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1

There's a lot of context that goes into deciding which digits are significant. I like to use the "anger management method" for deciding which digits are significant. Suppose you are shopping for a fancy television. You see an advertisement that there's one you like on sale for \$1369.99. You know that your area has 10% sales tax, so you predict that when ...


3

In physics, all numbers are imprecise. Significant figures are conceptually the digits that it is meaningful to include in the reported result. The concept When you read a gauge, the reading error is half the smallest interval on its scale - you basically take the value of the tick nearest to the hand's position (digital gauges do this for you). So, if ...


9

I teach high school in the United States. I want to preface with that, because conventions common in one context are not necessarily universal. That being said, it's pretty standard teaching practice here (at least in every class I've ever taken, taught, or known a colleague to teach) to assume that trailing zeroes are not significant unless otherwise ...


3

You are completely right, it is a confusing case you have. The number $1500$ does have 4 significant figures as it is. But, you are told it has only 2. This is strictly speaking not correct. But it is just a shorthand way of writing $1.5 \times 10^{3}$ (or $15 \times 10^{2}$). It is an easier way to write a number with not so high an order of magnetude. ...


13

This depends on the context. If you have 1500 of something and you counted them yourself and you are sure you have precisely 1500, then all four figures are significant. On the contrary, if you're guessing that you have 1500, implying a certainty of order 100, then only the leading two figures are significant. In scientific notation, one would write the ...


2

I was thinking how, since an object in our universe can move from one position to another, it must have passed through all the positions between those two positions. (I am thinking it moved it a straight line) This must mean that in actuality there are only so many positions between those two points doesn't it? There must be some maximum accuracy to ...


0

There are objects in the universe yet to be discovered as shown by constant new discoveries of new types of planets under conditions that were previously considered impossible. (Diamond Planet, Fire Ice Planet) I believe this works the same way with objects in motion as well. If the Big Bang theory is correct, and our current interpretation of how physics ...


1

From a trivial point of view and in line with what you say $x \propto y \iff x=k_{1}y$ for some constant $k_{1}$. On the other hand $x \propto \frac{1}{y} \iff x=\frac{k_{2}}{y}$ for some constant $k_{2}$. If none of these relationships holds, then there may well be an error in the assumptions in proportionality.


1

I think the error arises from the fact that it is common to use $$9.80 m/s^2$$ Perhaps this is what was meant.


3

Spectroscopic parallax is the technique whereby you estimate the absolute magnitude (i.e. the brightness it would have if it were placed at 10 pc) by estimating what "type" of star it is using information fro a spectrum. It can be applied to any kind of star where (a) you have a reasonable chance of determining the type of star from its spectrum and (b) ...


0

From you question it is not really clear how you calculated the values, so I makes some guesses here. From the 10 measurements for each point you get some statistical error for the measurement device. This includes, however, also fluctuations of the source etc. You don't know systematic errors, like an offset due to stray light or other sources. So for each ...



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