Please solve both 2.15 and 2.16* using MatLab

Problem 2.16 * with the following adjustments *Repeat problem 2.15 but compute P for only Copper and Aluminum. Generate a single plot of P versus length for Copper and Aluminum. Your plot should have appropriate axis labels, and grid lines. Submit your MATLAB code and your plot

2.15 In direct current applications, electrical power is calculated using Joule's law as P= VI where Pis power in watts, Vis the potential difference, measured in volts, I is the electrical current, measured in amperes, and Joule's law can be combined with Ohm's law vironmere V=IR to give P= IPR where Ris resistance measured in ohms. The resistance of a conductor of uniform cross section (a wire or ron for example) is 7 REPA where p is the electrical resistivity measured in ohm-meters, lis the length of the wire, and A is the cross-sectional area of the wire. This results in the equation for power p=1'p = 1 Electrical resistivity is a material property that has been tabulated for man materials. For example, Material Silver Copper Gold Aluminum Iron Resistivity, ohm-meters (measured at 20C) 1.59 x 10-8 1.68 x 10-8 2.44 x 10-8 2.82 x 10-8 1.0 x 10-7 Calculate the power that is dissipated through a wire with the following dimensions for each of the materials listed. diameter 0.001 m length 2.00 m Assume the wire carries a current of 120 amps. 2.16 Repeat the roli I'd