Question

Report ? with the correct associated uncertainty

Two values were measured. x = 4.6 ±0.3 s and y = 3.2 ± 0.2 s2. You then need to calculate the quantity, 4 2 2 Report ? with the correct associated uncertainty.

Answer 1

Given that x = 4.6 plusminus 0.3s y = 3.2 plusminus 0.2s^2 From that x = 4.6 and Delta q x = 0.3 s y = 3.2 and Delta y = 0.2s^2 Here alpha is combination alpha plusminus Delta alpha Use the following relation to find the value of alpha is is given below alpha = x^4/y^2 + ln (y/x^2) = (4.6)^4/(3.2)^2 + ln (3.2/(4.6)^2) = 41.84 Use the following relation to find Delta alpha is given below Delta alpha = (Delta x)^4/(Delta y)^2 + ln (Delta y/(Delta x)^2) = (0.3s)^4/(0.2s^2/(0.3s)^2) = 1.0010 Therefore alpha plusminus Delta alpha is 41.84 plusminus 1.0010 Use the uncertainly measurement the value of alpha plusminus Delta alpha is 41.8 plusminus 1.0