Been working on this for hours

  • Thread starter chattkis3
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  • #1
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Been working on this for hours!!!

Ive honestly given this problem 3 or more hours to no avail.

"An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is 38 kg, and the volume enclosed by the outside surface of the chunk is 0.025 m3. What is the radius of the spherical cavity?"

This is what I have set up which is not working:
density of concrete = M / (Voutside surface - Vcavity)

what's wrong with that?? thankkks!!
 

Answers and Replies

  • #2
Pyrrhus
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shouldn't it be?

[tex] \rho_{concrete} = \frac{M}{V_{outside}-V_{cavity}} [/tex]
 
Last edited:
  • #3
Doc Al
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What makes you think it's wrong?
 
  • #4
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Im getting a negative number if I do M / (Voutside + Vsphere) ... and the value I get (using a density for concrete of 2200) is .12m And the computer program is telling me that is wrong. I feel like I am missing something??
 
  • #5
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I get a different answer. Check you calculations. Remember that
[tex]V_{sphere}=\frac{4\pi r^{3}}{3}[/tex]
 
  • #6
Doc Al
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I don't think you're missing anything. Your original method is correct; your answer seems correct also.
 
  • #7
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Sirus, did you use my original equation with Voutside - Vcavity or Voutside + Vcavity

??
 
  • #8
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Ah, that's why! Yes, I used Cyclovenom's equation with a plus sign, which I think is incorrect. I think Doc Al is right. Try entering the answer with various precisions (round to a different number of decimals) and see if the program accepts your answer.
 
  • #9
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WOW, thanks for that. Let's just say I want a couple hours of my life back. The program is not suppose to be set up to worry about sig figs. But apparently my professor changed that option? **annoyed** thanks for the advice though everyone.
 
  • #10
Pyrrhus
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Oh sorry, actually it was suppossed to be a minus, the question on the top was just because i was :confused: because it wasn't working.
 

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