Resistance against length for three metals

Three wires, each $0.2\ \text{mm}^2$ in area, are made of copper ($1.7 \times 10^{-8}\ \Omega\,\text{m}$), iron ($1.0 \times 10^{-7}\ \Omega\,\text{m}$) and nichrome ($1.1 \times 10^{-6}\ \Omega\,\text{m}$). You will make the program find each wire's resistance for four lengths, how many times nichrome resists more than copper, and the length of nichrome that gives $10\ \Omega$. Then it draws resistance against length.

The idea. $R = \dfrac{\rho L}{A}$, with $A$ in square metres.

Your tasks.

1. Inside the loop, work out R for the length L.
2. Find times, the resistivity of nichrome divided by that of copper.
3. Find L10, the length of nichrome wire with a resistance of $10\ \Omega$, in metres.