course phy 121
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem
along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in
completing the assignment, before you look at the given solution.
003. `Query 3
Question: **** `qQuery 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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Your solution: Adjust the significant figures such that the units are all the same. 1.80m + 1.425m + .0000534m = 3.23 m. the
final result should be no more than the least significant number
Confidence Assessment: 2
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Given Solution:
`a** 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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Self-critique (if necessary):I disagree with your conversion of 142.5 cm into .01425 m. since 1 meter = 100 cm then 142.5 cm would
be 1.43 m if the .05 is rounded up to maintain 3 significant places
You are correct; I believe a previous edition had the measurement as 1.42 cm, and I failed to catch the change.
Question: **** `qUniversity Physics #34: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45
deg turn by components, verify by scaled sketch).
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Your solution:
Self-critique Rating:
Confidence Assessment:
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Given Solution:
`a** 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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Self-critique (if necessary):
Self-critique Rating:
&#This looks very good. Let me know if you have any questions. &#