D=P/gv^3 what units is the D

Question

(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?

Could you also explain why you think the units of D are wrong? — Alex R., Oct 17 '13 at 14:46

score 2 · Accepted Answer · answered Oct 17 '13 at 14:46

2

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.

answered Oct 17 '13 at 14:46

amWhy

209,954

Thank you. That helped a lot. So the units of D should be kg*s^2/m^2 multiplayed by 1000 because of kW. – Denis Oct 17 '13 at 14:55
You're welcome, Denis! – amWhy Oct 17 '13 at 14:55
@amWhy: Needs a TU to go with the green. +1 – Amzoti Oct 18 '13 at 00:35

D=P/gv^3 what units is the D

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