course Phy 121 ????]??????????assignment #003003. `Query 3
......!!!!!!!!...................................
22:43:17 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)
......!!!!!!!!...................................
RESPONSE --> My answer was 53.4. This is the correct number of significant figures, however I don't think this is the right answer. I tried to convert everything to micrometers using the conversion factor 1x10^-6 and then added the answers and then converted back to meters to get 53.4. confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:45:29 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I didn't realize you just put the ""micro"" part onto the end of the 5.34 when it says micro meters. I thought there was some sort of conversion you had to do to get this. Now it makes sense just to convert everything to meters and just add it up. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:45:38 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).
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
22:45:45 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
"