course PHY 231 L|cz祲|Ќgassignment #003
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03:14:35 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> not me confidence assessment: 0
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03:14:41 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> ok self critique assessment:
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w[xɶޛ}|j assignment #003 003. `Query 3 Physics I 07-12-2008
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18:24:18 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> not me confidence assessment: 0
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18:24:23 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> self critique assessment: 0
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18:31:36 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> Question 34 asked to find the x and y component of 3 seperate Vectors. a) 9.3m at 60 degrees X-component cos 60 = x/9.3m 9.3cos60 = x x = 4.65 m Y-Component sin 60 = y / 9.3m 9.3sin60 = y y = 8.05 m b) 22.0km at 135 degrees Y-Component sin45 = y / 22 km 22sin45 = y y = 15.56 km X-Component cos45 = x / 22km 22cos45 = x x = -15.56 km Note that the x value is negative because the vector is in the second quadrant(135 degrees) c) 6.35 cm 307 degrees X-Component cos53 = x/6.35cm 6.35cos53 = x x = 3.79 cm Y-Component sin53 = y/6.35 cm 6.35sin53 = y y = -5.07 Note that y is negative because it is in the 4th quadrant. confidence assessment: 3
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18:33:29 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE --> We are discussing 2 completely different problems. However, I gave you the answer to problem 34 in Chapter 1 of University Physics. Did I end up showing you the incorrect problem? self critique assessment: 0