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Please check if my solutions are correct or if I am totally wrong. Thank you!
1. Homework Statement [/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?
1. Homework Statement [/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?