A wire of radius R and length L has a resistance of 14 Ω. What is the resistanceof a wire made from the same material that has twice the radius and five timesthe length?

The resistance (R) of a wire can be calculated using the formula:

R = ρL/A

where: ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

The cross-sectional area (A) of a wire with radius r is given by the formula:

A = πr²

So, the resistance of the wire can be rewritten as:

R = ρL/(πr²)

Given that the resistance (R1) of the wire of radius R and length L is 14 Ω, we can write:

14 = ρL/(πR²) ----(1)

We are asked to find the resistance (R2) of a wire made from the same material (so ρ is the same) that has twice the radius (2R) and five times the length (5L). We can write:

R2 = ρ(5L)/(π(2R)²) = 5ρL/(4πR²) = 5/4 * ρL/(πR²) ----(2)

From equation (1), we know that ρL/(πR²) = 14. Substituting this into equation (2), we get:

R2 = 5/4 * 14 = 17.5 Ω

So, the resistance of a wire made from the same material that has twice the radius and five times the length is 17.5 Ω.