course mth 158 ???H?????????v?assignment #021021. `query 21
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01:46:52 **** query 2.5.8 / 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> y=k/x 4=k/9 36=k y=36/4 y=9
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01:47:10 ** 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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01:49:41 query 2.5.12 / 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^3+y^2) 1=k(8+9) 1=17k k=1/17
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01:49:52 ** 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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01:55:36 query 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)
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RESPONSE --> P=k(sqrt L)
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01:56:53 ** 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 --> ok I didn't know that you were saying that k=2 pi/ sqrt(32) I see T = 2 pi / sqrt(32) * sqrt(L)
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01:57:37 **** What equation relates period and length? ****
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RESPONSE --> p=kl
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02:02:56 query 2.5.42 / 2.7.42 (was 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 --> 1.24=k 432/16 1.24=27k .046=kor 1.24/27=k 1.44=.046 L/3^2 12.96=.046L L=281.74
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02:03:07 ** 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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