An engineer cuts a 1.0-m-long, 0.33-mm-diameter piece of wire, connects it across a 1.5 V battery, and finds that the current in the wire is 8.0 A.

  ρ = 1.6 0 10⁻⁵ Ω/ m

Explanation:

In this exercise we are asked to find the resistivity of the wire, it indicates the length L = 1.0 m, the diameter of 0.33 mm.

For this exercise we start by using ohm's law

           V = i R

           R =

           R = 1.5 / 8.0

           R = 0.1875 Ω

now we use the relation of resistance

           R = ρ L / A

            ρ = R A / L

the area of ​​the wire is

           A =π r²

           r = d/2

           A = π d²/ 4

           A = π (0.33 10⁻³)²/4

           A = 8.55 10⁻⁵ m²

we substitute

           ρ = 0.1875  8.55 10⁻⁵ / 1.0

           ρ = 1.6 0 10⁻⁵ Ω/ m

found in the nucleus

has mass of one amu