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3

It appears that the results of this paper are dependent on the precise value of the Hubble constant, but the only effect of this is to scale the different measured quantities by different amounts. The initial approach is to simply assume a reasonable value, which they take as $$H_0=50\:\mathrm{km\:s^{-1}Mpc^{-1}},$$ but a slightly more sophisticated ...

1

The units of these quantities vary with the coordinate system. Consider Minkowski space with the usual Cartesian spatial coordinates. We have $$ds^2 = -c^2 \mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2.$$ The nonzero metric coefficients are \begin{align} g_{tt} = -c^2 & \sim \frac{L^2}{T^2} \\ g_{xx} = g_{yy} = g_{zz} = 1 & \sim ...

2

The definition of force is under the assumption that there is no friction involved in the system. This means you can imagine the surface it is measured on as some kind of "super slippery ice" that has no frictional properties whatsoever. About the question whether the forces required to produce the same acceleration on different surfaces are the same: No ...

1

Let's do natural units the other way around. Suppose that we've always worked with natural units, we measure time and distances in the same units and then some crazy physicist comes along who puts in factors of c in equations, e.g. $$ds^2 = dt^2 - dx^2 - dy^2 - dz^2 \longrightarrow c^2 dt^2 - dx^2 - dy^2 - dz^2$$ He then defines a meter and a second such ...

1

The point of natural units is to rescale your units so that $c = 1$ and $\hbar = 1$ and $k_B = 1$. This is technically a type error because the quantities on both sides have different dimension, but it means "in the dimensions that give this the appropriate size." So this means that you have a $\text{cm time}$ unit, for example, which is the time it takes ...

0

Simply multiply by $d/d$, where $d$ is a arbitrary distance, then the unit of $d/(hv)$ will be (Joule)^(-1), and $\frac{Z_1Z_2e^2}{d}$ will be in $J$. Then the product is dimensionless.

0

All are dimensionless constants. With $\frac{e^2}{\hbar c} \approx \frac{1}{137.036}$ Similarly there are constants for weak and colour charge. These basically are the probability over time of a particle emitting a photon, W (or Z)-boson or gluon respectively. The weak constant is of the same order as the electromagnetic constant. The colour constant ...

1

I learned to keep track of the conversion from SI to Gaussian units for electromagnetism as \begin{align} \frac{e^2}{[4\pi\epsilon_0]} &= \alpha \hbar c \end{align} where the factor in [brackets] is unity in CGS units and isn't in SI. This is a nice way to remember things because it makes clear that Coulomb's law for two fundamental charges, $$\vec F = ... 4 You've been done a disservice if your earlier teachers didn't even mention the existence of Gaussian units (a cm-gram-sec system with "unrationalized" E&M). Not that I like them, but simply because they were very common in the mid twentieth century and they still have their adherents (some even on Physics SE). The unit of charge goes by several names ... 0 This isn't a derivation, but you might be interested in the fact that there is experimental evidence for the relation E=mc^2, and it is very simple. This is due to what is known, in nuclear physics, as the “mass defect” (cfr. [1]). Using methods to determine the masses of the nuclei (the so called "mass spectroscopy") one finds that the sum of the masses ... 1 In the 1800's the AU was connected to: the time average of 1 divided by the Earth-Sun, and that is the reason Gauss's constant occurred (until 2012) in the calculation of the AU. The reciprocal of the distance was used because it has less of a linear tread and also because it is not as dependent on the eccentricity (which has a large ~linear trend over the ... 1 However, say we interpret that last number as ft * lb A kilowatt hour per rpm is not a foot-pound - you cannot "interpret" it as that.$$W=\tau\cdot\thetaP=\frac{W}{t}P=\frac{\tau\cdot\theta}{t}=\tau\cdot\omega, where $\omega$ is your angular frequency in rad/s (about 500 rad/s, in the case of 5000 RPM). Plug in 100 horsepower for your ...

5

Just an accident. You've discovered that the ratio of $1\text{hp}$ to $1\text{kW}$ (0.7457) is pretty close to the ratio of a $\text{N m}$ to a $\text{ft lb}$ (0.7376). So if you apply one of them and the reciprocal of the other, the answer doesn't change much. As the value for horsepower isn't derived from other units (it's a measured quantity), there's ...

0

In the SI system of measurement, one Newton of force accelerates a 1 kg mass at 1 m/s^2. This is very convenient, as there is no "factor" that you have to worry about when doing force, work, and energy calculations. In the U.S. customary units of measurement, the unit of force is the lbf, or pound-force. The unit of mass if the lbm or pound-mass. One lbf ...

0

I'm unclear on exactly what you're asking, but most (all?) of the US customary units are ultimately defined in terms of SI units. As an example a US inch is defined to be exactly 2.54cm. Mass in US units is still just mass. The only real complication is the fact that the "pound" has historically been used both as a mass and as a force (where it is the ...

-1

According as you say, I can assume the function $\phi$=k $\frac{\mu}{\sigma}$ where k has the units of ms. Hence by partial differentiation, we get $\frac{\partial \phi}{\partial \mu}$ = $\frac{k}{\sigma}$ with units $ms/mV$. and for $\frac{\partial \phi}{\partial \sigma}$ = -k$\frac{\mu}{\sigma^2}$ with units $ms*mV/(mV)^2$ = $ms/mV$ Hence the units of ...

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