course Phy 231 ʉƻnassignment #003
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22:51:36 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 --> N/a confidence assessment: 3
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22:51:46 ** 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 --> n/a self critique assessment: 3
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22:55:52 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 --> Vector A = 2.6km in the y direction Vector B = 4.0km in the x direction Vector C = 3.1km To break C down into its components we use the sine and cosine functions. Cx = 3.1km * cos(45deg) = 2.19 Add this to the displacement from vector Bx (4) = 6.19km Cy = 2.19 + 2.6km = 4.79km This gives components Dx = 6.19km and Dy = 4.79km. The magnitude of this vector is the square root of the sum of the components squared. D = sqrt( 6.19km^2 + 4.79km^2 ) = 7.3km We use the formula theta = arctan( Dy/Dx) to find the angle. Subbing this info in we get the angle to be about 37.7 degrees. confidence assessment: 3
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22:55:57 ** 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 --> ok self critique assessment: 2
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