Intuition on Different SI Units and Squaring Decimals (Beginner Question) So this is a very basic question. I am certainly not in grasp of some key aspect of physics equations here. I hope someone will be able to help me out.
Here is the equation for drag force:
$$F_{D}=\frac{1}{2} \rho v^{2} C_{D} A$$
If $\rho$, $C_{D}$,$A$ = 1, The equation changes to $F_{D} = \frac{1}{2}v^{2}$
If I plug in $v$ = 1 $m/s$
The equation evaluates to $1/2 \times 1$
But Instead if I change the SI unit to cm the equation suddenly evaluates to $1/2 \times 100\times100 
 $ cm/s (100 m/s) which is a different and much bigger value compared to the first result(1 m/s). Is this move invalid in Physics equations? Are the equations strictly restricted to the primary SI units and converting them to other forms makes the equation invalid somehow?
Another related question:
Why do drag force or any force that is $F$ $\propto$ $v^{2}$ goes on to decrease below $v = 1$, but while $v > 1$ they increase exponentially.
Ex: If $v = 0.5$,
$0.5 * 0.5 = 0.25$ (Which is way smaller, almost half the size of the input)
and If $v = 5$,
$5 * 5 = 25$ (Which is way larger than the input)
If this maps onto real world, it feels to me like the force detects if the velocity is < 1, it makes the force way smaller and when it is greater than 1, way bigger. However again if I switch these to a completely different SI unit like cm (assuming I was talking about m/s). Suddenly the forces are exponentially bigger. I know this sounds kinda silly to an expert, but what is wrong with this formulation? I can't seem to put my finger on it.
 A: Well, few problems here, which all seem to come down to a disregard for units of measurement.
First, the quantities you take to be equal to $1$ in the beginning still have dimensions (apart from the drag coefficient, let's keep that at $1$ for simplicity), so they're not just "$1$". In fact, you see that shortly after this, you're writting results for a force in units of velocity, which is nonsense.
If $\rho=1\, kg/m^3$ and $A=1\, m^2$ then, for $v=1\, m/s$, you get $F_D=0.5\, N$, where $N$ is the force unit Newton given by $1\, N=1\,kg\cdot m/s^2$.
Now let's use $v=100\, cm/s$. Then $A=10^4 cm^2$ and $\rho=10^{-6}\, kg/m^3$. Substituting all of these, you get $F_D=50\,kg\cdot cm/s^2=0.5\,kg\cdot m/s^2$, which is the same as what we got above.
As for your second question, again, you have to keep track of units. It doesn't make sense to say that $v^2$ is "larger" or "smaller" than $v$. They have different units of measurement, so that comparison does not make sense.
In fact, notice how, if $v=5\, cm/s$, then $v^2=25\, cm^2/s^2$, but we can also write these as $v=0.05\,m/s$ and $v^2=0.0025\,m^2/s^2$.
This would be the equivalent of saying that the area of a square is somehow smaller  or bigger than the lengths of the sides.
