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0

Yes, you can assert it, and in one sense you are obviously correct - but only so long as you insist that Celsius is the "real" way to measure temperature. 10 degrees C is 50 degrees Fahrenheit. 20 degrees C is 68 degrees F. 68 / 50 is 1.36. So, are you claiming her to be simultaneously double the temperature AND 1.36 times the temperature? Or, more ...


3

If you want absolute temperatures, then the answer would be no. If you don't care about scales, but just numerical value, I'd say yes. The Celcius scale is a 'relative' scale, based on the freezing and boiling points of water, and the temperatures in $^oC$ do not have much meaning. Whatever they tell, they only tell it with respect to freezing and boiling ...


2

See also Simple Harmonic Motion - What are the units for $\omega_0$ ? and https://en.wikipedia.org/wiki/Joule#Confusion_with_newton-metre Here's a somewhat shorter explanation reflecting my own (possibly incorrect) intuition: Radians aren't "real" units; they're just a trick to keep track of which quantities involve angles and which don't, since it's ...


3

Your mistake is your definition of the angular velocity $$ \omega=rv \tag{not correct} $$ is incorrect. We know that $\omega$ has units of $1/s$, but your assertion gives it units of $m^2/s$. The correct definition is $$ \omega=\frac vr \tag{correct} $$ which gives the correct units. Using this: $$ \left[L\right] = [m]\left[r^2\right]\cdot\left[v\cdot ...


1

Why do we set x0=ct instead of x0=t? Nobody holds us to set x0=t and measure the time in centimeters of the travel of the clock's pointer. Or measure it by the change in the circumference of the earth during the day night cycle, or... It is a matter of calibrating time to centimeters in terms that are universal and unique. When the Lorenz ...


8

But that's exactly the deeper meaning! Setting things up so that all coordinates are in the same units (besides being a reasonable requirement for $x^\mu$ to be considered a four-vector) is a constant reminder that time is really not that different from space. In fact, if it weren't for that sign in the metric, spacetime would be completely symmetric in its ...


2

Argue by dimensional analysis. The force on a charged particle is usually taken to be $$\mathbf F = q(\mathbf E + \mathbf v \times \mathbf B)$$ and this defines the $\mathbf E$ and $\mathbf B$ fields. With this definition $[\mathbf E] = [c][\mathbf B]$. However you could take as definition $$\mathbf F = q(\mathbf E + \frac{\mathbf v}{c} \times \mathbf B)$$ ...


5

The difference has to do with the units in which $\vec{B}$ is measured in. In SI units Faraday's Law reads as, $$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$ In Gaussian and Heaviside-Lorentz Units it reads as, $$ \nabla \times \vec{E} = - \frac{1}{c}\frac{\partial \vec{B}}{\partial t} $$ Basically this amounts to redifining ...


0

It's just a result of Gauss theorem applied in symmetry in this case surface area of sphere with charge placed at it's center. Which is 4 Pi R^2


0

Torque could be measured in joules per radian. Torque by angle gives energy.


2

The simplest way to approach this is to note that the molar volume of an ideal gas (helium and air are close to ideal at STP) is $22.4$ litres. This means that $22.4$ litres of helium weighs $4$g and similarly $22.4$ litres of air (average $M_W = 28.8$) weighs $28.8$g. Archimedes' principle tells us that the upthrust is equal to the weight of fluid ...


1

Good question. You should certainly be able to solve this in SI units, where $c$ and $v$ have units and $c\ne 1$ The way to find the problem is to put units on things and make sure they are consistent. The $10^8$ at the start is then in light-years, the 1 is in years, so the $v$ outside the square root should be in light-years/year, not in m/sec. You can ...


1

They don't have different units. If you follow things carefully you see that the $10^{16}$ has units of $(lyr/yr)^2$, so provided you use the same units for $c^2$ then everything works. I would recommend explicitly sticking the units in at the beginning. Your first, equation, for example, should read \begin{equation} 1 yr = (10^{8} lyr) \frac{1}{\gamma v} ...



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