The resistivity of a material is how much the material resists the flow of charge through it. Metals have low resistivities (which makes them good conductors), while things like glass or plastic have high resistivities.

The equation for resistivity is as follows:

$\displaystyle R=\frac{\rho L}{A}$

$\displaystyle L$ is the length of the wire, $\displaystyle A$ is the cross-sectional area, $\displaystyle R$ is the resistance, and $\displaystyle \rho$ is the resistivity. We have values for $\displaystyle L$ and $\displaystyle R$, and we're given the radius of the wire so we can find $\displaystyle A$, so we're trying to solve for $\displaystyle \rho$. If we rearrange the equation, we end up with this:

$\displaystyle \frac{RA}{L}=\rho$

We're given the radius of the wire, so to find the area, we square the radius and multiply it by $\displaystyle \pi$.

$\displaystyle A=\pi r^2$

$\displaystyle A=\pi (0.0005)^2$

$\displaystyle A=7.854\cdot 10^{-7}$

Now, we can plug in our values.

$\displaystyle \frac{RA}{L}=\rho$

$\displaystyle \frac{(0.8709\cdot 10^{-3}) \cdot (7.854\cdot 10^{-7})}{0.04}=\rho$

$\displaystyle 1.71\cdot 10^{-8}=\rho$

Therefore, the answer is $\displaystyle 1.71\cdot10^{-8}\Omega\cdot m$.

Use the equation for resistivity:

$\displaystyle R= \rho \frac{L}A{}$

First, find the cross sectional area.

$\displaystyle A=\pi r^{2}$

$\displaystyle A=3.14cm^2$

Plug in known values.

$\displaystyle R= \rho \frac{L}A{}$

$\displaystyle \rho = \frac{25\Omega\cdot 3.14cm^2}{1cm}$

$\displaystyle \rho =78.5\Omega \cdot cm$ 

Since we have a cube of volume $\displaystyle 1cm^3$, the length of the material must be $\displaystyle 1cm$. Use the equation for resistivity. 

$\displaystyle R=\rho \frac{L}{A}$

Rearrange the equation and plug in known values.

$\displaystyle \rho = \frac{RA}{L}$

$\displaystyle \rho=\frac{3\Omega\cdot 1cm^2}{1cm}=3\Omega\cdot cm$

Use the resistivity equation:

$\displaystyle R= \rho \frac{L}{A}$

$\displaystyle R= 3500\Omega \cdot m \cdot \frac{0.01m}{0.01m^2}$

$\displaystyle R=3500\Omega$

We will use the relationship

 $\displaystyle \rho=R\frac{A}{l}$

Where $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is the surface area of the face the current is coming in or out of, and$\displaystyle l$ is the length from one face to the other.

Remember that the area of a circle is

$\displaystyle A=\pi\cdot r^2$

Combining our equations we get

$\displaystyle \rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\displaystyle \rho=935\frac{\pi\cdot{.0265}^2}{.045}$

$\displaystyle 45.8\Omega\cdot meters=\rho$

We will use the relationship

 $\displaystyle \rho=R\frac{A}{l}$

Where $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is the surface area of the face the current is coming in or out of, and$\displaystyle l$ is the length from one face to the other.

Remember that the area of a circle is

$\displaystyle A=\pi\cdot r^2$

Combining our equations we get

$\displaystyle \rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\displaystyle \rho=9.53\frac{\pi\cdot .015^2}{.025}$

$\displaystyle 2.70\times 10^{-3}\Omega\cdot meters=\rho$

We will use the relationship

 $\displaystyle \rho=R\frac{A}{l}$

Where $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is the surface area of the face the current is coming in or out of, and $\displaystyle l$ is the length from one face to the other.

Remember that the area of a circle is

$\displaystyle A=\pi\cdot r^2$

Combining our equations we get

$\displaystyle \rho=R\frac{\pi\cdot r^2}{l}$

We then need to plug in our values

$\displaystyle \rho=.22\frac{\pi\cdot.002^2}{.022}$

$\displaystyle \rho=1.26\times10^{-4}\Omega\cdot meters$

Use the equation for resistivity: 

$\displaystyle \rho=R\frac{A}{L}$

Here, $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is the cross-sectional area, and $\displaystyle L$ is the length of the resistor.

Begin by finding the cross-sectional area of the material, which is circular: 

$\displaystyle A=\pi r^2$

$\displaystyle A=\pi (0.005m)^2$

$\displaystyle A\approx 0.0000785m^2$

Now plug in all known values and solve for resistivity.

$\displaystyle \rho=0.3\Omega*\frac{0.0000785m^2}{0.03m}$

$\displaystyle \rho = 0.000785\Omega\cdot m$

$\displaystyle \rho = 7.85*10^{-4}\Omega\cdot m$

Use the relationship:

$\displaystyle \rho =R\frac{A}{l}$

Here, $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is cross sectional area, and $\displaystyle l$ is the length.

The area of a circle is:

$\displaystyle A=\pi r^2$

Substitute:

$\displaystyle \rho =R\frac{\pi r^2}{l}$

Plug in given values and solve

$\displaystyle \rho =.75\Omega \frac{\pi (.006m)^2}{.045m}$

$\displaystyle \rho = 1.88 *10^{-3} \Omega\cdot m$

Using the relationship

 $\displaystyle \rho=R\frac{A}{l}$

Where $\displaystyle \rho$ is the resistivity, $\displaystyle R$ is the resistance, $\displaystyle A$ is the cross sectional area, and $\displaystyle l$ is the length from one face to the other.

The area of a circle is

 $\displaystyle A=\pi*r^2$

 Combining equations

 $\displaystyle \rho=R\frac{\pi*r^2}{l}$

 Plugging in values

 $\displaystyle \rho=10.5\frac{\pi*.0065^2}{.0475}$

 $\displaystyle .0293 \Omega*meters=\rho$