problem40_15 - 40.15: E1 L 0.625E 0.625 2 2 ; E1 2mL2 2.00...

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40.15: J 10 3.20 eV 00 . 2 ; 2 625 . 0 625 . 0 19 1 2 2 2 1 - × = = = = E mL π E E m m 10 43 . 3 J) 10 kg)(3.20 10 109 . 9 ( 2 625 . 0 10 2 / 1 19 31 - - - × =
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