Asst_3_query

course Phy 121

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06-09-2006

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16:20:36

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 first converted all parts of the equation to meters.

1.80m + 1.425m + .534m = 3.760m

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06-09-2006

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16:30:28

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 -->

First we convert all the units to meters. Add them and give the answer to threee significant figures, which is what we have in 1.80m

1.80m + 1.425m + .00000534m = 3.22m

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16:31:39

** 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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RESPONSE -->

I think I get this except I did not convert 142.5cm to .01425m.

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16:31:43

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).

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RESPONSE -->

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Your work on this assignment is good. Let me know if you have questions.