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

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Ok, so first of all you know what is the dimension of force right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not really. This is my prephysics summer work.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Yes, dimension of force = \(\sf M^1L^1T2\) Do you understand how?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0F=ma, and acceleration=L/T^2

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1So, 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 \(\sf M^1L^1T^{2}\)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1RHS = \(\sf \Large \frac{mv^2}{d^2}= \frac{mass \times (metre^2)}{second^2\times distance^2}\) Agree?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what? Okay, I have: \[\frac{ mv^2 }{ d^2 }\] would equal \[\frac{ M(L/T^2) }{ L^2 }\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1\(\sf V=L/T \\ \Rightarrow V^2 = (L/T)^2=L^2/T^2\)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1So, that would give you for RHS \[\frac{ M(L^2/T^2) }{ L^2 }\] = \(\sf M/T^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry i meant \[\frac{ M \frac{ L^2 }{ T^2 } }{ L^2 }\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Yup, what can you say now, Is LHS = RHS?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes? They've got the same letters...

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1LHS=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
 one year ago
Best ResponseYou've already chosen the best response.0No, there's no L in the RHS. If there was, would they be equal?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1But 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
 one year ago
Best ResponseYou've already chosen the best response.0so the letters just have to be there or does the division, etc. matter as well?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Division matters, treat them like algebraic variables.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Do you want me to give you some examples for practice?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1You're most welcome c:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Abhisar Would L=2.5L ?
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