Determine if the equation given is dimensionally correct: \(\sf F=m(v^2)/(d^2)\)

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Determine if the equation given is dimensionally correct: \(\sf F=m(v^2)/(d^2)\)

Physics
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Ok, so first of all you know what is the dimension of force right?

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Not really. This is my pre-physics summer work.
Newtons?
ML/T^2
Yes, dimension of force = \(\sf M^1L^1T-2\) Do you understand how?
F=ma, and acceleration=L/T^2
\(\huge \checkmark\)
So, in order to determine if the given equation is dimensionally correct we just need to make sure that same dimension occurs on both hand sides of the equation. So in RHS of equation the dimension should also be \(\sf M^1L^1T^{-2}\)
Ok.
RHS = \(\sf \Large \frac{mv^2}{d^2}= \frac{mass \times (metre^2)}{second^2\times distance^2}\) Agree?
what? Okay, I have: \[\frac{ mv^2 }{ d^2 }\] would equal \[\frac{ M(L/T^2) }{ L^2 }\]
\(\sf V=L/T \\ \Rightarrow V^2 = (L/T)^2=L^2/T^2\)
So, that would give you for RHS \[\frac{ M(L^2/T^2) }{ L^2 }\] = \(\sf M/T^2\)
yeah sorry i meant \[\frac{ M \frac{ L^2 }{ T^2 } }{ L^2 }\]
Yup, what can you say now, Is LHS = RHS?
Yes? They've got the same letters...
LHS=Force= \(\sf M^1L^1T^{-2}\) RHS=\(\huge \frac{ M(L^2/T^2) }{ L^2 }\) = \(\sf M/T^2\) Are they both equal?
No, there's no L in the RHS. If there was, would they be equal?
Yes.
But since, RHS\(\neq\)LHS, the equation is dimensionally incorrect but if somehow they both were equal then the equation would have been dimensionally correct.
so the letters just have to be there or does the division, etc. matter as well?
Division matters, treat them like algebraic variables.
thank you!
Do you want me to give you some examples for practice?
I have a few more to do in the packet. Hopefully I'll get it before class tomorrow. I think I understand better now. Thanks again
You're most welcome c:
@Abhisar Would L=2.5L ?

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