problem35_38

problem35_38 - m1 818(6.06 10 7 m) = = 2.48 10 4 m. 2 2 2 m...

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35.38: a) For Jan, the total shift was m. 10 48 . 2 2 m) 10 06 . 6 ( 818 2 1 λ 4 7 1 - - × = × = = m x For Linda, the total shift was m. 10 05 . 2 2 m) 10 02 . 5 ( 818 2 λ 4 7 2 2 - - × = × = = m x b) The net displacement of the mirror is the difference of the above values:
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This note was uploaded on 01/31/2009 for the course PHYS 221 taught by Professor Hu during the Spring '08 term at Texas A&M.

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