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How do you use mu in physics?
What is the SI unit for mu in physics?
What is mu naught equal to?
Usually, people write the quantities in italics, unlike the units.
$$\begin{align}\text{GPE}&= mgh\\60\text{ J}&= m \cdot(9.8 \text{ m}/\text{s}^2)\cdot(10\text{ m})\\&= m\cdot(98 \text{ m}^2/\text{s}^2)\\m&\approx 0.6\text{ kg}\end{align}$$
In LaTeX, type the quantities in math mode: $m$ yields 
(meaning mass). To type units inside the math mode, use, for example, $10\text{ m} or $10\,\mathrm{m}$, which yields 
.
The authoritative reference for the SI is the official BIPM brochure on the SI:
https://www.bipm.org/documents/20126/41483022/SI-Brochure-9.pdf/fcf090b2-04e6-88cc-1149-c3e029ad8232?version=1.16&t=1632138652324&download=true
On p 147 it states “Unit symbols are printed in upright type regardless of the type used in the surrounding text”. So using any other typeface for variables will allow them to be distinguished.
On p 148 it says specifically “Symbols for quantities are generally single letters set in an italic font”, which is the standard practice and should generally be followed. This helps readers to understand at a glance which quantities mean what. Readers will instantly recognize that 
is five times some variable 
and that 
is 5 meters. Any other typeface conventions should be made explicit.
Finally, p 149 says “The numerical value always precedes the unit and a space is always used to separate the unit from the number”. That can also be helpful by grouping any units together with a space after the quantity to add visual separation.
Force and momentum Every explanation I've been given is that force is change in momentum so when you have ∆p/t you get ma. But when I ask why momentum is mv I just get the inverse which is that since force is the change in momentum and that force is ma, momentum has to be mv. This is basically circulatory logic. I understand that this is proven with experiments and observations but I want to know if there's a mathematical or logical reason.
Vectors and energy If I push a box at 1N in a circle the displacement will be 0, thus the work done is 0. And via the same logic the kinetic energy of an object in uniform motion traveling in a circle will be 0 if you measure it with displacement/time because if you measure it at its first revolution the displacement will be 0. Seems to make way more sense if you use scalar values because even if you take the absolute value of the vectors they still might not give you the full story if they give you negative displacement any point. For example if the circumference of the circle is 20m, and you travel 4/5ths of that you have a distance of 18m while the displacement will only be 2m which changes the work done. So again, it would make way more sense if we just used scalar values.