Asked by Hannah
I had to do an experiment where a tiny steel ball was measured using a micrometer. There was then a series of calculations that needed to be completed and then a percent error had to be calculated. I think that my data may be incorrect or maybe I am just not doing the calculations correctly but my percent error keeps coming out to be very high.
To start off I had to use the micrometer to find the diameter D(mm) of the steel ball. 5 trials were completed. For the first trial I got 15.90 mm. The volume V=1/6(pi)D^3 had to be calculated next so to do this I plugged 15.9 into the equation 1/6(3.14)(15.90^3) and got 2019. For each trial I plugged my D value into the equation and then to calculate the average Volume I added all the numbers up and divided by 5.
So far before I continue can you please tell me if I did this correctly. Thank you for your help!!
To start off I had to use the micrometer to find the diameter D(mm) of the steel ball. 5 trials were completed. For the first trial I got 15.90 mm. The volume V=1/6(pi)D^3 had to be calculated next so to do this I plugged 15.9 into the equation 1/6(3.14)(15.90^3) and got 2019. For each trial I plugged my D value into the equation and then to calculate the average Volume I added all the numbers up and divided by 5.
So far before I continue can you please tell me if I did this correctly. Thank you for your help!!
Answers
Answered by
Elena
1. Find the average magnitude of diameter
(D1+D2+D3+D4+D5)/5 ={D}
2. ΔD1 =|{D}-D1|
ΔD2=|{D}-D2|
…
ΔD5 =|{D}-D5|
Average absolute error is
{ΔD}= (ΔD1+,,,+ ΔD5)/5.
3. The relative error is {ΔD}/{D}
4. Since V~D³ the error of your result = 3• {ΔD}/{D}.
If you want it in %, multiply by 100.
(D1+D2+D3+D4+D5)/5 ={D}
2. ΔD1 =|{D}-D1|
ΔD2=|{D}-D2|
…
ΔD5 =|{D}-D5|
Average absolute error is
{ΔD}= (ΔD1+,,,+ ΔD5)/5.
3. The relative error is {ΔD}/{D}
4. Since V~D³ the error of your result = 3• {ΔD}/{D}.
If you want it in %, multiply by 100.
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