Answer
The equation is $p=\rho v^{2}$.
p is in $\frac{kg}{m\cdot s^{2}}$, $\rho$ is in $\frac{kg}{m^{3}}$ and $v^{2}$ is in $\frac{m^{2}}{s^{2}}$.
Inserting units for the physical quantities gives
$\frac{kg}{m\cdot s^{2}}=\frac{kg}{m^{3}}\times\frac{m^{2}}{s^{2}}$
or $\frac{kg}{m\cdot s^{2}}=\frac{kg}{m\cdot s^{2}}$.
The equation is dimensionally correct since the units on each side are $\frac{kg}{m\cdot s^{2}}$. This relation need not actually be physically true.
Work Step by Step
See the answer above.
