course Phy 241 Yw|]σʐb~assignment #003
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18:02:52 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
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18:03:01 ** 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 below .01 m can be distinguished. 142.5 cm is .01425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, good to within .00000001 m. Then theses are added you get 1.81425534 m; however the 1.80 m is only good to within .01 m so the result is 1.81 m. The rest of the number is meaningless, since the first number itself could be off by as much as .01 m. **
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18:12:22 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 --> For each set of magnitude and angle we first need to sketch the triangles to figure out the formulas used to find each component of the vectors. The first pair of information yields a line in the first quadrant 60 degrees up from the x axis, therefore we use the y coordinate as the opposite side and the x coordinate as the adjacent side. With trig we can use that the cos of the 60 degrees equals the adjacent over the hypotenuse. We know the hypotenuse to be 9.3 because it was given in the problem. Solving for A sub x gives 4.65. Now to find A sub y you can use the sin trig rule that the sin of the angle equals the opposite over the hypotenuse. This yields 8.05. This can be done with all pairs of angles and magnitudes. B sub x was found to be 15.56 and B sub y was -15.56. C sub x came out to be -5.07and C sub y, -3.82.
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18:21:48 ** 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 --> My problem 1.34 reads ""Use a scale drawing to find the x and y coordinates of the following vectors. Then it has three pairs of magnitudes and angles. I am not sure what you were asking. Was the information in the parenthesis another problem... In relation to that problem I understand how to get R sub x but I am not sure why R sub y comes out to 4.79, where does the 2.6 displacement come from. I know how to figure out the rest of this problem if I know both Rx and Ry.
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