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The nearest we have to a standard is International Atomic Time. This is: TAI as a time scale is a weighted average of the time kept by over 200 atomic clocks in over 50 national laboratories worldwide. The errors in individual clocks can be assessed by comparing them to the weighted average. Re the mention of pulsars in your comment, pulsars slow down ...


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The weighted mean of three values is given by $$ \bar{x} = \frac{ \sum_{i=1}^{3} x_i \alpha_i}{\sum_{i=1}^{3} \alpha_i},$$ where here $\alpha_i$ represents the weight that you give to each measurement. If you wished to just find the weighted mean from your data as you presented it originally, then the weight $\alpha_i = 1/\sigma_i^{2}$. If you then wish to ...


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The way that I would approach this is to calculate $\chi^2$ with $$\chi^2 = \sum_{i=1}^N\left({p - x_i \over \sigma_i}\right)^2$$ where $p$ is the best value for the fit and $N$ is the number of measurements $x_i$ with error $\sigma_i$. Now with a computer or Excel sheet I would plot $\chi^2$ as a function of $p$. As you know the value of $p$ for the ...


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If you are measuring in a laboratory with a ruler like the one in your diagram then I would say for a length of $9.5 cm$ you would be able to see with your eye that the length is say $9.5 \pm 0.2 cm$ and if it actually was on one of the markings, e.g. 6, then you might estimate that the measurement was say $6.0 \pm 0.1 cm$. Often when measuring length with ...


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I think what you could do is treat the quoted value as the most likely value and the two uncertainties as semi-gaussian probability distributions. So for example in error propagation you would have to propagate the upper and lower limits separately using the usual Gaussian error propagation formulae but keeping track of whether these propagate to the upper ...


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Now let's find out its volume: 16777.216. Now if I wanna report my result with the least significant figures (3), I would get an answer 168 which is plain wrong. Your problem is that you seem to be truncating the value! When employing the significant figure rule, you turn all other values to zero. Trailing zeros are placeholders, so they can be ...


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if error in wavelength is $\delta \lambda$ and error in frequency is $\delta f$ then $${\delta f \over f} = {\delta \lambda \over \lambda}$$ so $${\delta \lambda } = \lambda {\delta f \over f}$$ The formula above should give you something more reasonable for the error in the wavelength. More generally if $$ X=kA^n$$ where k is a perfectly known ...


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$\delta (nT ) = \delta n *T + n * \delta T $ $\delta n = 0$ So, $\delta (nT) = n* \delta T$ And therefore $\delta T = \delta (nT) /n$ In other words, your error in $T$ is just your measured error divided by $n$. So your calculations were right. This makes an intuitive sense because if we imagine that the error in the measurement is constant, then for ...


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This question confuses two frameworks, the classical thermodynamic one with the quantum mechanical one. A perfectly homogenous radiating body belongs to classical thermodynamics. Photons belong to quantum mechanics, as well as the uncertainty principle. Lets look at the problem classically. Electromagnetic radiation is continuous from a classical body ...


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The uncertainty in any particular measurement is $\sigma_E$. Resolution for these devices is almost always stated in relative terms as here, but take it like this because it depends on the energy measured. So just multiply by the energy. That is, express your signal in $\mathrm{GeV}$ and then find $$ \begin{align} \sigma_E = \left(\frac{0.1}{\sqrt{E}} ...



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