E=mc2


How do I solve this, or do it?

What is it?????!!!?????

I know the basics.....but really, I need help!!!!!!!


Thanks,

Try Physics & Math

E=mc^2

It says that mass and energy are related, by that much. You need to know the units everything should be in. I think it's useful for when you are considering a particle that changes velocity or for nuclear fuel related questions (eg. how much energy will you get out of this amount of mass).

I remember trying to use E=mc^2 in Physics 12 for a momentum question, for fun, but it didn't work... so I can't imagine what you expect to use it for (nuclear physicist? you should know this stuff!).

Thanks,

Can someone help me write a equation(problem)to it?

a system has as much internal energy as it has mass times a constant factor c². how do you solve it? well, either you are given the rest mass of a system and asked how much rest energy it has, in which case the answer is E=mc², or else you know how much energy it has, and are asked to solve for mass, in which case the answer is m=E/c²

ganz einfach.

Example question: Two deuterium nuclei fuse to form a helium nucleus. How much energy is released by this process?

Note:

deuterium = 1 proton + 1 neutron
Helium = 2 protons + 2 neutrons

Think of a light beam going in the x direction, polarized in the y, hitting a surface in the yz plane. The power absorbed by a charge q is Fvy , where F is Eq, electric field times fuse, and vy is the velocity section. The magnetic force on q is f = qvxB. Recall that the Poynting vector is S=(1/m o )ExB, and show that f is in the same direction as S, the x direction in this case. The magnitude of the force is f = qvy B. Eliminate vy from these equations and use B = E/c in a light wave, and show that the force f = power absorbed/c.

Note also that the impulse (force×time) is energy absorbed/c.
So light strikes a free object, the momentum given it is energy absorbed/c.

In other words I don't really know. O_O

I think the guy just wants a problem that you can solve using the equation.

A simple example: a main sequence star radiates 5000 Megajoules of energy every second (this figure may be ridiculous, I'm not an astronomer). Find the mass the star loses in a day.

Now 5000 MJ =5 x 10^9 J

One day = 24 hours = 60 x 60 x 24 seconds = 86400

Therefore the star loses 86400 x (5 x 10^9)

= 4.32 x 10^14 J/day

E =mc^2 ; m = E/(c^2) = (4.32 x 10^14)/(3 x 10^8)^2

= 0.0072Kg
(This may be in grams, but probably Kilos since 0.0072 seems a bit small)

However I believe that Einstein originally used the equation with the speed of light in vacuo in miles per hour and the energy in ergs.

BINGO...

Hit the nail on the head...

Thats perfect........................

Thanks mr. electron:cool: :D :cool:



-Guyute

Originally posted by Guyute
Thanks,

Can someone help me write a equation(problem)to it?

Here another one:
A system is composed of two photons, one with energy E₁ and the other with energy E₂. They are moving in opposite directions.

What is the mass of the system?

Answer:

Let us call the direction of the first photon x. Since a photon is massless, it will have momentum in the x direction equal to +E₁/c.
The other photon having energy E₂ , but going in the opposite direction, will have momentum: -E₂/c.

So the total energy of the system will be:
E = E₁ + E₂. (1)

and

P = E₁/c - E₂/c. (2)

Calling the mass of the system M and using the fact that for any system with mass M, energy E and momentum P we have:

E² = P²c² + M²c⁴,

We find that:
M²c⁴ = E² - P²c².

Using equations (1) and (2) we get:

M²c⁴ = (E₁ + E₂)² - (E₁ - E₂)²

= 4E₁E₂
or

M = 2√(E₁E₂)/c².

Guyute: The following is an example of how to use E = mc² The fission process in the interior of the sun converts 4 atoms of hydrogen into one atom of helium. This process is also the hydrogen bomb reaction. The atomic weight of hydrogen is 1.007 940 The atomic weight of helium is 4.002 602 Hence: 4*1007.94 = 4031.76 grams (a little more than 4 kilograms or about 9 pounds) of hydrogen is fused into 4002.602 grams of helium. The helium weighs 29.158 grams less than the hydrogen fused to create it. This discrepancy is called the packing energy, and is the amount of mass converted to energy via the fusion process.Applying E = mc² to the above gives a big number. I am not sure what units apply to the result, but the arithmetic is the following. (using the velocity of light in centimeters per second).

29.158*299,792,458² = 8,741,348,490.364 (about 8.74 billion), which is a lot of energy in whatever units.

Originally posted by Dinosaur

29.158*299,792,458² = 8,741,348,490.364 (about 8.74 billion), which is a lot of energy in whatever units.

If the mass is in grams and c is in cm/sec, your result is in energy units of erg.
If the mass is in kg and c in m/sec. the result us in Joules.

In your calculations you have taken the mass in grams and c in units of m/sec, and you also forgot to take the square of c.
The result that you get when you perform the correct calculations is 2.6 10<SUP>15</SUP> Joules!!!! (and just from 4 kg of hydrogen).

To understand what this means let us just calculate the amount of water at 20<SUP>o</SUP>C that you can evaporate with those 4 kg of hydrogen.

The heat capacity of water is 1 cal/g Kor 1000 cal/kg K.the heat of evaporation of water is 540 cal/g or 540000cal/kg. If you want to evaporate 1 kg of water at 20 <SUP>o</SUP>C you must give it (540+1*80)*1000 cal (80 is the 80<SUP>o</SUP>C which is the difference between 20<SUP>o</SUP>C and the temperature of evaporation), and since 1 cal = 4.18 J, you must give: 2.6*10⁶ J. Since you have from the 4 kg of hydrogen
2.6 10<SUP>15</SUP> J you can evaporate 10<SUP>9</SUP> kg of water or 10<SUP>6</SUP> tons.
Finally, in order to feel what are 10<SUP>6</SUP> tons of water, let us look at an olympic swimming pool which is 50 m long, about 20 m large and has an average depth of 2 m. It has a volume of 2000 m³, so it contains 2000 tons of water. So with your 4 kg of hydrogen, if you succed in transforming it into helium, you have enough energy to evaporate 500 olympic swimming pools.

100f: Thanx for telling me that the units are ergs.

BTW: A kilogram is about 2.2 pounds, so 10⁹ kg is about 2.2*10⁹ pounds.