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17

You've probably heard of Einstein's famous equation: $$ e = mc^2 $$ This states that mass and energy are equivalent, and indeed the LHC turns energy into matter every day. So to find the mass equivalent to an electron volt just convert eV to Joules and divide by $c^2$. 1 electron volt = $1.60217646 \times 10^{-19}$ joules, so 125 GeV is: $$ 125 \times ...


11

Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ...


9

An electron volt is just the energy acquired when an electron of charge $e$ falls through a potential of 1 volt, which means $$1eV = e \times 1 = 1.6 \times 10^{-19} J$$ When you lift up your $2.5Kg$ laptop (a 15-inch apple macbook pro, for example) by a foot, you do a work of approximately $2.5 Kg \times 10 ms^{-2} \times 0.3 m = 7.5 J$ which is about $4.7 ...


9

The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form $$(\text{number})(\text{unit}) = (\text{other number})(\text{other unit})$$ But of course, you can divide both sides of any equation by the same ...


9

The superscript $^2$ in $1750\text{ mm}^2$ refers to a squaring of the units, not the number $1750$. A more transparent way to write this is $1750\text{ mm}\cdot\text{mm}$. The idea is now to multiply by $1$, but $1$ written in a clever way: $$1=\frac{1\text{ m}}{1000\text{ mm}}$$. Can you see how that number is conceptually equal to $1$? The top and ...


8

Pure convention. There is no reason alternative conventions couldn't be used, apart from the need to avoid confusion. Newton introduced the constant to make the force law simple, whereas the electrostatic definition with the $4\pi$ is designed to make Poisson's equation (one of the equations for the electric field) look simple. You can write a Poisson ...


7

A picture, as they say, is worth a thousand words: See that, in fact, the ratio of Fahrenheit to Celsius is: $$ \dfrac{32 + \frac{9}{5}x}{x}$$ where $x$ is the temperature in degrees Celsius. Clearly, the reason the ratio is not constant is the presence of the constant offset 32. Only in the limit of arbitrarily high temperature does the ratio ...


7

Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in CGS Gaussian unit this is not (1 unit of current = 1 g1/2 cm3/2 s-2), by setting $\epsilon_{0,SI} = \frac1{4\pi}$. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is ...


6

Which units are fundamental and which are derived is pretty much a matter of arbitrary convention, not an objective fact about the world. You might think that the number of fundamental units would be well-defined, but even that's not true. Take electric charge for example. In the SI system of units (i.e., the "standard" metric system), charge cannot be ...


6

It's an energy. The SI unit for energy is a Joule. If the theory is going to predict an energy-mass correspondance, then it better give the energy in units of energy. If the units of $mc^{2}$ didn't work out, for the equation to make any sense, you'd have to include some constant $\alpha$ so that $E=\alpha mc^{2}$ would have units of energy on both sides. ...


5

As far as I can tell, you went wrong in a couple of places: instead of converting from $\mathrm{g}$ to $\mathrm{mg}$ your converting to $\mathrm{kg}$ in your first term, and in your fourth term, the units should be $\mathrm{g\, O_2}$ and $\mathrm{ nmol\, O_2}$, for consistency. This would've let you know that you've inverted your conversion from $\mathrm{ ...


5

It is the convention of setting the velocity of light $c=1$ that allows for this, the natural units, otherwise it is $\mathrm{GeV}/c^2$ The rest mass energy connection $$E^2=p^2+m^2$$ at rest then the mass is identified with energy in natural units.


5

As lurscher mentioned in a comment, you're using the wrong units for magnetic susceptibility. $\chi$ is actually a dimensionless number that is related to the magnetic permeability of a material relative to that of a vacuum. I think you were mixing it up with the molar magnetic susceptibility, which is $\chi_\text{mol} = \mathcal{M}\chi/\rho$, where ...


5

From Kepler's third law you can find that $$ \frac{GM_\odot}{4\pi^2} = 1 \frac{\text{AU}^3}{\text{year}^2} $$ where $M_\odot$ is the mass of the sun. For a solar system simulation these units will be more convenient than Earth masses.


5

I think there's a genuine and interesting physical point to be made here. Taking a slightly different example, the gravitational acceleration of a massive body on a test particle is $a = GM/r^2$. If you can measure $a$ and $r$ accurately then you can find $GM$ to equal accuracy. But to find $M$ you also need to know $G$, and $G$ is rather difficult to ...


5

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 ...


4

This is a typical "unit conversion" problem. Write $G$ in SI units: $$G=6.6738\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}.$$ Now find out how many kilograms are in an Earth mass, and how many meters are in an astronomical unit. Also consider converting seconds to some other more convenient measure of time so that $G$ comes close to unity. ...


4

Defining the symbol $k$ in Coulomb's law, $$F=k\frac{q_1q_2}{r^2},$$ to be $k=1/4\pi\epsilon_0$, is perfectly allowed when one understands it simply as a definition of $\epsilon_0$. The motivation for this definition is that when you work out the forces between two oppositely charged plates of area $A$ and charge $Q$ a distance $d$ apart, they come out as ...


4

If the height difference between the mercury level in the two arms is $h$ (it's called $\Delta h$ on the figure), then $$P_1 - P_2 = h\rho g$$ where $P_1,P_2$ are the pressures in both wings (called $P,P_{\rm ref}$ on the figure). One of them is the measured atmospheric pressure. The two pressures are being subtracted because the air pushes the liquid ...


4

The unit can be anything as long as you carry out the proper conversion. If you're using $KeV$ and $MeV$, there isn't any complete unit system that specifies the units of speed (which we need to convert to mass). Usually, particle physicists use $MeV$ as a mass unit as well as energy (thereby setting $c=1$ in this unit system). Later on stuff can be ...


4

I suppose (or hope, for I'd agree) he means equations where the variables are plain numbers rather than physical values, and the units are written out in the equations. Like $$ F\:\mathrm{N} = \frac{E\:\mathrm{J}}{s\:\mathrm{m}} $$ i.e. "$F$ Newtons equal $E$ Joules over $s$ meters". Which is a really horrible way of writing equations, and particularly ...


3

I think it has something to do with old electromagnetism units, which was vastly more complex than centimetres to inches conversion. I'm too young to have been exposed to the horrors of -cgs--units in electromagnetism so all the following is only an educated guess. My guess is due to some googling, with "measure equation" coming up in old papers associated ...


3

Because of the definition of Joules in terms of the base kg/m/s units. If you used lbs feet/second and calories you would need a conversion factor, because calories are defined in terms of heating water rather than mechanics, but the equation still works.


3

The simplest way to think of Avogadro's number is as a unit conversion factor. Just as there are 2.54 centimeters in an inch, there are Avogadro's number atomic mass units in a gram. Accordingly, if you know the atomic mass of an element you can count the number of atoms in one gram of it. Let's take carbon as an example. The most common isotope of carbon ...


3

When people say they are using CGS units, they mean: Centimeters are the implied units of length; Grams are the implied units of mass; Seconds are the implied units of time; Anything involving electric charge will be defined in accordance with $4\pi\epsilon_0 = 1$; and All other units are the same as in SI. Some might argue that this last point is not ...


3

This is difference in the unit system. The former uses SI units, where the latter uses cgs/Gauss system. In Gaussian units, unlike SI units, the electric field E and the magnetic field B have the same dimension. This amounts to a factor of c difference between how B is defined in the two unit systems, on top of the other differences. ...


3

Linearity does not necessarily mean a ratio exists. Linearity in this case means that a given change in one variable causes the same change in the other variable no matter what the value of the variable was before the change. As explained by the other answers, $F=32+\frac95 C$. $F$ is a linear polynomial in $C$ (i.e, $F(c)=aC+b$), but $F$ is not a linear map ...


3

Take a specific example: two charges of 1 coulomb separated by 1 meter. In MKSA, (now better known as SI) the force between them is given in Newtons, by:$$F=\frac{k q_1q_2}{r^2}=8.98\times10^9\text{ Newtons}$$since all the variables are 1. So now you want to do the same problem in cgs-electrostatic units. $k=1$, $r=100$, and most importantly, ...


3

The unit of the second is defined is the time duration of a certain number of periods of radiation emitted from a particular type of electron transition between energy levels in an isotype of Cesium (see here). It is an assumption that light travels at a constant speed $c$ independent of one's reference frame, so now that we have fixed a unit of time, we ...


3

By definition 1 N = 1 kg m/s^2 To remember this, remember F = ma. The left hand side has units of force. The right hand side is kg * (m/s^2). dividing both sides by m^2, we get 1 N/m^2 = 1 kg / (m s^2)



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