How long must a 0.62-mm-diameter aluminum wire be to have a 0.40 A current when connected to the terminals of a 1.5 V flashlight battery? Express your answer to two significant figures and include the appropriate units.

Explanation:

R = V / I

V = Voltage = 1.5 V

I = Current = 0.40 A

R = 3.75Ω

R = ρL / A

ρ = Density of Aluminium = 2.710

Radius of wire = 0.00031 m³, from the data given

A = Area of Wire = π×r²

A = 3.14159×0.00031

A = 0.00097 m²

L = (R*A)/ρ

L = (3.75 * 0.00097) / 2.710

L = 0.0013 meters

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