Please check if my solutions are correct or if I am totally wrong. Thank you!(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data[/b]

In an experiment to measure the density of a steel ball. The diameter of the ball is d= 1.76 +- 0.02 cm and the mass to be m= 22 +- 1g. (include error calculations and report values with uncertainties.)

a) The volume of a sphere is given by V = 4/3 x pie x r^3. What is the volume of the ball in m^3?

Attempt Solution: d= 1.76 +- 0.02 cm --> 0.0176 +- 0.0002 m

r= 0.00088 +- 0.0001 m

V = 4/3 x pie x r^3

= 4/3 x pie x (0.00088)^3

= 2.85 x 10^-6

Uncertainty:

V = V(3xr)

= (2.85 x 10^-6 )((3)(0.00010/0.0176))

= 4.86 x 10^-8

Final: V= 2.85 x 10^-6 +- 2.77 x 10^-13 m^3

b) Density is given by p= m/v. What is the density of the ball in kg/m^3/

0.022 +- 0.001 kg

p=(0.022/2.85x10^-6)

= 7719.3 kg/m^3

Uncertainty: P = P(m/m + V/V)

= 7719.3 ((0.001/0.022) + (4.86x10^-8)/(2.85x10^-6))

= 478.98

Final: P= 7719 +- 479 kg/m^3

c) Is your value consistent with a tabulated value of 7860 kg/m^3

How do I answer this scientifically?

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# Homework Help: Please check solutions to Error Uncertainty calculations!

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