When considering units and dimensions there's a simple rule you can remember.
A length has $1$1 dimension. Therefore the highest power for each unit is just $1$1. This can be kilometres, miles, inches, millimetres or any measurement of length. The key thing to note though is that the unit itself is measuring in one dimension and has a power of $1$1.
An area has $2$2 dimensions. Therefore the highest power is $2$2 - so $unit^2$unit2. You will also need to consider if there is a multiplication within an equation. For example, the area of a rectangle can be calculated used $A=l\times w$A=l×w. As two $1$1 dimensional measures are being multiplied together they will not be measuring in $2$2 dimensions. The area is a product of the $1$1 dimensional quantities.
A volume has $3$3 dimensions. Therefore the highest power is $3$3 - so $unit^3$unit3.
If $F=3q+r$F=3q+r, where both $q$q and $r$rrepresent lengths, what does $F$F represent?
Think: Consider each part of this equation. The power of each part is just $1$1 at the moment. Does the addition of $3q$3q and $r$r change the dimensions?
Do: As you are adding, there is no change to the dimension. Therefore this is still a measurement of length.
If $F=\frac{s^3}{p}$F=s3p where both s and p represent lengths, what does $F$F represent?
Think: $s$s has been multiplied by itself $3$3 times, so it would be a volume. However, it is then divided by a length ($F$F).
Do: If a volume is divided by a length, then the resulting property is an Area.
If $\text{gravitational potential energy}=\text{mass}\times\text{gravitational acceleration}\times\text{height}$gravitational potential energy=mass×gravitational acceleration×height , what are the dimensions of gravitational potential energy?
If $\text{work}=\text{force}\times\text{distance}$work=force×distance, what are the dimensions of work?
$\text{time}\times\text{length}^2\times\text{mass}^2$time×length2×mass2
$\frac{\text{mass}^2\times\text{time}}{\text{length}^2}$mass2×timelength2
$\frac{\text{mass}\times\text{length}}{\text{time}}$mass×lengthtime
$\frac{\text{mass}\times\text{length}^2}{\text{time}^2}$mass×length2time2