course phy 201
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15:00:05 ** 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 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .000534 m, good to within .000001 m. Then theses are added you get 3.225534 m; however the 1.80 m is only good to within .01 m so the result is 3.23 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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RESPONSE --> Yes, I converted cm to meters which is 100cm/m and then I found that um = 1000m so then I plugged into equation 1.80m +0.01425m + 5.34*10^-6 = 1.80
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15:00:13 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 --> didn't do
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15:00:19 ** 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
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