Assignment 21

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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12:15:12 **** query 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.

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RESPONSE --> y inv. sqrt(x) y = 4 x = 9 y = k / sqrt(x) 4 = k / sqrt (9) 4 = k / 3 k = 12 y = 12 / sqrt(x)

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12:15:20 ** The inverse proportionality to the square root gives us y = k / sqrt(x). y = 4 when x = 9 gives us 4 = k / sqrt(9) or 4 = k / 3 so that k = 4 * 3 = 12. The equation is therefore y = 12 / sqrt(x). **

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RESPONSE --> correct

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12:20:29 query 2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3.

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RESPONSE --> z = k(x^2 + y^2) 1 = k(2^2 + 3^2) 1 = k(4+9) 1 = k(13) 1/13 = k z = 1/3 (x^2 + y^2)

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12:21:23 ** The proportionality is z = k (x^3 + y^2). If x = 2, y = 3 and z = 1 we have 1 = k ( 2^3 + 3^2) or 17 k = 1 so that k = 1/17. The proportionality is therefore z = (x^3 + y^2) / 17. **

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RESPONSE --> According to your notes, I had the form right, but I didn't do 2^3. Instead I did 2^2.

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12:27:44 query 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)

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RESPONSE --> y = kx 5 = k * 2pi/sqrt(32) k = 5 / 1.11 k = 4.5 y = 4.5 x

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12:29:04 ** The equation is T = k sqrt(L), with k = 2 pi / sqrt(32). So we have T = 2 pi / sqrt(32) * sqrt(L). **

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RESPONSE --> I wasn't too sure how to do this one, but after seeing how you did it, I think I understand it.

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12:30:00 **** What equation relates period and length? ****

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RESPONSE --> T = 2pi / sqrt(32) * sqrt(L) it resembles M = kd^2 / sqrt (x)

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12:37:17 query 2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.

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RESPONSE --> K = k/l K = 432/4 = 108 p = kb 3 = k(1.44) k = 4.32

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12:38:46 ** We have R = k * L / D^2. Substituting we obtain 1.24 = k * 432 / 4^2 so that k = 1.24 * 4^2 / 432 = .046 approx. Thus R = .046 * L / D^2. Now if R = 1.44 and d = 3 we find L as follows: First solve the equation for L to get L = R * D^2 / (.046). Then substitute to get L = 1.44 * 3^2 / .046 = 280 approx. The wire should be about 280 ft long. **

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RESPONSE --> I was not sure how to do this one at all. I didn't know how to start it. After seeing how you set it up I think I understand.

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Be sure you are getting the proportionality problems right when you do the assigned problems--I suggest doing a number of odd-numbered problems and checking with the back of the book to make sure. Let me know if you have questions.