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You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched into incredibly long, thin wires. The company has developed a new technique that will stretch 1.00 g of gold into a wire of length L 5 2.40 km and uniform diameter. Your supervisor gives you the task of determining the resis- tance of such a wire at 20.08C.(the density of gold is 19.3 ✕ 103 kg/m3.)

Respuesta :

Explanation:

We know that the relation between volume and density is as follows.

      Volume = [tex]\frac{\text{mass}}{\text{density}}[/tex]

So,       V = [tex]\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}[/tex]

               = [tex]5.181 \times 10^{-8} m^{3}[/tex]

Now, we will calculate the area as follows.

      Area = [tex]\frac{\text{volume}}{\text{length}}[/tex]

               = [tex]\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}[/tex]

               = [tex]2.15 \times 10^{-11} m^{2}[/tex]

Formula to calculate the resistance is as follows.

         R = [tex]\rho \frac{l}{A}[/tex]

             = [tex]\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}[/tex]

             = [tex]2.71 \times 10^{6} ohm[/tex]

Thus, we can conclude that the resistance of given wire is [tex]2.71 \times 10^{6} ohm[/tex].