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29

You've taken this out of context. The context in which Tesla is writing, he's talking about kinetic energy and heat. Just a few sentences previously he states: ...according to an experimental findings and deductions of positive science, any material substance (cooled down to the absolute zero of temperature) should be devoid of an internal movement and ...


26

I think that sentence shows that Tesla did not quite grasp the results from relativity. This is not unusual, as many physicists required many years to fully accept relativity, but by 1932 (the date of the writing of that text) I would expect it to be already orthodox knowledge. In any case, Tesla is known to have been a self-taught experimental genius and ...


18

A lot of different forms, but mostly kinetic energy. A good table is given at Hyperphysics. The energy released from fission of uranium-235 is about 215 MeV. This is divided into: Kinetic energy of fragments (heat): ~168 MeV Assorted gamma rays: ~15-24 MeV Beta particles (electrons/positrons) and their kinetic energy: ~8 MeV Assorted neutrons and their ...


16

The problem is that the two calculations have hardly anything to do with one another - so it's no wonder you don't get the same result. The electron volt, as you say, measures the work you need to move an electron across a potential difference of one volt. On the other hand, if you want to calculate the mass of an electron using $E=mc^2$, what you need is ...


4

Yes, doubling the mass doubles the energy, while doubling the velocity quadruples it. Your question is basically about order of operations; the exponent only applies to the variable it's immediately on. As you note, you'd have to put parentheses to make it cover the $m$ as well. We say that energy is linear in mass, but quadratic in velocity.


4

One can answer this question by calculating the energy needed to shift half the Earth's mass so that it is infinitely far from the other half. Let's calculate the gravitational potential energy released as we create a planet: assuming a constant density $\rho$, when the planet is growing and of radius $r$ and thus of mass $M(r)=\frac{4}{3}\pi\,r^3\,\rho$, ...


3

The energy that is released when a the absorption of a neutron causes a heavy atom nucleus to fission into two daughter nuclei comes from the tighter binding energy of the two daughter nuclei compared to the weaker (smaller) binding energy of the original nucleus. This extra energy is mostly released in the form of the kinetic energy of the two daughter ...


3

The mass of an atom is always less than the sum of the masses of the particles that compose it. The lack of mass (or energy, from E = mc^2) is called binding energy and it is the energy expended by the particles to remain confined inside of the atom. When fission occurs, not more spending of energy to hold together the individual particles. So the energy ...


3

The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept. The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes ...


3

Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...


2

People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed $c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ...


2

You ask: If the clock is running slowly compared to a distant clock is this equivalent to the clock having a lower energy compared to a distant clock? but you have to very careful what you mean by energy in general relativity. As it stands your question too vague to be usefully answered. However in the weak field limit there is a sense in which time ...


2

You correctly identified there are two angles of interest, labeled $\theta$ and $\phi$. I actually want to pick two different angles for my analysis - see this diagram: First thing to note is that if the cylinder rolls without sliding, the length of the green arc and the red arc must be the same. Length of green arc: $(\phi + \theta) a$ Length of red ...


2

The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ...


2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.


2

In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ...


2

If you computed arc length you'd chop your curve into pieces, find the distance between the end points of each piece, add them up, and then take the limit as the pieces get individually smaller and hence more numerous. Now image that instead you took the square of the distance between the end points of each piece, that wouldn't work because you would get ...


2

Yes, more or less. You don't need the volume of the object, because temperature is what's called an intensive quantity--it doesn't matter how much "stuff" there is. What you need to know is what kind of molecules you're dealing with. According to the equipartition theorem each degree of freedom gets $\frac{1}{2} k_B T$ energy, where $k_B$ is Boltzmann's ...


2

In classical mechanics there is no distinction between free and bound as far as this relation is concerned. In relativistic quantum mechanics (i.e. QFT), a particle that satisfies this relations is said to be "on-shell" or a physical observable asymptotically free particle. It is certainly not satisfied for virtual particles, but they are as their name ...


2

Zeldredge's answer is great, answering the question from a mathematical point of view. Since you asked this question in a physics forum, I'll just add to that by answering from a physical point of view, clarifying why the energy is linear in mass and not quadratic: Suppose you have two objects of equal mass moving parallel to each other with velocity $v$, ...


1

Chemical vapor deposition is used to produce diamonds at low pressure https://en.wikipedia.org/wiki/Synthetic_diamond#Chemical_vapor_deposition


1

According to the phase diagram of diamond (see for example http://files.umwblogs.org/blogs.dir/6093/files/2011/10/carbon_phase_diagram2.jpg) there is a region where diamond is stable and graphite is metastable, at pressures and temperatures in the range you are asking about: Given the slope on the diagram, you would actually expect that this reaction is ...


1

You might find the wiki article on this topic helpful. Summarizing: When you have a 1-D box, the energy states of an electron can be given by $$E_n = E_0 + \frac{\hbar^2 \pi ^2}{2 m L^2} n^2$$ Now the things to note are this: Two electrons (with opposite spin) can occupy the same level The Fermi level is the energy of the last electron After each pair ...


1

Single particle energy eigenstates for a system of particles in a box are given by $$ E_n=\frac{\hbar^2 \pi^2}{2mL^2}\,n^2 + E_0. $$ The Fermi energy for a single particle is, by definition, the value of its energy that exhausts all the possible states given by $N$ indistinguishable particles; in the case at hand, for fermions (electrons), this is given by ...


1

break earth into 2 parts in which each part is nearly equal to half of volume of the earth Yes and no. No, as in you can't neatly split the planet in half. Most of the earth is liquid and will re-form once the cutting device has passed through it. Much like cutting pudding. Yes, if you want 2 half-earth balls when you are done and don't care about the ...


1

See for example this table which contains the excess energy for each nuclide. You can take this table to compute the number you are interested in. The answer depends not only on the atomic number, but on the number of neutrons as well. This is why you need to think about how you want to represent this. I recommend you study that table and then figure out ...


1

The total kinetic power of the system will be $\frac{1}{2}mv^2_1 +\frac{1}{2}mv^2_2$. The first equation that you mention is wrong, because this equation says that you have an object of mass $m$ with speed $v_1+v_2$. If you expand the squared term you will see that it is different. Now, what do you mean hit each other? Do they have opposite velocities? In ...


1

Mythbusters ran an episode on this one. When you learn of relative velocity, you learn that from the perspective of one of the drivers, the other driver will appear to come at you at 80 m/s. So the thought is that it should be the same as hitting a wall at 80 m/s. The energy doesn't add up though. Essentially, from the perspective of the driver, they are ...


1

The bond that holds water as a liquid is a simple static electricity bond. it has a strength and will 'break' with sufficient energy. this happens all the time. water evaporates, when a random chance of circumstances through thermal agitation and exterior pressure are at the right amount the molecule leaves the liquid and goes flying off as a gas. the higher ...



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