Question

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

Answer 1

I tried
V = W/Q

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

Answer 2

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

Answer 3

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

Answer 4

That's none of the choices :/

Answer 5

what is A?

Answer 6

A is the dimension of Electric current

Answer 7

Is the question a typo ?

Answer 8

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:
\[\frac{{work}}{{time}} = \frac{{ML{T^{ - 2}}L}}{T} = M{L^2}{T^{ - 3}}{A^0}\]

Answer 9

one*

Answer 10

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

Answer 11

:)