Question

If a quantity you calculated has units of (kg*m^2)/(s^2*c), what is that quantity?

Answer 1

The units of the quantity are:

`([kg] [m]^2)/([s]^2 [C])`
We can isolate one
`([m])/([s]^2)`, which corresponds to an acceleration, a:

`= ([kg][m] a)/([C]) =`
[kg] corresponds to a mass, m; [m] corresponds to a length, L; C corresponds to a charge, Q:

`= (maL)/(Q)`
But the product ma is a force, F:

`= (FL)/(Q)`
and the product FL is a work, so an energy, U:

`= (U)/(Q)`
and this ratio corresponds to an electrical potential. So, the quantity is an electrical potential.