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Determine the length of an aluminum wire ($ρ_{aluminum} = 2.82 \times 10 -8 Ω.m$) with a resistance of 0.04 Ω and a radius of 0.03 m.
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R=PL/A

R x A/P = L

0.04 x $\pi$ (0.03)^2 / 2.82 x 10^-8 = L

0.04 x $\pi$ (9 x 10^-4) / 2.82 x 10^-8 = L

0.04 x 2.827 x 10^-3 / 2.82 x 10^-8 = L

1.1308 x 10^-4/ 2.82 x 10^-8 = L

L = 4.010 x 10^3m
by Wooden (1,126 points)
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R = 0.04 Ω

A = 0.03

Paluminium = 2.82 x 10-8 Ω

R = pL/A

L = RA / p

= (0.04)(π(0.03)²) / (2.82 x 10^-8)

= 4010.5 = 4011m
by Wooden (786 points)
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R=pL/A

​​radius = .03 m Area = .03^2 x pie = 0.0028

L=RA/p

=0.04x0.0028/2.82x10^-8

=3,971 m

by Wooden (1,121 points)

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