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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    @dan815

  2. anonymous
    • one year ago
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    @Abhisar

  3. Abhisar
    • one year ago
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    Ok, so first of all you know what is the dimension of force right?

  4. anonymous
    • one year ago
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    Not really. This is my pre-physics summer work.

  5. anonymous
    • one year ago
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    Newtons?

  6. anonymous
    • one year ago
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    ML/T^2

  7. Abhisar
    • one year ago
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    Yes, dimension of force = \(\sf M^1L^1T-2\) Do you understand how?

  8. anonymous
    • one year ago
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    F=ma, and acceleration=L/T^2

  9. Abhisar
    • one year ago
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    \(\huge \checkmark\)

  10. Abhisar
    • one year ago
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    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 \(\sf M^1L^1T^{-2}\)

  11. anonymous
    • one year ago
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    Ok.

  12. Abhisar
    • one year ago
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    RHS = \(\sf \Large \frac{mv^2}{d^2}= \frac{mass \times (metre^2)}{second^2\times distance^2}\) Agree?

  13. anonymous
    • one year ago
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    what? Okay, I have: \[\frac{ mv^2 }{ d^2 }\] would equal \[\frac{ M(L/T^2) }{ L^2 }\]

  14. Abhisar
    • one year ago
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    \(\sf V=L/T \\ \Rightarrow V^2 = (L/T)^2=L^2/T^2\)

  15. Abhisar
    • one year ago
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    So, that would give you for RHS \[\frac{ M(L^2/T^2) }{ L^2 }\] = \(\sf M/T^2\)

  16. anonymous
    • one year ago
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    yeah sorry i meant \[\frac{ M \frac{ L^2 }{ T^2 } }{ L^2 }\]

  17. Abhisar
    • one year ago
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    Yup, what can you say now, Is LHS = RHS?

  18. anonymous
    • one year ago
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    Yes? They've got the same letters...

  19. Abhisar
    • one year ago
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    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?

  20. anonymous
    • one year ago
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    No, there's no L in the RHS. If there was, would they be equal?

  21. Abhisar
    • one year ago
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    Yes.

  22. Abhisar
    • one year ago
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    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.

  23. anonymous
    • one year ago
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    so the letters just have to be there or does the division, etc. matter as well?

  24. Abhisar
    • one year ago
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    Division matters, treat them like algebraic variables.

  25. anonymous
    • one year ago
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    thank you!

  26. Abhisar
    • one year ago
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    Do you want me to give you some examples for practice?

  27. anonymous
    • one year ago
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    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

  28. Abhisar
    • one year ago
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    You're most welcome c:

  29. anonymous
    • one year ago
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    @Abhisar Would L=2.5L ?

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