A rectangular parallelopiped is measured as $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If its specific resistance is $3 \times 10^{-7} ~\Omega \mathrm{m}$, then the resistance between its two opposite rectangular faces will be ___________ $\times 10^{-7} ~\Omega$.

<p>The resistance of a material can be calculated using the formula:</p> <p>$ R = \rho \frac{L}{A} $</p> <p>where </p> <ul> <li>$R$ is the resistance, </li> <li>$\rho$ (rho) is the resistivity or specific resistance of the material, </li> <li>$L$ is the length (or distance over which the resistance is being measured), and </li> <li>$A$ is the cross-sectional area through which the current flows.</li> </ul> <p>In the context of a rectangular parallelepiped, the &quot;length&quot; and &quot;cross-sectional area&quot; can vary depending on which faces of the shape are considered.</p> <p>In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:</p> The cross-sectional area $A$ should be: <br/><br/> $A = 100 \, \text{cm} \times 1 \, \text{cm} = 100 \, \text{cm}^2$ <br/><br/> Converting this to meters gives: <br/><br/> $A = 100 \, \text{cm}^2 = 1 \, \text{m} \times 0.01 \, \text{m} = 0.01 \, \text{m}^2$ <br/><br/> So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get: <br/><br/> $ R = \rho \frac{L}{A} = 3 \times 10^{-7} \Omega \, m \times \frac{0.01 \, m}{0.01 \, \text{m}^2} = 3 \times 10^{-7} \Omega $ <p>Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 \times 10^{-7} \Omega$.</p>