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TrojanPoem

  • one year ago

if you knew that: Pw = VI the dimensions are: a) ML^2A^-1T^-3 b) MLA^-1T^-2 c)ML^-2AT^-3 D)ML^2A^-1T^-2

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  1. TrojanPoem
    • one year ago
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    I tried V = W/Q pw = WI/Q Q = it pw = w/t Dimensions: ML^2T^-3 What's wrong ?

  2. Michele_Laino
    • one year ago
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    we have: volt* Amperes= (joule/ coulomb)*amperes=\[ = \frac{{joules}}{{amperes \cdot \sec }} \cdot amperes = \frac{{joules}}{{\sec }}\]

  3. Michele_Laino
    • one year ago
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    now, using dimensions, we can write this: \[\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}\]

  4. TrojanPoem
    • one year ago
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    That's none of the choices :/

  5. Michele_Laino
    • one year ago
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    what is A?

  6. TrojanPoem
    • one year ago
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    A is the dimension of Electric current

  7. TrojanPoem
    • one year ago
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    Is the question a typo ?

  8. Michele_Laino
    • one year ago
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    I think so, since as we can see from my steps above, amperes cancel out, since we have on amperes at numerator , and another one at denominator, so the right expression can be this one: \[\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}{A^0}\]

  9. Michele_Laino
    • one year ago
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    one*

  10. TrojanPoem
    • one year ago
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    Yeah. I believe so but was making sure :) thanks again

  11. Michele_Laino
    • one year ago
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    :)

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