course Phy 201 I did not complete the lab yet for the measurements because I don't have the other lab kit that I wasn't aware we are supposed to have. I tried to print the pages out and they didn't print out the way you were describing in the notes and I got confused so I will complete that this week if that is ok. Thank you. êL⪄îò÷û‰F˜°\øxƒê¶~†Ú¯ãXwÚˆassignment #003
......!!!!!!!!...................................
21:41:41 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 --> 3.76 m confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:44:21 ** 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 --> ok. The book had a slightly different answer. It also had the answer of 3.76 but there is obviously room for error. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:44:41 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 --> skip confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:44:54 ** 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 --> I am General Physics self critique assessment: 3
.................................................
"