Hello, please answer and explain :

You decide that it would be just as interesting to measure resistance versus length to find the resistivity value () of aluminum and brass. The wires are cylindrical and you measure the diameter of the aluminum wire at Al = (1.05 0.05) mm and the brass wire at La = (1.25 0.05) mm. You cut different length segments of each wire and measure their resistances.

Table 2 - Resistance values for aluminum and brass wire at different lengths aluminium (n) :l: 2% { NB : Replace the coma by a dot. For example 0,15 is 0.15 ; 0,51 is 0.51 ) 1.5 Linear Fit for: Data Set | Resistance Laiton R.Laiton = mx+b m (Slope): 0.04897 +/- 0.001139 0/m b (Y-Intercept): 0.004667 +/- 0.02218 0 Correlation: 0.9989 RMSE: 0.02382 0 1 - Resistance Aluminium (() Resistance Laiton () 0.5 Linear Fit for: Data Set | Resistance Aluminium R. Al = mx+b m (Slope): 0.03326 +/- 0.0007959 0/m b (Y-Intercept): -0.01200 +/- 0.01550 0 Correlation: 0.9989 RMSE: 0.01665 0 10 20 30 Longueur (m)