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You are working at a company that manufactures electrical 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 1.60 g of gold into a wire of length L = 2.20 km and uniform diameter. Your supervisor gives you the task of determining the resistance of such a wire (in MΩ) at 20.0°C. (The density of gold is 19.3 ✕ 103 kg/m3.)

Respuesta :

Answer:[tex]R=1.424 M\Omega [/tex]

Explanation:

Given

mass of gold [tex]m=1.6 gm[/tex]

Length of wire [tex]L=2.2 km[/tex]

Resistivity of gold [tex]\rho =2.44\times 10^{-8}[/tex]

density of gold [tex]=19.3\times 10^3 kg/m^3[/tex]

and [tex]mass=volume\times density[/tex]

[tex]1.6\times 10^{-3}=volume\times 19.3\times 10^3[/tex]

[tex]volume=8.29\times 10^{-8} m^3[/tex]

And [tex]Resistance R=\frac{\rho L}{A}[/tex]

also be written as

[tex]R=\frac{\rho L^2}{V}[/tex]

where [tex]L=length[/tex]

[tex]V=volume[/tex]

[tex]\rho =resistivity\ of\ gold[/tex]

[tex]R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}[/tex]

[tex]R=1.42\times 10^{6} \Omega [/tex]

[tex]R=1.424 M\Omega [/tex]

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