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

1. TrojanPoem

I tried V = W/Q pw = WI/Q Q = it pw = w/t Dimensions: ML^2T^-3 What's wrong ?

2. Michele_Laino

we have: volt* Amperes= (joule/ coulomb)*amperes=$= \frac{{joules}}{{amperes \cdot \sec }} \cdot amperes = \frac{{joules}}{{\sec }}$

3. Michele_Laino

now, using dimensions, we can write this: $\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}$

4. TrojanPoem

That's none of the choices :/

5. Michele_Laino

what is A?

6. TrojanPoem

A is the dimension of Electric current

7. TrojanPoem

Is the question a typo ?

8. 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: $\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}{A^0}$

9. Michele_Laino

one*

10. TrojanPoem

Yeah. I believe so but was making sure :) thanks again

11. Michele_Laino

:)