If a mile of 24-gauge copper wire has a resistance of 0.14 kω and the resistivity of copper is 1.7 × 10−8 ω ⋅ m, what is the diameter of the wire? (1 mile = 1.6 km)

D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.

In order to solve this problem we have to use the equation that relates resistance and resistivity:

R = ρL/A

Where ρ is the resistivity of the matter, the length of the wire, and A the area of the cross section of the wire.

If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.

First, we have to clear A from the equation R = ρL/A:

A = ρL/R

Substituting the values

A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)

A = 1.9x10⁻⁷m²

The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:

D = √4A/π

Substituting the value of A:

D = √4(1.9x10⁻⁷m²)/π

D = 497.4x10⁻⁶m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s