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Water poured into leaky container

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 5.3t0.8 - 3.2t + 22, with t ≥ 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? What is the rate of mass change at (c) t = 1.6 s and (d) t = 4.9 s? Change g/s to kg/min for part c and d.

    2. Relevant equations



    3. The attempt at a solution
    I already got part a and part b, but I'm having trouble with parts c and d.
    This is what I got for the derivative: dm/dt=4.24t^-0.2-3.2
    Part C: I plugged in 1.6 into my derivative and got 0.6595961104. then I multiplied by .06 to convert to kg/min and got 0.0395757666. But that's not the right answer.
     
  2. jcsd
  3. Jan 16, 2014 #2
    I get the same answer. Do you know what the correct answer for C is?
     
  4. Jan 16, 2014 #3
    No I don't, is it maybe a problem with significant figures? My online homework is kind of picky.
     
  5. Jan 16, 2014 #4
    If it's significant figures (which it may very well be) have you tried 0.040?
     
  6. Jan 16, 2014 #5
    Thank you so much! It worked :)
     
  7. Jan 16, 2014 #6
    Maybe they want the answer in grams/sec?
     
  8. Jan 16, 2014 #7
    Now I'm having trouble with part D. I plugged in 4.9 into the derivative and got -0.1144922571 g/s and then converted it to kg/min and got -0.0068695354. I tried putting in -0.007 for significant figures, but that didn't work.
     
  9. Jan 16, 2014 #8
    @Chestermiller No, they want it in km/min.
     
  10. Jan 16, 2014 #9
    For D, try -0.01. Depending on the exact precision rules it could also be 0.
     
  11. Jan 16, 2014 #10
    I tried both and neither one worked :(
     
  12. Jan 16, 2014 #11
    Then it must be -0.069.
    (I checked you calculation - it's right)
     
  13. Jan 16, 2014 #12
    That one didn't work either, maybe there's something wrong with that problem. I'll try e-mailing my professor about it. Thanks so much for your help!
     
  14. Jan 16, 2014 #13
    I get -0.0069 rather than -0.069
     
  15. Jan 16, 2014 #14
    Thanks Chester, that one worked!
     
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