Lab Report 5
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Q7
Q7
equation 7 for "error and uncertainty" gives me sqrt((.002/vi)^2+(.001/.165)^2)=.0111 for the correct error. Then my incorrect error is (.002/vi)=.0094. Then when i do ((correct-incorrect)/correct)*100 to get the ratio my ratio is ~16%. any ideas to what i am doing wrong?
equation 7 for "error and uncertainty" gives me sqrt((.002/vi)^2+(.001/.165)^2)=.0111 for the correct error. Then my incorrect error is (.002/vi)=.0094. Then when i do ((correct-incorrect)/correct)*100 to get the ratio my ratio is ~16%. any ideas to what i am doing wrong?
raven- Guest
Re: Lab Report 5
For Question 5
The missing image is for Δv/v which is equal to the relative error of the width. Just express the error of Δv/v in terms of width using w and Δw
Question 7
This question needs to be broken down. I forget my exact collections, although I followed one of the formulas posted previously. Our ratio is found using the formula (correct-approximate)/correct. All we are doing is finding some sort of value range, in rough terms. There is a precise number we want, we call it correct and there is also an estimated value, a round about, what we can expect value. This is represented by the word approximate in the equation. The words correct and approximate really just stand for certain formulas, one to find a precise value and the other is the estimated value.
The ratio is for the Δρ that is the momentum (interpreted as p). In Question 6 we found the formula to find the error of momentum. Question 6 represents the correct value because it is a bit more precise using more numbers. The question tells us that the approximate value isΔρ but not as precise. For the approximate value we can ignore the Δm/m because it is negligible.
So what is the formula for Δρ? Then find it again, but with out using the Δm/m and use these as your values for the (correct-approximate)/correct formula to get a final ratio.
Also you do not need to multiple by 100. It is not a percentile error, it is a ratio, so try that value without the x 100%.
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Δπ√αθΨΛδΓωχσρβ
The missing image is for Δv/v which is equal to the relative error of the width. Just express the error of Δv/v in terms of width using w and Δw
Question 7
This question needs to be broken down. I forget my exact collections, although I followed one of the formulas posted previously. Our ratio is found using the formula (correct-approximate)/correct. All we are doing is finding some sort of value range, in rough terms. There is a precise number we want, we call it correct and there is also an estimated value, a round about, what we can expect value. This is represented by the word approximate in the equation. The words correct and approximate really just stand for certain formulas, one to find a precise value and the other is the estimated value.
The ratio is for the Δρ that is the momentum (interpreted as p). In Question 6 we found the formula to find the error of momentum. Question 6 represents the correct value because it is a bit more precise using more numbers. The question tells us that the approximate value isΔρ but not as precise. For the approximate value we can ignore the Δm/m because it is negligible.
So what is the formula for Δρ? Then find it again, but with out using the Δm/m and use these as your values for the (correct-approximate)/correct formula to get a final ratio.
Also you do not need to multiple by 100. It is not a percentile error, it is a ratio, so try that value without the x 100%.
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Δπ√αθΨΛδΓωχσρβ
Guest01- Posts: 133
Join date: 2008-09-20
number 7
for #7 i understand the formula that we use but which w do we use (small or big?) and which v's or p's?
confused- Guest
Re: Lab Report 5
I forget which ones I used, however I'm sure if you are consistent, that is use just the big mass and the big width and the big momentum that should be good.
Guest01- Posts: 133
Join date: 2008-09-20
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