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April 17, 2014

Homework Help: Physics(Please help, thanks)

Posted by Hannah on Sunday, September 2, 2012 at 2:32pm.

An experiment was done where a tiny steel ball was measured with a micrometer. I know I already posted a question about this before and was given a respond but after trying to fix my answers I am still lost. If someone could please review my results and tell me where I went wrong. Thank you very much.

1) For the D(mm) my results were: Trial 1= 15.90, tr 2= 16.01, tr 3 = 16.0, tr 4 = 16.0 and trial 5 = 15.9 , average=15.9

2) For V=1/2(pi)D^3: trial 1= 2019 , tr 2 = 2061, tr 3 = 2058, tr 4 = 2058, trial 5 = 2019, average 2043. I just plugged in each D value into the equation.

3) |dvi|= |Vi - V(the mean of volume)|: Trial 1 = 24 , tr 2= 18, tr 3 = 15, tr 4 =15, trial 5 = 24, average= 19.2
For this I took the average volume and subtracted it from each volume for each trial.

4) m(g) : Trial 1 = 16.1 , tr 2 = 15.8, tr 3 = 16.0 , tr 4 = 15.6, trial 5 = 15.9, average= 15.9. For this I had to weigh the ball using a scale.

5) Density (g/mm^3) : Trial 1= 8.0, trial 2 = 7.7, tr 3 = 7.8 , tr 4 = 7.6 , trial 5 = 7.9, average = 7.8
For this I divided each mass by volume of every trial.

6) This is where I am confused. |dpi|= |density1 - the mean of density|. For this I took the average density which was 7.8 and subtracted it from each density for the 5 trials.
Trial 1 = 0.1, trial 2 = 0.2, tr 3 = 0.1, tr 4 = 0.1, trial 5 = 0.1, average= 0.12

7) Now it says to compare the measured mean of density (D with the line above) with accepted Density for Fe(7.8 X 10^3 kg/m^3) and calculate the percent error.

I converted 0.12 g/mm^3 into kg/mm^3 and then subtracted that from 7.8X10^3 and then divided by 7.8 X 10^3 and my error was extremely high.

Is my data just incorrect or did I do the calculations wrong?

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