Answer:
[tex]1.5 \Omega[/tex]
Explanation:
The resistance of a wire is given by the equation:
[tex]R=\rho \frac{L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
In this problem, we have a wire of platinoid, whose resistivity is
[tex]\rho = 3.3\cdot 10^{-7} \Omega m[/tex]
The length of the wire is
L = 7.0 m
And its radius is
[tex]r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m[/tex], so the cross-sectional area is
[tex]A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2[/tex]
Solving for R, we find the resistance of the wire:
[tex]R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega[/tex]
