Answer:
a. [tex]\bar{d}=4.34 cm[/tex]
b. [tex]\sigma=0.023 cm[/tex]
c. [tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]
Explanation:
a) The average of this values is the sum each number divided by the total number of values.
[tex]\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}[/tex]
[tex]\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}[/tex]
[tex]\bar{d}=4.34 cm[/tex]
b) The standard deviation equations is:
[tex]\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}[/tex]
If we put all this values in that equation we will get:
[tex]\sigma=0.023 cm[/tex]
Then the mean diameter will be:
[tex]\bar{d}=(4.34\pm 0.023)cm[/tex]
c) We know that the density is the mass divided by the volume (ρ = m/V)
and we know that the volume of a cylinder is: [tex]V=\pi R^{2}h[/tex]
Then:
[tex]\rho=\frac{m}{\pi R^{2}h}[/tex]
Using the values that we have, we can calculate the value of density:
[tex]\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}[/tex]
We need to use propagation of error to find the error of the density.
[tex]\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}[/tex]
Let's find each partial derivative:
1. [tex]\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089[/tex]
2. [tex]\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004[/tex]
3. [tex]\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071[/tex]
Therefore:
[tex]\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}[/tex]
[tex]\delta\rho=0.00058[/tex]
So the density is:
[tex]\rho=(0.0089\pm 0.00058) kg/cm^{3}[/tex]
I hope it helps you!
