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Classic E=mc2

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    a mass of 3.33x10^-28kg of uranium is converted into energy during nuclear fission (exploding) according to einstien's formular

    where energy is measured in joules (J), m is measured in kg and c = 3000000000m/s is the speed of light
    2. Relevant equations

    show that the amount of energy released during fission is 2.997x10^-1 J

    3. The attempt at a solution

    E = mc^2
    = 28kg x (3x10^8ms^-1)^2
    = 28kg x (3x10^8ms^-1)x(3x10^8ms^-1)
    = 28kg (9x10^16m^2s^-2)
    = 28 x (9x10^16)kg m^2s^-2
    = 252 x 10^16 J
    = 2,520,000,000,000,000

    thats my attempt.. i know nothing about this stuff and i was trying to do this for a friend.. im curious now what others come up with
  2. jcsd
  3. Feb 3, 2009 #2
    [itex] m = 3.33\times 10^{-28} [/itex]

    [itex] c = 3\times 10^8 [/itex]

    [itex] E = mc^2[/itex]

    [itex] E = (3.33\times 10^{-28})(3\times 10^8)^2 = 2.997\times 10^{-11} [/itex]
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