An aluminum wire of length 1.0 meter has a resistance of 9.0x10^-3 ohm. If the wire were cut into two equal lengths, each length would have a resistance of

Each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

Since resistance R = ρl/A where

Since ρ and A are constant for the aluminum wire, R ∝ l

So. R₂/R₁ = l₂/l₁ where

Now, since the length of the wire is cut into two, l₂ = l₁/2

So, R₂/R₁ = l₁/2 ÷ l₁

R₂/R₁ = 1/2

R₂ = R₁/2

R₂ = 9.0 × 10⁻³ Ω/2

R₂ = 4.5 × 10⁻³ Ω

So, each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

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since resistance depends on length, then the resistance will be shared equally since it was cut equally therefore the resistance is

4.5*10^-3 ohm

Explanation: