Comparing two sides of an equation to prove both has the same base SI units?

The process of verifying if your equation has the right units on the right and the left us called dimensional analysis, and is a very useful way of checking if your answers to more complex science problems make sense.

The first thing I always do in these kinds of problems is simply break down everything into standard SI units. Let's do this now:

#I# (as it says above) is power per unit area. Power is in units of #kg * m^2/s^3#, and area is simply #m^2#, so we'd have:

#I = (kg * m^2/s^3)/(m^2)#

#K# is a dimensionless constant, and hence can just be ignored in this context.

#v# is a speed, so I'm thinking it has units of #m/s#.

#p# is a density, which has units of mass (#kg#) over volume (#m^3#). This leaves:

#p = (kg)/m^3#

You'll usually use liters or something of that form for most calculations, but if we're making everything SI units it's a good idea to keep it as #m^3#.

#f# is frequency, measured in hertz, or #1/s#.

Lastly, #A# is an amplitude, measured in #m#.

Now, let's plug all of the above into our original equation:

#I = Kvpf^2A^2#
#=> (kg * m^2/s^3)/(m^2) = (m/s)((kg)/m^3)(1/s)^2(m)^2#

Now is the fun part: go in and cancel out everything that divides out!

#=> (kg * cancel(m^2)/s^3)/(cancel(m^2)) = (cancel(m)/s)((kg)/cancel(m^3))(1/s)^2cancel(m^2)#

#=> (kg)/s^3 = (kg)/s^3#

These dimensions match up, so we are good!

In these kinds of problems the final answer is not the big deal -- it's the process of plugging in those base units and seeing how they cancel out. Make sure you master this -- it will come in handy across science disciplines.

Hope that helped :)