## anonymous one year ago How do I calculate the absolute uncertainty when there is an exponent? (see question 2)

1. anonymous

1) when the number is (5.3322∗ 10^-5) would the uncertainty be (1*10^-9) or (0.0001)?

2. anonymous

2) how to calculate the absolute uncertainty when there is an exponent? For example $(-7.5899*10^{-6}) *(22.62 \pm 0.01)$.

3. anonymous

for example lets say u are given 10.1 you dont know if it was 10.05 to 10.15 so your uncertainty is 10.1 +/- 0.05

4. anonymous

We were told in class that when the uncertainty is implicit we will assume its the last digit and it will always be 1. For example: 0.0967 the uncertainty will be 0.0001.

5. anonymous

(5.3322∗ 10^-5) same thing this part 5.3322 you dont know if it was 5.33215 to 5.33225 so 5.3322 +/- 0.00005

6. anonymous

ok i guess that maybe fine too its like roudning up the uncertainty itself

7. anonymous

+/- 0.00005 = +/- 0.0001

8. anonymous

ooohhhh ok. so the uncertainty will be determined regardless of the exponent.

9. anonymous

yeah that is right

10. anonymous

(5.33220∗ 10^-5) if they want to go upto exponent

11. anonymous

you have to add that 0

12. anonymous

ok. thanks!!

13. anonymous

sure np

14. anonymous

for your question 2, absolute uncertainty is that the uncertainty that is based on proportion

15. anonymous

iwhen y=x^n, is the formula n* (relative uncertainty), relative uncertainty= (uncertainty/value)

16. anonymous

okay i see

17. anonymous

here is a more logical approach to calculating your absolute uncertainty

18. anonymous

suppose you have 10.0+/- 0.1 and 2.0+/- 0.2 you see 10.0*2.0 = 20.0 then you see the max and mix that happens with the uncertainty 10.1*2.2 = 20.2 + 2.02 = 22.22 and the min 9.9*1.8 = 18-0.18=17.82

19. anonymous

(22.22-17.82)/2 would be the uncertainty?

20. anonymous

ok

21. anonymous

let me check how u actually find absolute uncertainties hmm

22. anonymous

look at this http://web.uvic.ca/~jalexndr/192UncertRules.pdf

23. anonymous

that is probably best forget what i said so far

24. anonymous

ok. thank you so much!!