Question

What would the units be for Table 2?

e V m (Nue) (Ir) (N46) which gives us: e a V (Ir) m Rearranging to the form of the equation for a straight line y = mx +b, we get e aV = (Ir) + Therefore we will plot: aV vs. (Ir)2 Remember: y vs. X Using your data from Table 1, calculate the values for Table 2. Table 2 (1r)2 αν 1 2 3 4 5 Graph 1. Open Graphical Analysis from the dock. 2 Enter your x and y values. 3. Remember to add labels, units, and to unconnect the points. 4. Click the Linear Fit icon on the toolbar. 5. Double click in the fit info box and mark the box for uncertainty (standard deviation) From your graph, what is our experimental result for e/m? e + m

Answer 1

consider the dimension of er [ ] = MILOTA From the equation e (IM3) Vi potential Is current rs distance [] = av (Ir) ² [av] = [[Ir}] [ - [A² L²] [5 'LT'A' = M'LT'A 3 (2) .unit of av= ar=mst/kg in ci SI A: Ampere unit of chosen [([2] = 4²L² (2) unit of fr² = x²m² Only then ce will have the dimension of M'LT'A'