by t = 2 d em c = 2(3 . 82 × 10 8 m) 2 . 998 × 10 8 m / s =2 . 55 s . (b) We denote the uncertainty in time measurement as δt and let 2
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Unformatted text preview: δd es = 15 cm. Then, since d em ∝ t , δt/t = δd em /d em . We solve for δt : δt = tδd em d em = (2 . 55 s)(0 . 15 m) 2(3 . 82 × 10 8 m) = 5 . × 10 − 10 s ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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