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31

Both values are computed as a position weighted average. For the center of mass we average the mass in this way, while for the center of gravity we average the effect of gravity on the body (i.e. the weight). $$ \begin{align*} x_{com} &= \frac{\int x \, \rho(x) \,\mathrm{d}x}{\int \rho(x) \, \mathrm{d}x} \\ \\ x_{cog} &= \frac{\int x \, \rho(x)\, ...


26

A particle's rest mass never changes. It's mass is a natural constant, and one of the numbers which uniquely identifies it (like its spin). On the other hand, the invariant mass of the atomic system does increase as the electron becomes excited, bringing the atom into a higher energy state. In that sense, the atom (not the electron) gets "heavier" because of ...


26

This is really an extended comment to Geoffrey's answer, so please upvote Geoffrey's answer rather than this. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen ...


23

As a quick rehash of layman's terms definitions you've probably heard: The Center of Mass (CM) represents a single point where you could treat the object as a point particle, with the combined mass of the object. It's found by the average location of the mass of an object. The Center of Gravity (CG) is a point that represents the average pull of gravity on ...


12

I assume your question was asked with the implicit "and everything else is kept the same" (still GR + standard model, just with one parameter tweaked). This would have a large effect, because now the neutron would be much more stable! The neutron is already quite stable (~ 10 minute half life), due to the tight energy constraints in the reaction decaying ...


8

"According to Newton's law the negative mass should be repelled" -- Nope, in both Newtonian physics and in general relativity, negative mass would be attracted gravitationally to positive mass, although negative mass would exert a repulsive gravitational effect on positive mass (but if the negative mass is small compared to the mass of the black hole this ...


8

The centre of mass is the average point of the "mass" of the body, whereas centre of gravity is the average point of the "weight" that is mass times the local gravitational acceleration. For small objects both of them are almost same, but for large objects as the value of gravitational acceleration can change along the body (as the gravitational ...


7

Here you assume that the weighing machine does not measure the inertia of the object, but as is usually the case, it measures the normal normal force exerted on the object as it rests on the weighing machine. The total force exerted on the object of mass M in rest is zero, the force of gravity of M g acts downward. The buoyancy force acts upward and is due ...


7

Suppose the mass of your block is $m$, then the downward force due to gravity is: $$ F_g = mg $$ where $g$ is the acceleration due to gravity. This force is what we normally think of as the weight. But you're weighing the blocks in air, and Archimedes' principle tells us that if an object of volume $V$ is immersed in a fluid of density $\rho$ then there is ...


6

Yes, it is correct to say that the Higgs boson, just like other elementary particles, get its mass from the interaction with the Higgs boson – which means "with itself" in the case of this particle. More concretely, the mass may be derived from the Higgs potential (energy density) $$ V(h) = \frac a4 h^4 - \frac b2 h^2 + c$$ where the additive shift $c$ ...


5

Let bottomleft point be $(0,0)$ and assuming each small segment is a uniform square of side 1 unit. The $y$ coordinate of the center of mass will be $y_{cm}=\dfrac{28\cdot4.5+25\cdot3.5+27\cdot2.5+27\cdot1.5+26\cdot.5}{143}\approx2.34$ Similarly the x coordinate will be, ...


4

In the Standard Model, fermions are given their mass through yukawa terms, described by the Lagrangian: $$\mathcal{L}_{\mathrm{F}} = \overline{\psi} \gamma^{\mu} D_{\mu} \psi + y_{\psi} \overline{\psi} \phi \psi$$ Where $y_\psi$ is the yukawa coupling and $\phi$ is the Higgs field. At this stage, much like the gauge bosons, the fermions do not yet have ...


4

Yes, your expectations seem reasonable when thinking of the Higgs mechanism induced mass, as explained by @nmoy. However, note that one needs to be careful in defining what one means by "mass" at high temperatures. A theory at finite (non-zero) temperature breaks Lorentz invariance. There are multiple ways to think of this: There is a preferred frame, ...


4

Javier, brings up some interesting points. However, protons and neutrons get most their mass from relativistic quarks. If the quarks could be slowed down they would weigh a few electron masses. So what ever is responsible for giving the electrons mass its value seems to be giving the quarks their rest mass. I just checked the up and down,they are around 4 ...


3

Assuming you are talking about exoplanets, I'll offer this. To obtain a density you need a mass and radius. Masses come via two methods - either measuring the radial velocity variations of the star it orbits (the bigger the RV variations, the bigger the planet mass), or so-called transit timing variations. This latter works in multiple "transiting planet" ...


3

I think as time goes on, the layer on the down part of the hourglass becomes thicker. So as the result of free fall, each next grain will have less velocity as it has come a lesser distance.So as we consider that each grain stops after hitting the bottom, the amount of impact of each grain decreases.So at first, the scale will show a great (comparing to ...


2

You don't decide yourself to make $dv$ negative or positive, you only find out what it is! For example, take the equation $x < 0$. Well that means $x$ is negative, so (by your logic) I should replace the equation with $-x < 0$, but now it says $x$ is positive?? This is in essence what you're doing, so I hope you can see the flaw. The way derivatives ...


2

Let's take the things step by step, from linear momentum conservation. Of course, the rigorous treatment is by Tsiolkovsky's law. Here I am saying something less rigorous but intuitive. To increase the velocity from zero to $\Delta V$ you consume a mass of fuel $\Delta m$. When I say from zero, I mean that I consider a discrete series of quick fuel ...


2

So I am guessing you have a a 2 dimensional density? say, kg/m^2? In this case, you will need to get the area of your box -> say it is 2m^2. The mass of the box is 2 * 0.5 = 1 kg. But the mass of your person is 80kg. So your new mass is 81kg. This means the new density will be 81/2 = 40.5kg/m^2. The item will only float if the item has a lower density ...


2

One should always specify whether one is talking about rest mass per unit rest frame volume, $\rho_0 = m_0/V_0$, rest mass per unit observer-frame volume, $D = m_0/(V_0/\gamma) = \gamma\rho_0$, or relativistic mass per unit observer-frame volume, $(\gamma m_0)/(V_0/\gamma) = \gamma^2\rho_0$.1 (I can't imagine the fourth case, relativistic mass per unit rest ...


2

In our Moon, the center of gravity is not the same as the the center of mass. The result is that the Earth always sees the same side of the moon. This is becasue gravity pulls at the center of gravity, but the orbit is determined by the center of mass. The center of mass determines the kinematics - how an object will rotate, spin, revolve, and orbit. If ...


2

It sounds like that site is discussing the Standard Model neutrino. Neutrinos were presumed to be massless for a long time and the SM still models them as massless. We now know that this is not true, that (at least two of the three flavors of) neutrinos do have a small mass, although the SM is still a good approximation. So, like other massive particles ...


2

Exoplanet means those planets that arent in our solar system, and solar system planets are in our solar system . The two (blue)triangles must be earth and mars i suppose and purple points are exoplanets and yes most of the exoplanets are made of Hydrogen and helium and revolve around other stars from the graph given .


2

Everything in the equation is treated as a function of time. $p(t)$ is defined to be the momentum of the rocket at time $t$. $m(t)$ is defined to be the mass of the rocket at time $t$. $v(t)$ is defined to be the speed of the rocket at time $t$. ("Rocket" here is the system that is actually losing mass. This is where the whole discussion originates from; ...


2

The center of mass of an object is the point where the first moment of mass is zero. Put differently, when you support the object at that point, it will be balanced. Assume that point is $x_0$, then $$\int_0^\ell (x-x_0) \lambda(x) dx = 0$$ Substitute $\lambda$ and some simple manipulation will give you an expression for $x_0$. Let us know how far you ...


2

If it had a foundation in the laws of physics, I don't think we would call it magic... The word "magic" seems to be used for an impossible un-explainable process (if it was explainable, it might not be interesting enough to be used in a movie and to be called magic). You will have to close your eyes for many physical laws to accept it as such. Magic can, ...


1

Your claim that nothing will happen because they initially have no velocity is where things start to go wrong. In fact everything in relativity has the same speed (the speed of light). A particle that looks to be "at rest" in some reference frame simply has all of its velocity pointing in the "time direction". This is an intuitive reason why you would see ...


1

The center of mass (or gravity) is given by the formula (see Wikipedia, http://en.wikipedia.org/wiki/Center_of_mass) $$ (1) \ \Sigma_{i=0}^N \ m_i (\vec r_i - \vec R) = 0$$ If someone has a problem with the uniformity of the shapes, then we can replace the sum by an integral, and inside each shape consider the mass as a function of $\vec r$, i.e. $m(\vec ...


1

A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite? Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite. ...


1

In the standard model the mass for the quarks and other elementary particles comes from from the Higgs mechanism. In the case of the quarks, the masses given in the table of the link are calculated using convoluted theoretical models and data input from scattering experiments.4 At the moment chiral symmetry breaking does not contribute to the quark masses ...



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