anonymous
  • anonymous
Determine if the equation given is dimensionally correct: \(\sf F=m(v^2)/(d^2)\)
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@Abhisar
Abhisar
  • Abhisar
Ok, so first of all you know what is the dimension of force right?

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anonymous
  • anonymous
Not really. This is my pre-physics summer work.
anonymous
  • anonymous
Newtons?
anonymous
  • anonymous
ML/T^2
Abhisar
  • Abhisar
Yes, dimension of force = \(\sf M^1L^1T-2\) Do you understand how?
anonymous
  • anonymous
F=ma, and acceleration=L/T^2
Abhisar
  • Abhisar
\(\huge \checkmark\)
Abhisar
  • Abhisar
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 \(\sf M^1L^1T^{-2}\)
anonymous
  • anonymous
Ok.
Abhisar
  • Abhisar
RHS = \(\sf \Large \frac{mv^2}{d^2}= \frac{mass \times (metre^2)}{second^2\times distance^2}\) Agree?
anonymous
  • anonymous
what? Okay, I have: \[\frac{ mv^2 }{ d^2 }\] would equal \[\frac{ M(L/T^2) }{ L^2 }\]
Abhisar
  • Abhisar
\(\sf V=L/T \\ \Rightarrow V^2 = (L/T)^2=L^2/T^2\)
Abhisar
  • Abhisar
So, that would give you for RHS \[\frac{ M(L^2/T^2) }{ L^2 }\] = \(\sf M/T^2\)
anonymous
  • anonymous
yeah sorry i meant \[\frac{ M \frac{ L^2 }{ T^2 } }{ L^2 }\]
Abhisar
  • Abhisar
Yup, what can you say now, Is LHS = RHS?
anonymous
  • anonymous
Yes? They've got the same letters...
Abhisar
  • Abhisar
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?
anonymous
  • anonymous
No, there's no L in the RHS. If there was, would they be equal?
Abhisar
  • Abhisar
Yes.
Abhisar
  • Abhisar
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.
anonymous
  • anonymous
so the letters just have to be there or does the division, etc. matter as well?
Abhisar
  • Abhisar
Division matters, treat them like algebraic variables.
anonymous
  • anonymous
thank you!
Abhisar
  • Abhisar
Do you want me to give you some examples for practice?
anonymous
  • anonymous
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
Abhisar
  • Abhisar
You're most welcome c:
anonymous
  • anonymous
@Abhisar Would L=2.5L ?

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