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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
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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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I tried V = W/Q pw = WI/Q Q = it pw = w/t Dimensions: ML^2T^3 What's wrong ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have: volt* Amperes= (joule/ coulomb)*amperes=\[ = \frac{{joules}}{{amperes \cdot \sec }} \cdot amperes = \frac{{joules}}{{\sec }}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, using dimensions, we can write this: \[\frac{{work}}{{time}} = \frac{{ML{T^{  2}}L}}{T} = M{L^2}{T^{  3}}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0That's none of the choices :/

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0A is the dimension of Electric current

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Is the question a typo ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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: \[\frac{{work}}{{time}} = \frac{{ML{T^{  2}}L}}{T} = M{L^2}{T^{  3}}{A^0}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. I believe so but was making sure :) thanks again
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