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0

I am not sure what you mean by if something has more energy it should "slow time" more than something with low energy But even if I just neglect that for a moment. I'll use a description of waves for visual aid and simplicity. Imagine in "not slowed down time" a wave has a period (and frequency) of 1 in this time. So there is one wave crest and one ...


0

It's probably best to not think of single photons as sources of gravitational energy. For one thing, most bulk electromagnetic fields are not eigenstates of the photon number operator. For another thing, the thing that couples to the gravitational field is the energy density of the field. This density is proportional to the intensity of the field, ...


2

Its the mass of the atom with atomic number Z and mass number A. If you study the right side of the equation you will see that the first two terms are the total mass of the protons and the total mass of the neutrons. All other terms are modifying this sum of the masses of the constituent particles. So to calculate the mass defect, simply take the first two ...


2

See the Wikipedia article on the semi-empirical mass formula. The formula is for the mass of the nucleus, so you would need to add on the mass of the electrons (and subtract their binding energy) to get the mass of the atom. However, as previous questions have mentioned, the mass of the electrons is a small correction and in fact generally smaller than the ...


3

Have a look at this table, which shows the binding energy (i.e. the ionisation energy) of electrons in various atoms. The highest energies are a few hundred eV, and those are for the core electrons (though admittedly it doesn't show $1s$ energies for the heavy atoms) so the average electron binding energy will be lower. By contrast, the average binding ...


0

The binding energy of the nucleus is an energy (or equivalent mass via $c^2$) which in some sense accounts for the "extra" (or lacking) energy which is not accounted for by the mass of its constituent particles. The binding energy per nucleon is exactly what it says, the binding energy divided by the number of nucleons (a nucleon is a proton or a neutron).


0

binding energy per nucleon means the nuclear energy between nucleons to attract each other.whereas binding energy of atoms is the bond energy of atoms to react with others.


1

To understand how light is affected by gravity, it helps to think of light as energy. So let's ask a basic question: when it come to light, what is energy? By the Planck-Einstein relation, we know $$E = h \nu$$ where $\nu$ is the frequency of the light and $h$ is Planck's constant. So when we talk about the energy of light, keep that in mind. Also note ...


0

I presume you are asking will heated gas weight more? Remember E=mc^2, mass-energy equivalence. The gravitational force will depends on the stress energy tensor in general relativity. The stress energy tensor 00 component is the total energy of the body, which includes the rest mass plus the kinetic energy of the object. Note that the rest mass does not ...


0

It will weigh more as long as nothing boils off the surface. Why? Because $E = mc^2$. The heat energy applied adds an extremely small amount of mass (waaaaay smaller than you could measure with the equipment available to you).


-1

Firstly we can discuss the definition of the kinetic energy. Because, we have v^2 in the kinetic energy so a new definition for the speed of light leads to new definition for whole unit system. I mean if you write c instead of c^2, you get energy in a different unit system so you should redefine velocity, energy, etc otherwise they are not compatible....


1

Where $E_0 = mc^2$ comes from: From elementary physics, the work $W$, done on an object by a constant force of magnitude $F$ that acts through distance $s$, is given by $W = Fs$. By work-kinetic energy theorem, $$KE = \int_{0}^{s} F ds$$. In non-relativistic physics, the KE of an object of mass $m$ & speed $v$ is $$KE = \dfrac{mv^2}{2}$$ To find the ...


0

It's what we get through the derivation(s). If you look it up, we derive the equations and they don't give you $mc^3 \quad \text{or} \quad mc$.


4

All these kinds of questions boil down to a pure question of convention: there is no in-principle reason why one couldn't think of your proposed $c^2$ as the "fundamental" constant: it's simply that generations of people have found it less awkward to use $c$ instead because one needs a parameter in special relativity that makes a time interval into an ...


2

Since unit of $c$ is $\text{ms^{-1}}$ and unit of $m$ is $kg$. hence to get energy in $kg \frac{m^2}{s^2}$ we assume $c$ to be as it is, $\approx 290000~\text{km/s}$ $c$, $E$ and $m$ are not unitless values. By redefining mass unit (say, you start to measure everything tomorrow in $\sqrt {kg}$'s) you can change given formula to $E = kg^2c^2$


11

There is only one mass. Lets make this clear. The concept of "relativistic mass" is not really a useful concept in my opinion. The invariant mass, or simply the mass, is defined as (in natural units, so $c = 1$): $$E^2 - p^2 = m^2$$ The reason this is a much more useful definition for a mass, is because this quantity is Lorentz invariant, meaning it has ...


5

Defination of mass is : $$m^2=E^2-p^2$$ Where $m$,$E$ and $p$ are mass,total energy and momentum of the object respectively. Its not necessary that if a photon has energy it should have mass because an object or a particle can have energy due to its momentum alone or its mass alone or due to its both mass and momentum and thus a photon has energy due to ...


4

Photons never have zero mass. As you point out, they have energy, therefore they have mass - usually referred to as "relativistic mass". Photons are said to have zero "rest mass", ie no mass when they are at rest (stationary). But photons are never stationary, so this really has no physical meaning. That they would have no mass at rest is required by ...


1

The problem with this is that $$E = mc^2$$. Density, however, is given by $$density = \dfrac{mass}{volume}$$ Thus, if volume = 0, then density is infinite. Black holes have a finite mass. It is there density which is finite because all the mass is at a single point (singularity, volume = 0).


1

No, the energy of a black hole is not infinite. It depends on its mass, angular momentum and charge. Infinite density at a point does not translate to infinite energy in the $E=mc^2$ sense. It is in fact possible to extract energy from black holes by exploiting certain properties of accretion disks or ergospheres, but this is a finite process.


3

The semi-empirical mass formula is derived from a fit to the measured masses. If you know the numbers of protons and neutrons then the idea is that the SEMF should give you a good estimate of that mass (there are of course small residuals of the order 0.2 MeV to the fit). $$M(A,Z)c^2 = (A-Z)m_n c^2 + Zm_pc^2 - AE_b,$$ where $A,Z$ are the mass number and ...


2

The semi-empirical mass formula (SEMF) is much more involved than this, and tries to take into account various other factors that will affect the mass/energy of the nucleus. For any nucleus, of course if you KNOW its mass beforehand then simply subtracting the individual masses of protons and neutrons will give you ($\pm$)the binding energy, but the SEMF at ...


3

There are plenty, plenty and plenty of equations with constants under the different degrees. The main reason for this: the $c^2$, $\hbar^2$ and other doesn't have any direct physical meaning; contrary the $c$, $\hbar$, $e$, various masses and other.


1

einstein did not give e=mcc at first rather he had given l=mvv where l is energy and v is velocity o f light. e =mcc is written in this way to show the unique dependence of energy to the speed of light.


2

From a relativistic point of view, the energy of a particle with mass $m$ and momentum $p$ is given by: $$E^2=m^2c^4+p^2c^2$$ where $c$ is the speed of light. You can clearly see that a massive particle will have some mass energy $E_0=mc^2$, but also some kinetic energy. The photon being massless, the above equation reduces to: $$E_\gamma=pc$$ One ...



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