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Help with Energy unit Joules EASY

  1. May 1, 2008 #1
    [SOLVED] Help with Energy unit ... Joules... EASY!!!

    1. The problem statement, all variables and given/known data

    The sun burns up 3.6 x 10[tex]^{9}[/tex] kg of matter each second.
    Using the formula E = mc[tex]^{2}[/tex], determine how much energy the
    sun releases each second. [Speed of light: 3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s]


    2. Relevant equations

    E = mc[tex]^{2}[/tex]


    3. The attempt at a solution

    E = mc[tex]^{2}[/tex]
    . .= (3.6 x 10[tex]^{9}[/tex] kg)(3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s)[tex]^{2}[/tex]
    . .= (3.6 x 10[tex]^{9}[/tex] kg)(9 x 10[tex]^{16}[/tex] m[tex]^{2}[/tex][tex]/[/tex]s[tex]^{2}[/tex])
    . .= 32.4 x 10[tex]^{25}[/tex] [tex]\frac{kg \cdot m^2}{s^2}[/tex]



    Now I'm stumped with what to do with the units... I know it should be Joules per second. But how do I show in my solution that Joules will come out? Joules is N[tex]\cdot[/tex]m!!!

    Do I have to change the kg into N? .... My textbook says that the magnitude of c[tex]^{2}[/tex] is 9 x 10[tex]^{16}[/tex] Joules per kilogram.

    Please check my calculations also!!!
    Thank You!!! :smile:
     
  2. jcsd
  3. May 1, 2008 #2

    malawi_glenn

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    First hint: How is "Joule" defined?
    Second hint: If E = mc[tex]^{2}[/tex] is valid, and you have SI units to the right, then you must obtain SI units on the left hand side as well.

    The second hint would not be true if you was working with the natural units in sub-atomic phyiscs, where both c and h-bar = 1 i.e E = m, then one must do a more careful unit analysis.
     
  4. May 1, 2008 #3

    HallsofIvy

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    A Joule is N m and a Newton, N, is kg m/s2. So a Joule is kgm2/s2. That's exactly what you have!
     
  5. May 1, 2008 #4
    Hmmm.....

    well Joule is the unit of work. N*m
    Newton x meter
     
  6. May 1, 2008 #5

    So that's understood? I'll then just add....

    E = mc[tex]^{2}[/tex]
    . .= (3.6 x 10[tex]^{9}[/tex] kg)(3 x 10[tex]^{8}[/tex] m[tex]/[/tex]s)[tex]^{2}[/tex]
    . .= (3.6 x 10[tex]^{9}[/tex] kg)(9 x 10[tex]^{16}[/tex] m[tex]^{2}[/tex][tex]/[/tex]s[tex]^{2}[/tex])
    . .= 32.4 x 10[tex]^{25}[/tex] [tex]\frac{kg \cdot m^2}{s^2}[/tex]
    . .= 32.4 x 10[tex]^{25}[/tex] Joules

    Simple as that?
    Yipee! :biggrin:

    Are my calculations correct? exponents....
     
  7. May 1, 2008 #6

    malawi_glenn

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    Yes, since F [N] = m*a [kg * m/s^2 ]
     
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