Assignment 3

course Phy 201

The program only showed 2 questions for this part of the assignment. I hope that is all that was required.

?}???c??|?????assignment #003

003. `Query 3

Physics I

03-05-2008

......!!!!!!!!...................................

18:12:07

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 -->

For this problem, I converted the measurements to a common measure, added them, and rounded to get the average number of significant figures.

confidence assessment: 1

.................................................

......!!!!!!!!...................................

18:15:15

** 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 did not realize that the final zero in 1.80 was significant. I just averaged the number of what I thought was significant figures, but I understand now that that is wrong. Am I correct in assuming the correct number of significant figures is determined by whichever figure has the fewest?

self critique assessment: 2

That is pretty much the case; it would be better to say that the number of significant figures determines the best possible resolution of the result. In this case the resolution of the 0 in 1.80 is .01 meters. The resolution of the 5 in 5.34 * 10^-6 meters is well beyond the precision of the .01 meter resolution, so none of the figures in that number will be significant to the sum (i.e., 1.80 + 5.45 * 10^-6 is 1.80000545; since even the second 0 is not significant, the 545 doesn't make any difference at all).

.................................................

......!!!!!!!!...................................

18:15:46

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 --> I am not in university physics."

&#

Your work looks good. See my notes. Let me know if you have any questions. &#