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0

Just write it out like this: $t = \frac{e}{\frac{m}{shc}}= \frac{J}{\frac{kg}{\frac{J}{kg\cdot{}^\circ{}C}}} = \frac{J}{J \cdot kg \cdot \frac{1}{kg\cdot^{\circ}C}} = \frac{1}{\frac{1}{^{\circ}C}} = ^{\circ}{\rm{}C}$

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When multiplying or dividing units, all you need to do is put the units in the numerator or denominator (wherever they appeared) of the answer. So: $$[e/M]={J\over kg}$$ $$[M/shc]={kg\over{J\over kg^oC}}={kg^2\,^{\circ}\rm C\over J}$$ But this is not the correct way of analyzing your units. You have $t = e / M / shc = e / (M * shc)$ The units of this are: ...

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I think that it is called 'packing efficency'. In euclidien 3D space the sphere minimizes the surface. With a fractal object you can maximize the surface without limit, I think.

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If you divide a volume by an area you get a length (as you have found), this length is physically just the length of a cylinder, using the XKCD example (you could use any n-sided prism) where the circle face has an area equal to the surface area (of your original shape). you can see this image that demonstrates it: NOTE: The scale between the sphere and ...

1

let's consider some simple examples: a sphere, a cube, and a rectangular parallelepiped. Let's denote the radio of the volume to the surface area of a given object by $\ell$, then we have \begin{align} \ell(\mathrm{sphere}) &= \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{1}{3} R = \frac{1}{6}D \\ \ell(\mathrm{cube}) &= \frac{L^3}{6L^2} = ...

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For the case of a sphere the ratio you found is: $$\frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$$ We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus. One approach is then to ask "what is a function divided by its derivative". This is really similar to ...

1

The physical representation depends on the geometry of the system. In the case of a sphere, then we have the simple result $$\frac{V}{A} = \frac{(4\pi/3)R^3}{4\pi R^2} = \frac{R}{3}.$$ That is, the ratio is one-third of the radius. Now spheres are special in that they maximize this ratio. For example suppose you had a cube of side length $s$. Then $$... 1 Converting units is simple if you simply work with the units as algebraic symbols. Thus start with the equalities$$1\textrm{ cal}=4.18\text{ J},1\text{ day}=24\times60\times60\text{ s}=86,400\text{ s, and}1\text{ cm}=0.01\text{ m}$$and just plug the numbers in:$$ 1\text{ cal cm}^{-2}\text{ day}^{-1} =\frac{4.18\text{ J}}{(0.01\text{ m})^2 \times ...

2

Bonus points for checking your dimensions. Your forumlas omit a lot of constants and assumes no atoms in the ground state, but that does not really matter for getting the units right. For that you need to realize that: The unit for cross section is meter squared [m]x[m], not the reciprocal. When you take e to the power of something, the "something" has to ...

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Many questions here. Well, units have a formal definition in the sense that every physical quantity has two parts: A quantity and a unit. Take "distance". How would you describe distance using just a number? That's impossible. Stating just a number would immediately lead to the question "... of what?". Think about it like this: Everything you measure is ...

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The colour charge of quantum chromodynamics is, as far as we can tell, not experimentally measurable, because of quark confinement. More specifically, quantum chromodynamical systems are always colour neutral. Quarks do have a color charge, but they are always observed in groups of two (color + anticolor) or three (red + green + blue) for which the total ...

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A coupling constant in the interaction Langangian plays the role of the charge. For example, for electrodynamics the interaction term of Langangian has the form $$\mathcal{L}_\text{int EM}=-ej^\mu A_\mu=-e\bar{\psi}\gamma^\mu\psi A_\mu$$ where $j^\mu$ is a current of electron (similarly $\bar{\psi}\gamma^\mu\psi$ is the same current calculated from the ...

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There are various systems of units that fall within the family colloquially referred to as the metric system. The SI (formerly known by the more descriptive term mks) is based on the meter, kilogram, and second. The cgs system is based on the centimeter, gram, and second. As far as I know, nobody actually uses an mgs (meter-gram-second) system, although it ...

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Among the base units of the International System, the kilogram is the only one whose name and symbol, for historical reasons, include a prefix. Names and symbols for decimal multiples and submultiples of the unit of mass are formed by attaching prefix names to the unit name "gram", and prefix symbols to the unit symbol "g" (CIPM 1967, Recommendation 2). ...

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Degrees and radians are just different units for the same quantity, angular displacement. So your question is fundamentally the same issue as whether you should use, say, meters or feet to represent a distance. You just have to convert the quantity in each case to the unit that your code expects. It's a little confusing in this case because in some cases, ...

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I am not sure about this, but I think a “measure equation” is something astronomers seem to like a lot: $$\frac\Gamma H \approx \left( \frac T{1.6\cdot 10^{10} \, \mathrm K} \right)^3$$ Or for absolute, relative magnitude and distance (although I am sure I mixed something up): $$m - M = 5 - 5 \log\left(\frac{R}{10 \, \mathrm{pc}}\right)$$ So equations ...

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There were historically several systems of units (ancestors to modern SI, CGS electric, CGS magnetic, CGS Gaussian, CGS by Heaviside), and the ultimate choice in favour of Gaussian CGS was made when Special Relativity has united electric and magnetic fields into one electromagnetic field tensor. Only in Gaussian (and Heavisidian) versions, these fields take ...

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The natural way to write it in this notation is $$F = q(E + \beta \times B)$$ where $\beta$ is the velocity measured in natural units - the velocity as a fraction of the speed of light. In the CGS system, we instead write $\beta = \frac{v}{c}$ and the equation becomes $$F = q(E + \frac{v}{c} \times B)$$ That's not silly enough, though, so we go really ...

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Anyone see my mistake? The third line: k = 0.3906 lbf-in/rad = 0.044135 N-m/rad is IMHO mistaken; better might be instead: k = 0.3906 lbf-in/rad = 0.013452 N-m^2/rad where 0.013452 == 0.3048 * 0.044135. Now good luck finishing your homework ... [Edit May 5th -- correction: for dimensional consistency "lb f in / rad" is converted to "N m^2 / ...

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Let's square root 0.1kg: expressed in kg, we get $\sqrt{0.1}\approx 0.316$. expressed in g, we get $\sqrt{100}=10$. So obviously the unit changes. If it stayed the same, we'd have $0.316\mbox{kg} = 10\mbox{g}$ which is clearly false.

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Take the root of the unit of area (Eg: 4 m$^2$ ) We get the unit of length (Eg: 2 m) which is the unit for different physical quantity So it definitely changes

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As the other answers (and dmckee's comments) note, yes, if you take the square root of a dimensional quantity then you need to take the square root of the units too: $$\sqrt{4\;{\rm kg}} = 2\;{\rm kg}^{\frac12}$$ And no, I can't think of any meaningful physical interpretation for the unit ${\rm kg}^{\frac12}$ either. However, in the comments you say ...

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Yes, the dimension of a quantity changes if it is square-rooted. If $m$ is a mass with dimension $[m]=\textrm{kg}$, $\sqrt{m}$ is not a mass, but another quantity with dimension $[\sqrt{m}] = \textrm{kg}^{1/2}$. More generally, if $[a] = A$ and if $[b]=B$, then $[a^n b^m] = A^nB^m$ etc.

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It becomes the square root of the unit. Think of energy: $$E = \frac{1}{2}mv^{2}$$ If I solve for $v$, I have $v = \sqrt{\frac{2E}{m}}$. Since $\rm 1 J = 1 kg \cdot m^{2}/s^{2}$, we see that the units have to obey the square root, or we will end up with our velocity equalling something other than m/s.

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A simple googling (takes barely 30 secs max.) can tell you the answer. Moment is something that has the "distance" along with it. Hence, moment of mass is $\mathrm{m\times kg}$. If you require another moment, $\mathrm{m^2\times kg}$. Probably, these first & second moments are engineering parameters. Whatever, it's simply the first moment of mass..!

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First moment of a mass distribution.

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