HW_5_solution_2008

HW_5_solution_2008 - EGM 5533 HW#5 – Solution Page 1 of 9...

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Unformatted text preview: EGM 5533 HW#5 – Solution Page 1 of 9 EGM 5533 HW#5 – Solution Page 2 of 9 EGM 5533 HW#5 – Solution 2πR3t 1.005 X 10‐7 m4 1.005 X 10‐7) 2714.33 π 4[ (0.042 )]2 4A 4 A22 4 = dl π (0.04) π (0.04) dl ∫∫ tt { 2 } { 2 } + 2 4 ( ) ( ) 1000 1000 ‐7 4 b = 1.34 X 10 m J Page 3 of 9 EGM 5533 HW#5 – Solution 1.34 X 10‐7 3618 1.33 Page 4 of 9 EGM 5533 HW#5 – Solution 33 % Page 5 of 9 EGM 5533 HW#5 – Solution Page 6 of 9 EGM 5533 HW#5 – Solution Page 7 of 9 EGM 5533 HW#5 – Solution Page 8 of 9 EGM 5533 HW#5 – Solution Page 9 of 9 ...
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This note was uploaded on 01/03/2012 for the course EGN 5533 taught by Professor Staff during the Spring '08 term at University of Florida.

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