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A frame is one image (produced by some imaging device such as a computer monitor). Frames per second is therefore the measure of how many unique images are produced in one second (i.e. the frequency of the frames). Hertz is the SI unit of frequency, typically used as a measure of cycles per second. When you're referring to cycles or frames, you're not ...


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"Frames" is not a unit of anything. A frame is a thing. FPS in Hertz measures frames in one second. More generally Hertz can be used as the unit of any "thing" per second. In the case of an oscillating wave we measure cycles per second in Hertz. When I was young, there was no "Hertz", and the units were "cps" and "fps". Those old designations were ...


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What you have listed is not a formula: It's the gravitational constant itself. Getting an approximate for the gravitational constant should be fairly simple, experimentally. All you need to do is jump. Here's the equation you need: $$F = \frac{G M m}{r^2}$$ Where $F$ is the magnitude of the force exerted on each of the two objects in question, $G$ is ...


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The symbols $\text{m}$, $\text{kg}$, $\text{s}$ aren't variables that depend on a particular situation. They are the units. You can write $G = 6.67384\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}$, which means that $G$ is measured in cubic meters per kilogram-squared second, exactly the same way that you can say the Earth's radius is $6370 ...


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You're confused. The number you're referring to, $6.6738410\times 10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, is Newton's gravitational constant, $G$. This is not a formula or equation. The 'variables' are actually units: meters (m), kilograms (kg), and seconds (s). The formula where this constant most commonly appears is for Newtonian gravity: $$ ...


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This isn't an equation so there isn't anything to "solve". In terms of what the symbols mean "m3 kg-1 s-2" is just giving you the units of the gravitational constant, which in SI are meters cubed per kilogram per second squared. E^-11 is just a short hand for $\times 10^{-11}$ . The whole thing is just a number (indeed a constant!) \begin{equation} G = ...


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I like alemi's suggestion in a comment of crash density, by analogy with the linear "mass density" for a rod. Among other advantages, this frees up "crash rate" to mean the number of crashes per million hours driven. Alternatively you could invert the crash density to talk about the "mean distance between crashes."


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You could use the generic/boring term ratio, and precede it with CMK for crashers per million kilometers. The entire thing then becomes the CMK ratio. Though CMK rate can also work as others have said so long as your definition is clear. Then there's CMK factor if you want to use a fancier but more ambiguous term.


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Collider physicists actually use a quantity called "luminosity" which has inverse areal units, and they just quote as such (note that a "barn" is a (very small) measure of area), so when they say "inverse-femtobarns" they mean the inverse of an exceedingly small area which equates to a very high luminosity. There is nothing fundamentally wrong with "inverse ...


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The term "Crash Recurrence" comes to mind. http://dictionary.reference.com/browse/recurrence an act or instance of recurring. return to a previous condition, habit, subject, etc. It captures the concept of an interval without implying that it is regular or predictable, which seems to be what you're looking for.


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You feel uncertain because what you're trying to do is dimensionaly unsound. What unit is $F$ ? There is no answer to that question. And here's how you can convince yourself of this. The value of $F$ depends on the units of $E_0$. In one case you can have a purely imaginary $F$, in another a purely real one, and in yet another a linear combination of both ...


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$E_0$ better be dimensionless as the exponential function is defined for dimensionless arguments. If it has some dimension, it needs to be multiplied or divided by a constant to fix that. Yes, you can take that constant to make the exponential $-1$, but usually $F$ has units as well and there is some constant multiplying the exponential. In that case, the ...


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That's not how unit changing works. If $E_0 \neq 1$ in the units in which your equation is written, then to change to units in which $E'_0 = 1$ you have to introduce some (possibly dimensionful) conversion factor $k$ so that $E_0 = k E'_0$. So all you do with such a change is $$ F = \mathrm{e}^{E_0\mathrm{i}\pi} = \mathrm{e}^{kE'_0\mathrm{i}\pi} ...


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You know that in GR you need a locally Minkioski spacetime. This, in each point of your manifold you can change the coordinates so that the metric is diagonal, and the square of the infinitesimal displacement is $ds^2=\left(ct\right)^2-x^2-y^2-z^2.$ So here is where the $c$ come from. Then, when you want to compute the coupling constant $k=\frac{8\pi ...


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I'd have used $\Omega$ rather than $W$, and I'd say either way that it's dimensionless. Additonally the Boltzmann constant serves only to define the temperature unit (in the case of SI that's kelvins). So your presmise is incorrect and that's the origin of your problem. Read a textbook, a good one is Schreoeder.


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Since this has been open for a while now, here's the solution. The thing with unit conversions is to remember to be systematic and to double-check for typos and power-of-ten slipups. Even after fifteen years of unit conversion problems I find stupid little mistakes when I write the things out in full like I have below; if I try to "save time" by not writing ...


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As a metrologist, I am glad of this interest in correct notation, often not enough pondered also among metrologists but essential for understanding with each others. I would say first that any numerical value of an experimental result is always expressed as a rational, not irrational, number, because the number of digits is always bounded by the position of ...


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To obtain a limit $$f_k=\int^\infty_{-\infty}f(x)\exp\left(-2\pi i k x\right) dx$$ from a discrete Fourier transform, you must actually look what you mean by the formula $$f_k=\sum_{i=-N/2}^{N/2}f_i \exp\left(-2\pi i k x_i\right).\;\; (*)$$ This formula is taking samples of undetermined step in the $x$ direction and it will not converge to a Fourier ...


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The formula for $N$-DFT should be: $$\tilde{F}[k] = \sum_{n=0}^{N-1} \tilde{f}[n]\exp(-2\pi ikn/N),$$ where $\tilde{f}[n]$ is the discrete input and $\tilde{F}[k]$ is the discrete frequency output. One can optionally scale by $N^{-1/2}$. The indexes $n$ and $k$ are dimensionless. Usually $\tilde{f}[n]$ is obtained from $f(x)$ by sampling, which involves ...


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You're mistakingly assuming that a sequence must have exactly the same unit to be considered the approximation of a signal. In general, a sequence $f_k$ can be considered as an approximation of $f(x)$ if $f_k=\int^{x_{k+1}}_{x_k} f(x) dx $. Clearly the units differ here. Note that by this rule, $f_k=0$ is an approximation of $f(x)=0$, $c*f_k$ is an ...


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I don't understand why the time component of the 4-vector [ $m~\gamma(|\vec v|)~(c, \vec v)$ ] is being denoted as $E/c$. So the underlying question is two-fold: Why is "energy" considered the time component of some 4-vector at all?, and Why this specific time component expression, among time components of all different 4-vectors imaginable? (Where ...


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In short, up until now, the $kg$ was arbitrary, but now people are trying to define it based on universal constants There is a ongoing process to try and link all the units to universal constants. This has been done already for the second (using Cesium) and the meter (using the speed of light in a vacuum) However, the kilogram is a little less ...


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Talking about a "viable" source, we can talk about time, as in how long does it take to go from my place to the grocery store? 10 minutes by bicycle. How long does it take from LA to Boston? 0.04 light seconds, or 4 car hours, or 5 weeks on a boat, etc. So we see that we can use time to define other units, not to mention we really do use time to define many ...


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The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom. How is this period calculated? The sensor has probably some equations that imply transformations using other units like Kg etc... The number of periods of anything isn't calculated, it's ...


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Your book may be treating things a little backwards from the way they are usually done. The usual way is to define the momentum four-vector as the combination $(E/c, \vec{p})$, where $E$ is already known to be the total energy (the thing that reduces to $mc^2 + \frac{1}{2}mv^2$ for $v\ll c$) and then go on to show that it satisfies the properties expected of ...


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Variables, units and dimensions are fundamentally different concepts. When writing symbols it is conventional to use different fonts to avoid confusion. Quantity symbols (variables) are written in italic font, e.g. $A$. Units are written is upright serif roman font, e.g. $\textrm A$. And Dimensions are written is sans serif roman capital font, e.g. ...


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When you're doing Fermi estimation, it's convenient to use a "system of units" where $$ \sqrt{10} = \pi = 3 = 1 = -1 = c $$ though this is probably more tongue-in-cheek than you were hoping for. For some reason I think of these as "pauper's units," though I'm not sure whether that phrase is my own invention or not. You might be interested in this other ...


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Ok, here's why rescaling units won't get rid of $\pi$: The circumference of a circle is $C = 2\pi r$. Note that $C$ and $r$ are measured in the same unit, both are units of length. Now rescale lengths as $l' = c l$ with an arbitrary constant $c$ (might be $\pi$, but doesn't matter). Plugging in yields $$C' = cC = c 2 \pi r = 2 \pi c r = 2 \pi r'$$ So you ...


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Planck units is almost (except coefficients like $\pi$) exactly the unit system you want, and it is frequently used in quantum field theory. According to its definition: $c = G = \hbar = k_\text{B} = 1 \ $


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Yes, "3 apples" is a value of a dimensionless (unitless) quantity, the number of apples. Quite generally, quantities that are either integer, or very special if they are integer, are unitless. All 7 base units of the SI system, or any product of their powers (derived units), has the property that it measures an intrinsically continuous quantity such that ...


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ACuriousMind's comment in answer form: $kT$ is something that appears all the time in equations in thermodynamics. You can always choose suitable units to ensure that the quantity $kT = 1$ (think of it as some kind of normalisation if you like!). To convert back to an answer that includes those $kT$'s we simply perform some dimensional analysis and insert ...



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