TrojanPoem
  • TrojanPoem
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TrojanPoem
  • TrojanPoem
I tried V = W/Q pw = WI/Q Q = it pw = w/t Dimensions: ML^2T^-3 What's wrong ?
Michele_Laino
  • Michele_Laino
we have: volt* Amperes= (joule/ coulomb)*amperes=\[ = \frac{{joules}}{{amperes \cdot \sec }} \cdot amperes = \frac{{joules}}{{\sec }}\]
Michele_Laino
  • Michele_Laino
now, using dimensions, we can write this: \[\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}\]

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TrojanPoem
  • TrojanPoem
That's none of the choices :/
Michele_Laino
  • Michele_Laino
what is A?
TrojanPoem
  • TrojanPoem
A is the dimension of Electric current
TrojanPoem
  • TrojanPoem
Is the question a typo ?
Michele_Laino
  • Michele_Laino
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: \[\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}{A^0}\]
Michele_Laino
  • Michele_Laino
one*
TrojanPoem
  • TrojanPoem
Yeah. I believe so but was making sure :) thanks again
Michele_Laino
  • Michele_Laino
:)

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