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0

Since the scales of the graph are in Newton and meters, your answer is fine. Think of integration as simply adding small individual work for small displacements. For a displacement ds, work done will be dw=F.s Its unit will be in J (or Nm), right. Integrating dw will not change the unit, as adding doesnt change the dimensions. Hope the explanation is ...


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The slope you calculated is $\frac{10 \rm N}{8\rm m}=1.25\,{\rm \frac Nm}$, so the force function becomes $F=1.25\,{\rm \frac Nm}*x$. Hence, the work is $$ W=\int{(1.25\,{\rm \frac Nm}\times x)dx}=1.25\,{\rm \frac Nm}\int{xdx}=0.625\,{\rm \frac Nm}x^2 $$ (If $F(0)=0$). Because the unit of $x^2$ is $\rm m^2$, then the unit of the work is, as expected, $\rm ...


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Kelvin is the SI unit. It is far more common than Rankine. I cannot recall ever encountering Rankine temperature units, except in historical or humorously-backward contexts. Note that these measure temperature, not heat. The SI and "imperial" measures of heat is are the joule and the BTU, respectively. To avoid causing headaches, use SI for everything — ...


2

It depends on which logarithm you choose. To demonstrate (but not prove) the information theory equation: the definition $S = \log W$ (uncertainty is the logarithm of multiplicity, Boltzmann's famous relation) becomes equivalent, if $p_i = 1/W$ ($W$ equally probable outcomes) to $$S = -\sum_{i=1}^W p_i \log p_i.$$But since $\log_a(b) = \log_c(b) / \log_c(a)$ ...


1

Let $x$ be dimensionless and Using the property $\delta (ax)=\frac{1}{|a|}\delta (x)$ we see that indeed the dimension of a Dirac delta is the dimension of the inverse of its argument. One reoccurring example is eg $\delta(p'-p)$ where $p$ denotes momentum, this delta has dimension of inverse mass in natural units.


3

The unit langley is amount of energy distributed over an area found in solar radiation, named after Samuel Pierpont Langley. The conversion is straight-forward: $$ 1\frac{\rm ly}{\rm min}=697.3\,\frac{\rm W}{\rm m^2} $$ So you'll have to do some more unit conversions to get the ly/day to W/m$^2$ conversion. See also this site for other unit conversions ...


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I believe the original article is http://research.smp.uq.edu.au/kassal/papers/kassal003.pdf and the section concerned (clipped from the article) is Note that the units of $\gamma$ can be derived by looking at the components whose dimensions are known (following @Omry's comment): $$kT = \mathrm{energy}\\ E_R = \text{energy}\\ \hbar \omega = ...


2

Supposedly, this constant should be in units of cm^-1 but when I do calculations with different values of temperature (in K), my units end up in cm/s^2. You did something wrong. $kT$ has units of energy. Assuming $E_R$ also has units of energy and $\omega_c$ has units of inverse time means $2\pi kT/\hbar\, E_R/(\hbar\omega_c)$ has units of inverse time ...


2

It's simple dimensional analysis. The theory has two fundamental parameters, Newton's constant $G$, which determines the strength of gravitational attraction. it has units $\frac{N\cdot m^{2}}{kg^{2}} = \frac{kg\cdot m}{s^{2}}\left(\frac{m^{2}}{kg^{2}}\right) = \frac{m^{3}}{kg\cdot s^{2}}$. Secondly, you have the speed of light, which tells you how much ...


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I think about these things in microscopic units, like eV/atom. However measurements are made with macroscopic amounts of material and tend to be tabulated in macroscopic units. The two units aren't that different ($\rm 1\,eV/atom \approx 100\,kJ/mol$) but pointless uncertainty can be introduced in the conversion.


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The mole essentially denotes "number of molecules". When you say "per mole", you are implying "per molecule", which is highly desirable since it directly converts a macroscopic property such as 'total heat released during reaction' to a microscopic property such as 'heat released per molecule'. Since $\epsilon$ is an intensive function as it is per ...


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You should probably still use kJ/mole for consistency. It is clear what you mean, just like you might use "1 g/L" even when dealing with quantities less than a litre. In general, when working with multiples of units derived as ratios of other units, it is better to scale the denominator than the numerator. Compare "km/s" and "m/ms". Both have a thousand ...


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Note on notation: I use $[\cdot]$ do denote the units of the quantity in brackets. Derivatives always have units of $1/[\text{differentiation variable}]$. This can be clearly seen from the definition of the derivative in terms of difference qutionts: $$ \partial_x f(x) := \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$ So if $x$ has some unit $[x]$ then ...


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The $\nabla$ operator is a spatial derivative of the $\frac{\delta}{\delta x}$ etc kind. This has units of $1/m$


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The possible choices of sets of units for your simulations is probably infinite, so the answer is ultimately going to be choose a set of units that fit what you need & run with it. For instance, suppose you want to study the Argon interacting via the Lennard-Jones potential, an appropriate choice of units could be mass, $\mu$, length, $\sigma$, and ...


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LAMMPS is a well known molecular dynamics simulation code, here is their documentation on the pages where they discuss units. http://lammps.sandia.gov/doc/units.html


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As a general definition, the solid angle is a three dimensional angle subtended by a 2-D or 3-D object at a certain point in the space. It measures that how closer an objects appears to the observer (at a certain point) in the space. Higher the solid angle closer will the object appear to an observer. Solid angle is generally measured in SI unit ...


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a) No, it has $N$ sets of parameters. Usually, you need two parameters per particle the position vector $\vec r_n$ and the spin projection $\sigma$. $\sigma$ is a discrete variable taking $2S + 1$ values. b) No, $\Psi$ encodes all information about the system (just as the positions and momenta of the particles encode all information in classical point ...



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