course Phy201 ̓ö¬‚^܃¤µÑ—•ÆáñÙ¯ÛöeýÕýassignment #003
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21:26:37 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)
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RESPONSE --> I know the answer to the equation is 3.76m. I can break the first two segments of the equation down and get 1.80m + 142.cm =3.225m then we have 5.35 x 10^5, some how that ends up being 3.225+.535=3.759 which equals 3.76m. I'm not exactly sure 100% on the 5.34'micro m conversion part. confidence assessment: 1
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21:30:27 ** 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. **
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RESPONSE --> I understand what I did wrong here....i was on the right track on the first two segments of the problem....on the last part I thought the question was referring to the problem in the book. I actually did get the .00000534, but i thought that coudln't be correct because i thought the answer should have been 3.76, but i understand that it should be 3.23m. self critique assessment: 2