karlaLTR
  • karlaLTR
the energy output of the sun is approximately 4.0 x 10^26 J/s. If all this energy results from mass-energy conversion in the fusion process, calculate the rate at which the sun is losing mass.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I think we can approach this using differentials, which will introduce the rates for us. We know that\[\huge E=mc^2\]Differentiating will yield:\[\huge \frac{dE}{dt}=c^2\frac{dm}{dt}\]You're given the rate of energy (should be negative because it's an output) and c is just the speed of light. Solve for the rate of mass (which should also be negative because it's losing mass)!
anonymous
  • anonymous
This is with respect to time, of course
karlaLTR
  • karlaLTR
Is there any way I could do this only using the E=mc^2 formula??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
That is using the formula, just indirectly. The issue with using E=mc^2 is that those are exact instantaneous values. We're given RATES, which are not instantaneous. In order to calculate this, we have to find differentials which are also rates. To avoid the clutter, you can use: \[\huge \text{E}'=c^2m'\]The ' (called "E prime" or "m prime") refers to the derivative/rates of each variable.
anonymous
  • anonymous
It's different notation, but the same meaning as what I wrote before. It just looks nicer.
DariusX
  • DariusX
M = E/c^2 E = 4.0 x 10^26 J = 4.0 x 10^26 / 10^-7 = 4.0 x 10^33 erg c = 3 x 10^8 m/s^2 M = 4.0 x 10^33 / 9 x 10^16 = 4.4 x 10^16 g/sec sun loses 4.4x10^16 grams per second
anonymous
  • anonymous
There is an issue with doing that. Mathematically it works out, given the linearity. Conceptually, E or m is not a rate -- they're given instantaneous values.
anonymous
  • anonymous
We are asked to find rates, so we must alter the equation so that our variables are also rates. We do that by differentiating. dm/dt can be spoken as "the rate of change of mass with respect to time" and similarly with dE/dt. This is what we want because this is what the problem asks for.
anonymous
  • anonymous
I think @IrishBoy123 can explain in better detail for you
karlaLTR
  • karlaLTR
thank you! I'm just a little confused because I haven't learned all this in class yet, the furthest I've learnt is E=mc^2
IrishBoy123
  • IrishBoy123
i'm with you on this @CShrix hello @karlaLTR !!
anonymous
  • anonymous
Have you learned any sort of calculus yet?
karlaLTR
  • karlaLTR
not yet, I've only taking grade 11 physics at the moment
anonymous
  • anonymous
Mmm, this really what this problem requires. The curriculum might let you use E=mc^2 because the relationship between energy/mass and the rate of change of energy/mass is still linear. E' = c^2 m' E = c^2 m That is why DariusX's method works. Conceptually, this is wrong because mass and energy are not rates. They're specific given values. A rate is given by a change during a time interval. I think for your class' sake, you can just use E=mc^2. If you learn calculus later in HS/College, you'll see why this is like this.
karlaLTR
  • karlaLTR
I understand the logic in what you're saying, I think your way is a lot more efficient in terms or accuracy but It's easier to do it the other way for now since I'm not required to get too complicated for now, thank you for your help! You're really smart :)
anonymous
  • anonymous
You are very welcome X) Best of luck!

Looking for something else?

Not the answer you are looking for? Search for more explanations.