Suppose you were determining the density of a sphere with a mass of 9.00g. You m

Question

Suppose you were determining the density of a sphere with a mass of 9.00g.  You measure the diameter to be 2.90cm but fail to include a zeroing error of .035cm.  That is, the correct value is .035cm larger than the measured value.  Calculate the density using the original, incorrect diameter and then re-calculate it using the corrected diameter.  The absolute error in the density is the absolute value of the difference between the measured  (incorrect) value and the "true" (correct) value.   The percent error is the absolute error divided by the "true" value, all multiplied by 100%, i.e.

What is the percent error in the original, incorrect density?

Explanation / Answer

DENSITY = mass/volume

error density, raidus =1.45*10^-2 m
densty = 9*10^-3/4pi* 1.45^3 *10^-6

density = 2350 Kg/m^3

original density   radius r = 2.935 /2 = 1.4675 *10^-5

density = 9*10^-3/4pi *1.4675^3 *10^-6

=2267 kg.m^2

% error = 3%

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