How can a large force result in a relatively small power?
1 answer:
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The given data is tabulated below
Measurement Method A Method B
-------------------- -------------- --------------
1 2.2 2.703
2 2.3 2.701
3 2.7 2.705
4 2.4 5.811
A graph of the data reveals that 5.811 from measurement B is an outlier, and should be rejected.
Part a: Calculate average density.
Method A:
Avg. density = (2.2+2.3+2.7+2.4)/4 = 2.4 g/cm³
Method B:
Avg. density = (2.703+2.701+2.705)/3 = 2.703 g/cm³
Answer:
Average densities are
2.4 g/cm³ by method A
2.703 g/cm³ by method B, with the outlier removed.
Part b: Calculate the percent error.
The true value is 2.702 g/cm³
Method A:
%error = 100*(|2.4 - 2.702|/2.702) = 11.2%
Method B:
%error = 100*(|2.703 - 2.702|/2.702) = 0.04%
Answer:
The percent error is
11.2 by method A,
0.04% by method B.
Part c:
Method B is more accurate and more precise because its average value is closest to the true value and its %error is extremely small.
Measurement Method A Method B
-------------------- -------------- --------------
1 2.2 2.703
2 2.3 2.701
3 2.7 2.705
4 2.4 5.811
A graph of the data reveals that 5.811 from measurement B is an outlier, and should be rejected.
Part a: Calculate average density.
Method A:
Avg. density = (2.2+2.3+2.7+2.4)/4 = 2.4 g/cm³
Method B:
Avg. density = (2.703+2.701+2.705)/3 = 2.703 g/cm³
Answer:
Average densities are
2.4 g/cm³ by method A
2.703 g/cm³ by method B, with the outlier removed.
Part b: Calculate the percent error.
The true value is 2.702 g/cm³
Method A:
%error = 100*(|2.4 - 2.702|/2.702) = 11.2%
Method B:
%error = 100*(|2.703 - 2.702|/2.702) = 0.04%
Answer:
The percent error is
11.2 by method A,
0.04% by method B.
Part c:
Method B is more accurate and more precise because its average value is closest to the true value and its %error is extremely small.