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18

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: $$125 \times ... 13 The joule is the amount of energy needed to apply one newton of force for a distance of one meter:$$ \rm J=N\cdot m=\frac{kg\,m^2}{s^2}\tag{1} $$Where the 2nd equality comes from the definition of the newton (mass times acceleration): \rm N=kg\,m/s^2. The pascal is defined as one newton of force applied to a one-square-meter area:$$ \rm ...

12

The angle $\theta$ might have units of degrees, yes. It could also have been in e.g. radians. But putting it through the sine function removes the unit. The $\sin(\theta)$ is unitless. The sine and cosine functions are defined as the "distance" vertically and horizontally, respectively, to the point on the unit circle. That is, a distance per unit lenght. ...

12

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 ...

11

It's a unit conversion: $$1\,{\rm yr}=\frac{365\,{\rm days}}{\rm year}\times\frac{24\,{\rm hrs}}{1\,{\rm day}}\times\frac{3600\,{\rm sec}}{1\,{\rm hour}}=3.1556926\times10^7\,{\rm sec}$$ Since $3.1557$ is (somewhat) close to $\pi\sim3.1416$, we use the approximation you cite. Technically, the year is actually 365.25 days long, so using that gives a ...

10

It's just an artifact of using different units for the same types of quantities in the same equation. Suppose we have an ideal pump that puts a force $F$ on a fluid to have it move at a steady velocity $v$. The power required to do this is $$P = Fv.$$ Since pressure $p$ is force per unit area, then a flow with cross sectional area $A$ has $F = Ap$. At the ...

10

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 $4.7 ... 10 The superscript$^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 ... 10 This is a good question - as in the example with$L,\lambda$you provide, not every rescaling and not every set of constants is valid. The recipe for the set of good natural units is the following: take all the units that appear in your theory and create a space with one dimension for every one of them. Say we have a theory with time, length and energy - ... 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 ... 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 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 ... 7 The answer to your question lies in simple dimensional analysis. Joules are the units of energy, so also the units of work. Work, as we all know, satisfies the relationship $$W=\vec F \cdot \vec s$$ Meanwhile, $$\vec F= m \vec a$$ Substituting in this relationship gives us $$W=m\vec a \cdot\vec s$$ Now, let's look at the units of this equation: ... 7 The kinetic energy is given by$E_c=\frac12mv^2$, as the factor$1/2$is dimensionless, you can see that$\mathrm{[m^2.s^{-2}]=[J.kg^{-1}]}$. Dimension analysis remains correct if the velocity$v$takes the value$c$, because$c$is also a velocity. 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. ... 6 There is a conversion factor, but it isn't just a multiplication because the two scales do not have their zero at the same point. The celcius scale is defined as having its zero at the freezing point of water. The fahrenheit scale, on the other hand, appears simply arbitrary To convert from celcius to fahrenheit, you multiply by$9/5$(the ratio to ... 6 1 horse power = 33000 foot-pounds per minute (by definition) 1 US gallon = 231 cubic inch (by definition) 1 psi = 1 pound per square inch (by definition) In the equation $$HP = k \Delta P F$$ where$F$= flow rate in gallons per minute,$\Delta P$is pressure difference in psi, and$HP$is power in HP, you need a conversion factor. Doing everything in ... 6 If a constant pressure of$1\,\rm Pa$is exerted on a piston, and pushes it back so as to liberate a volume of$1\,\rm m^3$, then the work done by pressure on the piston amounts to$1\,\rm J$. 5 In the second formula,$c$is the speed of light. Both formulas use different system of units. The first one uses the SI:$q$in coulombs,$\vec{E}$in volts per meter and$\vec{B}$in teslas. The second one uses gaussian units:$q$in statcoulombs,$\vec{E}$in statvolts per centimeter and$\vec{B}$in gauss (being statvolts per centimeter and gauss ... 5 The important part is that it works "between two temperatures whose difference is 100°C". Celsius and Kelvin are not the same, but their degrees measure the same. You can see that by direct substitution$T(ºC)=T(K)+273.15$therefore$T_2(ºC)-T_1(ºC)=T_2(K)+273.15 -(T_1(K)+273.15)=T_2(K)-T_1(K)$5 The most general form of Maxwell's equations are (setting$\mu_0 = \varepsilon_0 = 1) \begin{align} \vec{\nabla} \cdot \vec{B} &= 0 \\ \vec{\nabla} \times \vec{E} &= - \frac{ \partial \vec{B} }{ \partial t} \\ \vec{\nabla} \cdot \vec{E} &= \rho \\ \vec{\nabla} \times \vec{B} &= \vec{J} + \frac{ \partial \vec{E} }{ \partial t} \end{align} ... 5 I reject the notion that\epsilon_0=4\pi\times10^{-7}\,{\rm F/m}$and$\mu_0\sim9\times10^{-12}\,{\rm H/m}$, they are precisely$\epsilon_0=1$and$\mu_0=c^{-2}$. This obviously takes care of any issue with defining any unit with$\pi$in it because, $$\sqrt{\frac{1}{\epsilon_0\mu_0}}=\sqrt{\frac{1}{1\cdot c^{-2}}}=\sqrt{c^2}=c$$ We use the MKSA system ... 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 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 ... 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.

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