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Area of a silver sheet

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Accomplished silver workers in India can pound silver into incredibly thin sheets, as thin as 3.00x10-7 m (about one hundredth of the thickness of this sheet of paper). Find the area of such a sheet can be formed from 1.39kg of silver? (For the density of silver use the value 10500 kg/m3.)

    2. Relevant equations
    Idk i feel like this should be really easy...
    area - L x W
    dont really know where to go from here...



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2009 #2

    cepheid

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    Because the density of silver is a constant, a given mass of silver *has to* occupy a certain volume of space. Since you know the thickness of the sheet, and the volume it must take up, can you think of how to arrive at the area?
     
  4. Nov 15, 2009 #3
    As a starting point, I would flip the density (find the reciprocal of it), this will give you the number of cubic meters per kg [m3/kg]. You can multiply this by your silver mass, which will give you the volume of your silver....

    See if that gets you on the right track...
     
  5. Nov 15, 2009 #4
    @Sean -Ahhh that works perfectly. I have no idea what i did tho. lol
     
  6. Nov 15, 2009 #5

    cepheid

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    Well that's no good at all, the point is to learn physics, not to do mindless calculations. That's why I was hoping you would deduce what needed to be done conceptually from my post (before Sean gave away further information, more than what I would have revealed). If you re-read what I wrote, does it now make sense? For an object of a given density, a given mass takes up a given volume. Using the equation for density, rho = m/V, you can calculate what V must be for a *given* rho and m.

    Again, conceptually, if I have "m" kg of silver, it is going to take up a certain amount of space, and the larger m is, the larger the amount of space will be taken up. I hope that that, at least, is intuitive. The exact amount of space taken up by m is determined by the density, rho, which tells you how densely - packed or compact the matter in a given substance is.
     
  7. Nov 15, 2009 #6
    Thanks cepheid...that makes sense. I had totally forgotten the density = m/v formula.
    I have another question that has been troubling me regarding specific gravity...
    An aluminum cylinder weighs 1.02 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.87 × 10-5 m3. The apparent weight of the cylinder when completely submerged is 0.758 N. What is the specific gravity of the alcohol?
    i have the formula...
    specificgravity= density object/ density water
    ...this can be further deduced to M/V / density of water...
    where M = 1.02 N
    V=3.87 × 10-5 m3
    and density of water = 1024
    however im not gettin the correct answer :/
     
  8. Nov 15, 2009 #7

    cepheid

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    That's because M is not equal to 1.02 N. Mass is not the same thing as weight. Mass is measured in kg. Weight is a force, and forces are measured in N.
     
  9. Nov 15, 2009 #8
    ok...soooo since weight = m x g
    do i divide by 9.8?
    sorry im retarded when it comes to conversions...
     
  10. Nov 15, 2009 #9

    cepheid

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    Yes, that's right. But this question is more complicated than that. It's not asking you for the specific gravity of aluminum. It is asking you for the specific gravity of alcohol. So you have to use the fact that the cylinder has an apparent weight of 0.758 N in alcohol in order to figure out how much more or less dense alcohol is than water.
     
  11. Nov 15, 2009 #10
    hmm. so i do i subtract out the mass of the aluminum cylinder (.104) from the apparent wieght in alcohol and then use the new mass in my calculation (m/v)/density of water?
     
  12. Nov 15, 2009 #11

    cepheid

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    Whether or not an object floats in a fluid depends upon whether that object is more or less dense than that fluid (because of Archimedes' principle). I'm not sure how much I need to explain here and how much you already know. The fact that the aluminum sinks in alcohol tells you that alcohol is less dense than aluminum. Exactly how much less dense can be deduced from the fact that 3.87 x 10-5 m3 of aluminum weighs 0.758 N more than that same volume of alcohol does. Once you have used this fact to compute the density of the alcohol, you can compute its specific gravity.
     
  13. Nov 15, 2009 #12
    so shouldn't that just mean that I should be able to use .0773 kg / 3.87 × 10-5 m3 to get the density of alcohol??
     
  14. Nov 15, 2009 #13

    cepheid

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    Hi,

    0.0773 kg is not the answer for the mass of alcohol displaced that I am getting.

    The given volume of aluminum weighs 0.758 N more than the same volume of alcohol (which is what was displaced). This suggests that that volume of alcohol weighs 1.02 N - 0.758 N = 0.262 N. From that you can figure out what mass of alcohol was displaced. If you know what mass of alcohol was displaced, and what volume it occupied, then you know its density.
     
  15. Nov 15, 2009 #14
    ok that's what i was thinking.
    So i did .262/9.8 to get the mass as .0267. Then I divided this by the volume (3.87 × 10-5 m3 ) to get the density of alcohol. Then i divided that density by the density of water (1024), to find the specific gravity. ....its not correct...:(
     
  16. Nov 15, 2009 #15

    cepheid

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    Hmm. Are you sure that your units are correct? Also, the density of water is not 1024 in kg/m^3...it is closer to 1000 kg/m^3 (maybe slightly less, and depends on temperature).
     
  17. Nov 15, 2009 #16
    OMG THANKYOU. That did the trick. They wanted us to use density of 1024 for water in a previous problem so i was still using that. Thanks for all your help!!
     
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