EGM4313HW4

# EGM4313HW4 - Page 567, problem 5: B1 1.04 cm 2 10.6 0.056 s...

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Page 567, problem 5: 2 22 2 1 1 0.01752 1.04 1.752 10.6 0.056 1 0 ( , ) sin 0.1 n t s cm cm c ss B B otherwise u x t x e Page 567, problem 7: 10 0 10 2 3 3 2 2 3 3 0 3 3 3 3 1 3 5 7 3 3 3 3 3 3 3 2 2 10 sin 10 10 1 10 100 2000 1000 200 cos sin 5 10 10 400 400 1 cos 1 1 800 800 800 800 , , , , 3 5 7 0, n n nx B x x dx n x x x n x x n n n n n nn B B B B BB 2 2 2 2 2 46 0.01752 0.01752 3 0.01752 5 3 3 3 0, 0, 800 1 3 1 5 , sin sin sin 10 3 10 5 10 ttt B x x x u x t e e e Page 567, problem 10:

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Results computing the temperature at the midpoint for various times using 6 terms in the series solution and Excel: time = 1 # terms Value 0 50 1 103.553 2 103.553 3 99.07698 4 99.07698 5 99.24583 6 99.24583 The temperature is approximately 99.25 at the midpoint at t = 1. time = 2 # terms Value 0 50 1 95.04926 2 95.04926 3 94.10514 4 94.10514 5 94.10738 6 94.10738

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## This note was uploaded on 12/15/2011 for the course EGM 4313 taught by Professor Mei during the Fall '08 term at University of Florida.

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EGM4313HW4 - Page 567, problem 5: B1 1.04 cm 2 10.6 0.056 s...

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