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Mass Energy Equivalence - E=mc2
Transcript of Mass Energy Equivalence - E=mc2
"Does Inertia of a Body Depend Upon Its Energy Content?"
released by Einstein as a part of a four-part paper series
This was the first time energy, change in mass, and
the speed of light were united in this way
The concept of space-time was born.
The Derivation of the Formula
First, imagine a cat in space (in a vacuum) emitting light.
Its velocity does not change because the emission of energy
is uniformly emitted.
Imagine that you now travel away from the radioactive
cat in a spaceship. The cat will appear as if it is moving
past your spaceship.
The cat would also have lost energy,
due to conservation of energy.
Therefore, the cat has some kinetic energy
relative to the spaceship. But its energy
decreases when it emits that flash of
energy at the beginning of the video.
His preceding paper was titled "On the Electrodynamics of Moving Bodies",
or in German, "Zur Elektrodynamik bewegter Körper"
E' = E(1 + )
So to go back to the cat's energy, the energy the cat has is
its kinetic energy (as it 'moves' relative to the spaceship),
but with the energy from the flash of light it emitted deducted,
which is E' = E(1 + )
On the other hand, if you witness the flash of light that
the cat emits, and then you fly away, you will see it lose
energy (-E) and then gain KE relative to the spaceship
Then, KE - E(1+ ) = -E + KE .
Applications of E = mc
With some rearranging, we get E = mc .
When radioactive elements such as radium are decaying,
they are emitting calories of heat.
These calories were converted from the mass of
the radium, so that mass is being directly converted
The equation could be used to power
spaceships without the use of gallons
and gallons of fuel.
i.e. photons from the sun
A lesser known version of the equation, E = m c + p c ,
is used frequently in kinematics (p = momentum)
It is applied in neutrino detection in the Arctic.
For example, when two hydrogen atoms fuse,
they form a helium atom, but the mass of the
helium atom is less than the mass of the
two hydrogen atoms. The missing mass has been
converted to energy.
Here are just a few fields of physics to which
E = mc is vital.
Reactions between atoms have also been explained
in similar ways
How does this thing work anyways?
Wikipedia tells us:
"Mass–energy equivalence is the concept that the mass of a body is a measure of its energy content."
Mass is not converted directly to energy.
Conservation of mass applies, and so
does conservation of energy.
The conversion results from the elimination of quarks and antiquarks or any other fundamental particle and its corresponding antiparticle; their annihilation releases energy.
Antiquarks are essentially the antiparticle
to quarks, which make up the mass.
Cheng, Chae, Henry
YES, THAT DUDE -->
Antiquarks and quarks are split up
when energy is taken from the matter.
The energy is re-introduced when the
antiquarks and quarks cancel each other out.
Energy is emitted as photons, an elementary
particle with no known substructure.
All matter is made of fundamental particles.
One naturally occuring fusion process in nature is the fusion that occurs in stars. It is the fusion of four protons into one alpha particle, while two positrons, two neutrinos, and energy are released. Two neutrinos turn two of the protons into neutrons.
A few questions for your enjoyment!
If the earth (mass of 5.28 x 10^24 kg) is moving at the speed of 50,000,000m/s (1/6 c).
and anti-earth (se mass)is moving at the speed of 200,000,000m/s (2/3 c) in the opposite direction
what will be the energy produced when they collide?
Use E = m c + p c.
The Earth, of mass 5.98 x 10 kg, moves along its solar orbit at an average speed of 2.96 x 10 m/s. How much mass, if converted into energy with an efficiency of 80%, could accelerate the Earth from rest to that speed in 940,000,000 km?
Hint: use W=Fd.
E is measured in joules,
m is measured in kilograms,
and c, the speed of light, is measured in m/s.
The combustion energy of coal is approximately
3.2x10 J/kg. If 1.0 kg of coal is burned, what
percentage of the coal is lost in the
V =V +2ad
(2.96x10 ) = 0 + 2(a)(9.4x10 )
(5.92x10 )=(1.88x10 )a
E=(5.28x10 )(3.1489x10 )(9.4x10 )
E=mc , with 4/5 (80%) efficiency
1.5629x10 =(4/5)m(3x10 )
m=2.1706 x 10
3.2 x 10 = m(3 x 10 )
m=(3.2x10 )/(9 x 10 )
m= 3.56x10 kg
((3.56 x 10 ) kg / 1 kg) x 100% = 3.56 x 10 %
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