You probably can still put in an uncertainty from measurement or something which is definitely significant here since there are no bigger uncertainties in the repeats.
You probably can still put in an uncertainty from measurement or something which is definitely significant here since there are no bigger uncertainties in the repeats.
I don't understand.
the values are for the time it takes for a ball to go past a light gate. The Light gate measures at a resolution of 0.001 seconds.
I was taught for digital readings if you had a singular reading or several readings that were all the same, you would do ± the resolution of the instrument/the smallest measurement the instrument could take. So in this case it would be ±0.001s
I was taught for digital readings if you had a singular reading or several readings that were all the same, you would do ± the resolution of the instrument/the smallest measurement the instrument could take. So in this case it would be ±0.001s
You probably can still put in an uncertainty from measurement or something which is definitely significant here since there are no bigger uncertainties in the repeats.
I'm guessing you could, but the uncertainty equation is: range/2
I would do as carpetguy suggested and explain why. I don't know exactly what you're doing but it's almost certain that you have more significant uncertainties on other readings. I don't think you can ignore it altogether or assume that it's zero since your lightgate is effectively reading only to 2 sig fig.
I was taught for digital readings if you had a singular reading or several readings that were all the same, you would do ± the resolution of the instrument/the smallest measurement the instrument could take. So in this case it would be ±0.001s
If the gate spits out answers with only 0.001 set intervals between values, then half this and make it the +- uncertainty. This is because you can round any value within that uncertainty range back to the same number with the same precision as the original (of course the upper bound doesn't really work but it's on the boundary). What that really is saying is that the measurement device wasn't good enough to tell me anything more accurate than this range - and the real value could indeed be within here. However, you do know that, according to the equipment, the value is not outside of this range.
I'm surprised at all of these "I was taught" and "formula" stuff. Intuition is good sometimes.
Also (and sorry to be a pedant but you can lose marks for these things) remember that if the reading is 0.032 then the uncertainty would have to be 0.000 rather than 0.00 - ie the same number of dp for both.