Answer: Calculating Relative & Absolute Uncertainty of R^-2

  • Thread starter Thread starter gcsm
  • Start date Start date
  • Tags Tags
    Relative
AI Thread Summary
The discussion focuses on calculating the absolute uncertainty of R raised to the power of negative two, given a distance measurement of 4.000±0.002 m. Participants clarify that the absolute uncertainty is 0.002 m, but there is confusion about whether the question pertains to absolute or relative uncertainty. It is suggested that to find the relative uncertainty, one must divide the absolute uncertainty by the measured value. The conversation highlights the need for clarity in distinguishing between absolute and relative uncertainties in calculations. Understanding these concepts is crucial for accurately interpreting measurement uncertainties.
gcsm
Messages
4
Reaction score
0
Absolute Uncertanity

Homework Statement


A distance R is measured to be 4.000±0.002 m
What is the absolute uncertainty in R to the power of negative two?


Homework Equations





The Attempt at a Solution



4^-2 = 0.00625
0.002^-2 = 250000

Im stuck at this question I have no idea where to go, any help would be appreciated!
 
Last edited:
Physics news on Phys.org
umm you already gave the absolute uncertainty, it is 0,002m...
 
but I tried that, meaning 0.002 and that's not it, I also tried 250000 (using SI) and that isn't correct either...not really sure where to go...
 
Are you looking for the absolute or relative uncertainty? as in the thread title you say relative, but then in the problem description you say absolute.

The absolute is 0.002m that is 100%, then most probably you are looking for the relative uncertainty.
That is divide the absolute by the average.
 
sorry for the confusion, the question is as follows:
What is the absolute uncertainty in R to the power of negative two?

but wouldn't you have to do something to the +-0.002m?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Homework help: Uncertainty with negative power
Replies
9
Views
2K
What Is the Relative Uncertainty in the Square Root of a Measured Distance?
Replies
4
Views
10K
How to calculate the uncertainty for weight?
Replies
4
Views
5K
How do I determine the uncertainty value of the star's absolute magnitude?
Replies
6
Views
3K
Relative uncertainty (negative exponent)
Replies
2
Views
4K
Calculating Relative and Absolute Error: How to Find f(r,h,m)
Replies
3
Views
1K
Measurement Uncertainty Problem in MIT OCW 8.01x
Replies
7
Views
2K
Absolute uncertainty of calculated density
Replies
3
Views
7K
Error propagation (tape shorter than measured length)
Replies
6
Views
2K
Absolute Uncertainty of 1/d^2 Homework
Replies
21
Views
5K

Hot Threads

Recent Insights

Back
Top