Acertain laboratory experiment requires an aluminum wire of length of 39.0 m and a resistance of 2.10 ω at 20.0°c. what diameter wire must be used? (the resistivity of aluminum at 20.0°c is 2.82 10-8 ω · m.)

817.02 x 10⁻⁶ m

Explanation:

L = length of the aluminum wire = 39.0 m

R = Resistance of the wire = 2.10 Ω

ρ = Resistivity of aluminum wire = 2.82 x 10⁻⁸ Ω·m

A = area of cross-section of wire

Resistance of the wire is given as

A = 5.24 x 10⁻⁷ m²

d = diameter of the wire

Area of cross-section of wire is given as

A = (0.25) πd²

5.24 x 10⁻⁷ = (0.25) (3.14) d²

d = 817.02 x 10⁻⁶ m

d) the number of neutrons plus the number of protons