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So I'm doing a physics problem on conversions, and having a bit of trouble with the intermediate steps (since I keep getting the wrong answers. The problem is as follows:

Suppose that it takes 18.2 hours to drain a container of 76.0 $m^3$ of water. What is the “mass flow rate,” in kg/s, of water from the container? (Assuming that each cubic centimeter of water has a mass of exactly 1g)

My process went something like this (after dividing 76 by 18.2):

$$\frac{4.17582m^3}{1hr} \cdot (\frac{1cm}{0.01m})^{3} \cdot \frac{1g}{1cm^3} \cdot \frac{1kg}{1000g} \cdot \frac{1hr}{60min} \cdot \frac{1min}{60sec}$$

which got me the answer: 1.16 kg/sec (rounded to 3 significant figures). Is this the right process?

secondubly
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  • Looks good to me. You want to make sure the undesired units cancel, and you have. And I guess you mean average flow rate, since instantaneously the flow rate satisfies Torricelli's Law. :-) – Matthew Leingang Jan 19 '15 at 19:08
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    @MatthewLeingang yeah, I do mean average flow rate, but thank you! Maybe I just typed it in wrong lol – secondubly Jan 19 '15 at 19:09
  • Measurement units multiply and divide the same way as numbers do, so you want to arrange your conversion factors so that the ones to be eliminated will cancel. It looks like that's what you have done, so it should be correct. (A quick way to check is that 1 cubic meter of water has a mass of 1000 kg., so the mass flow rate is about 4200 kg/hr = 4200/3600 kg/sec = 7/6 kg/sec .) – colormegone Jan 19 '15 at 19:09
  • Heh, I typed in 1.116, not 1.16, so I kept getting it wrong. I think I need a break heh – secondubly Jan 19 '15 at 19:10

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