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Physics problem bugging me!

  1. Sep 23, 2003 #1
    Ok, this is the problem.

    Grains of fine sand are approximately sphere of an average radius of of 50 ìm are made of silicon dioxide. A solid cube of this material with a volume of 1m3 has a mass of 2600kg.

    What mass of sand grains would have a total surface area (the total area of all the individual grains) equal to the surface area of a cube 1.1 meters on an edge?

    Ok. I figured out the surface area of the cube by taking 1.1m*1.1m*1.1m*6= 7.986m^3

    I figured out the surface area of a grain of sand by first converting micrometers into meters, and then plugging that number into the surface area forumula for a sphere.
    SA of sphere= 4*pi*(50*10^-6m)^2= 3.14E-8.

    Now, somehow the SA of a cube and SA of the grain are related, but I can't get the answer to be accepted by the online homework system. Any ideas?
  2. jcsd
  3. Sep 23, 2003 #2


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    Science Advisor

    You probably can't get the answer accepted by the online computer system because it isn't correct!

    First of all, surface AREA is not measured in CUBIC meters- it is measured in square meters. You calculated the volume of the cube and then multiplied by 6 to get area? There are, indeed, 6 faces on a cube: each one is a square with edge length 1.1 m: the Area of such a square is ??? That's what you want to multiply by 6.

    The surface area of the cube and the surface are of the spheres (not just one) are related because the problem tells you they are to be the same!!! The problem asks for the mass the number of spheres whose surface area is the same as the surface area of the cube.

    Once you have found the surface area of the cube (correctly) and the surface area of an individual sphere, determine HOW MANY spheres it would take to have the same surface area as the cube. You can also calculate the volume of each sphere and you are told the density (indirectly) so you can calculate the mass of each sphere. Finally, multiply the mass of each sphere by the number of spheres required.
  4. Sep 23, 2003 #3

    Thank you so much. Thank God I found this forum today! :)
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