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1

The problem with your question (and figure) is that the resolution is well beyond the gradation of the instrument, and that is misleading. So let's suppose can't really distinguish 3.1 from 3.5, all we know is that 3.1 is closer to 3 than 5, and while 3.5 is in the middle, we'll round it to 4. In this case the uncertainty is the standard deviation of the ...


1

This is because the first part of the value represent the significant digits that are certain. The answer must contain extra significant digits on top of the ones that are certain. Having taken the reading, you are 100% sure that the ones digit is $3$ and not $4$ or $2$. Writing $3.0$ would mean that you are sure of the length to be $3.0$ which you are ...


0

Velocity (actually here speed) is given by: $$v= \frac {s}{t}$$ Now the percentage error in the measurement of velocity is given by: $$ \delta v = \delta s + \delta t$$ The percentage error of say a quantity $a$ is given by: $$ \delta a = \frac { \Delta a}{a} \times 100\%$$ So for finding the percentage error of velocity you must have velocity $v$, ...


0

From a comment. So I have no way of evaluating the accuracy of data? The graph is from the Handbook of Chemistry and Physics if that means something I said that you should look in the source material for hints. The CRC handbooks are authoritative sources (well, for the year they were printed, anyway) and are prepared by metrologists to a very high ...


0

If this is for homework, your best bet is to do the old "manual digitization" and then use the produced data itself as a source of error in the result. Scan the image into your favorite electronic device. Then put the graph into your favorite spread sheet program. Then build a polynomial graph that fits over the scan as closely as possible. Then use how much ...


2

I think you're grossly over-complicating things here. I've listed the convention below for most high-school and undergraduate level treatment of uncertainty. $$\\$$ When a measurement is made with an analogue device, such as a ruler, or a thermometer, then our error is taken to be half of the smallest division. However, this doesn't mean you can't record ...


1

You have used values that are given to up to two significant figures to calculate the same quantity twice. The results are consistent to two significant figures. I don’t see any problem here, and it certainly doesn’t imply that $c$ and $h$ are not constant.


1

You were done after the first two lines of math. There is no quantum mechanics involved in this relationship. It's purely classical. The discrepancy at the end is because you rounded excessively.


1

Maybe some examples would help. $$arccos(0) / pi = 0.5$$ $$arccos(0) / pi = 0.500000000000$$ ... does that second one make any sense? Not really. I mean, yeah, it's exactly 0.5, and so is 0.50000000000 - but those zeroes don't really add anything to the picture. But what about: The event took 0.5 seconds to occur The event took 0.5000000000 seconds to ...


3

It's Experiment Time! (I was starting to see both points of view on whether to drop the 1, and was curious if there was some objective way of tackling the problem... so I figured it might be a good opportunity for an experiment. For Science!) Assumptions: Significant Digits are a way of signifying precision on a number - either from uncertainty of ...


4

First, if I'm measuring something that has an anticipated error of $\pm 0.01$ and I get $2.50$ five times in a row, I'm going to worry that my measuring equipment is stuck on $2.50$, or that the process that I'm measuring has variation of $\pm 0.01$ that's really slow, or that my experiment is otherwise broken, etc. But let's say I'm measuring something ...


21

The first thing to realize is that sig fig rules are just a rule of thumb. They're a quick and dirty way of doing error propagation, but if you want a totally correct error propagation, do real error propagation. In your example, however, the rules can be interpreted in a way that might make sense, depending on the context. In a computation like $2.50+2.50+...


2

Keeping track of "significant digits" is a heuristic for indicating approximately the precision of a number. It's not a substitute for a real uncertainty analysis, but it's good enough for many people and many purposes. When some people run up against the limitations of significant figures, they have enough background (or colleagues with enough background) ...


15

This isn't an actual rule. And as some people point out in the comments, it's not even mentioned in the Wikipedia article on significant digits. The rule applies to $0$, not to $1$. Simple counter-example: $10$. Would the authors claim that this number has no significant digits? You can verify this by doing a search for "sig fig counter." All of them ...


80

Significant figures are a shorthand to express how precisely you know a number. For example, if a number has two significant figures, then you know its value to roughly $1\%$. I say roughly, because it depends on the number. For example, if you report $$L = 89 \, \text{cm}$$ then this implies roughly that you know it's between $88.5$ and $89.5$ cm. That is, ...


13

Truncating numbers to a certain precision is completely arbitrary. There's no reason not to make it more arbitrary. It seems like someone didn't like the step in precision between 9.99 and 10.0 so they moved it to between 19.99 and 20.0. In any field where results are clustered around a power of 10, doing this may be beneficial.


1

The statistical error is calculated from the size of the sample and various measures of how it might (statistically) vary. There's no single formula for that, but there are a few well-recognized methods: Poisson statistics for small numbers of events, normal (Gaussian) statistics for large numbers of samples from a single distribution, etc. The statistical ...


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