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Homework Help: Exam question, simple physics calculus.

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A full bowl cointains water, its mass is 395g.An iron screw is placed on the bowl and the mass is now 565g.
    This seemed simple, but I asked my teacher to clarify and he said that upon placing the screw on the water an equal volume of water is removed so the volume remains the same.
    Find the screw's volume.
    2. Relevant equations

    Iron's density = 7,86g/cm^3.
    water's density = 1g/cm^3
    screw volume + water volume = 395cm^3

    3. The attempt at a solution
    water volume = 395g/1g/cm^3
    Water volume =395cm^3
    565g-395g =170g
    7,86g/cm^3 - 1g/cm^3 = 6,86g/cm^3
    (it's pretty hard for me to explain why I did this but basically if I don't minus the water's density the volume, the answer won't compute the water that was removed)

    Teacher said it was wrong because this made no sense/was wrong and that I was too complicated, the part was wrong for him was this one:
    7,86/cm^3 - 1g/cm^3 = 6,86/cm^3
    so the solution was:
    21,something cm^3

    I tried to proove that I was right and I did this
    x= water's volume
    y= screw's volume
    dx= water's density
    yx= screw's density



    x+7,86y =565
    x+7,86y-6,86y = 565 -6,86y
    x+y = 565 - 6,86y
    395 = 565 - 6,86y
    395-395 = 565 -395 - 6,86y
    0 = 170 - 6,8y
    0+6,86y = 170 -6,86y +6,86y
    6,86y = 170
    6,86y/6,86 = 170/6,86

    Where did I go wrong.
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2
    Haha, all righty, first of all 595 - 325 = 270!!

    http://img26.imageshack.us/img26/2167/screenshotdocument1micr.png [Broken]

    I sometimes find drawing a picture simplifies things. The first picture on the left shows the amount of water (LIGHT BLUE WATER) that will be displaced AFTER the screw is dropped in - in the 2nd picture the screw is in and that LIGHT BLUE WATER there before is now displaced.

    So before, the mass was
    mass of water that stays + mass of light blue water = 325g
    After, the mass is
    mass of water that stays + mass of steel = 595g

    I guess this makes it clearer that mass of steel - mass of light blue water = 270g

    I hope you can take it from here.

    All you need now is to remember that DENSITY = MASS DIVIDED BY VOLUME!!!

    For some reason I got the screw's volume as being around 39cm3 or something. Might be wrong, sorry tired, 5am over here!
    Last edited by a moderator: May 5, 2017
  4. Dec 20, 2011 #3
    You have a wrong number in your first equation.

    You wrote: x+y=395

    when it ought to be x+y=325 because 325 is the volume in cm^3 of the bowl. You will get the correct answer if you correct this.
  5. Dec 20, 2011 #4
    I am sorry the numbers are supposed to be 395 and 565.
    Do I need to minus the water's density and divide 170 by that? or just divide 170 by the screw's density?
  6. Dec 20, 2011 #5
    Wow it's unbelievable, this is why I failed the exam and thus the whole subject.
    He kept telling me that to find the mass of the screw I just had to do
    I tried to explain him why I thought that was incorrect, is it possible that if I complain about this he can unfail my test?
  7. Dec 21, 2011 #6
    I agree with you. The problem as you state it says that "and he said that upon placing the screw on the water an equal volume of water is removed so the volume remains the same."

    Thus if you divide the difference as above by the density of the bolt, you do not get the volume of the bolt.

    I agree with your answer of 24.78 cm^3. Here's why:

    mw + mb = 565
    where mw is the mass of water after a mass of water equal to the volume of the bolt multiplied by the density of water comes out and where mb is the mass of the bolt.

    mw = 395 - rhow * vb
    where rhow is density of water and vb is volume of bolt.

    mb = rhob * vb
    where rhob is density of bolt


    565 = 395 - rhow * vb + rhob * vb

    vb = 24.78 cm^3

    So, once again, I agree with you.

    PS Sorry about my misread yesterday on the numbers. I was using a Kindle that has a very small print so I misread the number.
  8. Dec 21, 2011 #7
    Thank you very much all of you.
    One more question.
    I had a problem on uniform acceleration.
    I had to calculate the distance a car traveled in a given time given its starting speed and acceleration.

    I used the formula:
    x=Vo.t+ (Vf-Vo)/2.t
    Where Vf= a.t +Vo
    Would this be the same as the traditional formula
    x= Vo.t+ a/2.t^2
    Said teacher corrected me
    x= Vo.t+ (Vf-Vo)/2.t^2
    That's wrong right?
    Should I roundhouse kick him? I'm mad.
  9. Dec 21, 2011 #8
    According to forum rules you should start a new thread with your second, unrelated question. But I shall tell you this: If you are using a formula, terms that are added absolutely must have the same dimensions.
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