# Tag Info

0

Perhaps you are confused between dimension and unit. Note that $cm$ and $m$ are different units but have same dimension of length. See? It's simple. They have only different magnitudes. You have to understand that you cannot subtract or add 1 kg from 1 metre. Makes no sense, right? Suppose you want to know about speed. You know that it is $\frac ... 0 I'm not sure exactly what you are asking, so I will simply write some relevant facts that might answer your question. Dimensional analysis is a powerful tool for solving problems in physics. If we want the formula for a quantity$Q$, we can guess the formula for$Q$by first writing a product of all relevant dimensional quantities raised to unknown powers ... 3 Yes, physics covers sound. The SI unit of frequency is Hertz ($\mathrm{Hz}$), which is the same as$\mathrm{s^{-1}}$. The intensity of sound is measured in decibels ($\mathrm{dB}$), which are not an SI unit, but they are in common use and accepted by many standards bodies. 0 I only wanted to comment, but I don't have enough reputation, so excuse me for taking up a whole answer for this. The statement "The mole is the SI unit for amount of substance" doesn't mean anything to me. I think of a mole as the ratio of the SI standard unit of mass, the gram, to another fundamental mass, that of the proton. That doesn't match the true ... 0 I would describe it as (example) 120 joules per coulomb (120 volts) divided by 60 coulombs per second (60 amps) equals 2 (ohms) of resistance "which means you have 1/2 or 2 times less the amperes then voltage". so maybe an ohm can be n of VpA (# of volts[SI] per amp[SI] or in this case, # of N Kg per charge for every charge per second). But that's still ... 2 Not sure whether this is correct, but if you have to do it, I think you can say that it is: the work done by the conductor per unit charge per unit current through the conductor, or in terms of SI units,$\mathrm{\frac JC\cdot \frac1A}$which is the same as: the work done by the conductor per unit current per unit time per unit current,$\mathrm{\frac ...

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I think the short answer is, you don't. The reason we call the unit of force a Newton and not a kg m/s$^2$ is because it is convenient and it expresses the relation you want to convey when used elsewhere (e.g., $F=-kx$ for a spring). Similarly, it is convenient to "hide" the MKS base units into a single term, the potential $V$ in this case, so that the ...

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I'm not sure there's much of a point to what you're asking. The intuitive way to understand an Ohm is to use $\Omega = V/A$. If you want to use SI units, you can, and the math indeed tells you that your other definition is correct, but you're not gonna get much out of it. Indeed, the most you could do is to separate it like this: \begin{align}\Omega ...

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