(Sorry for my bad english)
I got a task to do:
In the research, scientists noticed that the necessary power P is obtained from P=Dgv^3, where D is constant, g is the earth's standard acceleration due to gravity and v is the speed. What is the D and what units it is, if the power is 75.0kW, speed 1.41m/s and acceleration of gravity 9.81m/s^2?
So it should go D=P/gv^3
D=75kW/(9.81m/s^2)*(1.41m/s)^3
where D is about 2.73kWs^5/m^4
But looks like the units are wrong, could someone please explain me how it should go and what units D has?
Hint: a Watt is given by $$\mathrm{W = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}} $$
So a $\mathrm{kW}$ is given by: $\;\;?$
Now, try to determine a different representation of units for $D$, with appropriate cancellations. The units may not necessarily be wrong, as you listed, but they may simply be represented differently in the solution.