11 IV V VI known HZJ For 1= 0 m* AZ. TO -5 18 mm Observed In*10*-7 F=0 n mass [kg weight [N) Current | [A) 2 current I [A) 12 UO error =4.18mm 1 7.90E-06 7.74E-05 2.16 4.6656 2.00E-06 1.26E-06 59.18 Weigh 1 2 4x- 10-7 Tm/A, 2 1.55E-04 3.85 14.8225 .26E-06 1.26E-06 0.21 3 2.32E-04 4.72 22.2784 1.26E-06 1.26E-06 0.01 3. 10E-04 5.78 33.4084 . 12E-08 1.26E-06 -11.08 5 3.87E-04 5.57 43.1649 1.08E -06 1.26E-06 13.97 4.65E-04 7.32 53.5824 1.04E-06 1.26E-06 -16.84 F, = ILB, Hol 2 7 r=5.18mm Weight 1+2V. Data Analysis 19. 20. 21. Use the spreadsheet program to set up a data table with following columns: (i) mass (kg), (ii) weight (N), (iii) current I (A) and (iv) its square 12 (A%, for r=4.18 mm, and (v) current I (A) and (vi) its square 7% (A%), for r=5.18 mm. Note: The mass m of the individual sequins is given on their storage vial. The weight of each sequin is mg. Plot the weight vs. I* at a fixed spacing of r=4.18 mm. Fit the data with a straight line and use the slope to determine o, as explained in the theory section. Noles: 1. We can include a data point at the origin (F=0, I*=0). 2. In order to display the slope with enough accuracy, you may need to format the trend line label in scientific notation, using 3 significant figures. Plot the measured data as points and make the line of best fit a solid line. Compare your value of iy to its defined value of 4710 T Tm/A, and calculate the percent error. Repeat step 20 using the data for r=5.18 mm. Add the curve to the same plot, to allow comparison of the two cases