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The statement that photons are massless means that photons do not have rest mass. In particular, this means that, in units where $c=1$, the magnitude of the photon 3-momentum must be equal to the total energy of the photons, rather than the standard relationship where $m^{2} = E^{2}-p^{2}$. But, you can create multi-photon systems where the net momentum ...

5

Yes, it's possible. A static setup like this will work as long as any small motion of the parts would increase the potential energy. In this case, it looks like there is only one possible motion - rotation of the entire ruler-hanger-hammer piece about the axis where the ruler touches the table. If the ruler were to rotate down a little bit, the entire ...

4

Yes, you do need to add in the mass of dark matter if it's present, however on small scales the dark matter is almost uniformly distributed. To see this, consider formation of the Solar System from the original dust cloud. If you take some test particle far from the Sun and let it fall towards the Sun it will accelerate towards the Sun, then pass it and ...

4

The fraction of baryonic matter to dark matter is not deduced only from galactic dynamics. It is also derived from big bang nucleosynthesis and from the higher multipole acoustic peaks in the CMB spectrum. I would say that the element abundance is a far more important indicator of the fraction between baryonic and dark matter. Big-Bang nucleosynthesis ...

3

If you are close to a 'bell' you will be pulled towards it, assuming it's mass is far greater than that of the bar and that the other 'bell' is not far far greater. The centre of mass of a system of objects is separate from where the force of gravity pulls you towards. For instance, the centre of mass of the Earth-moon system lies somewhere in space ...

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You have 3 obvious weak points in the system. Where the (longer) arm is attached to the existing bracket Where the bracket attaches to the pole Where the pole is attached to the ground / base. Notation I'll use: $W$ = weight of monitor $L$ = maximum arm length $=> M = WL$ = moment at point of pole attachement $x$ = distance between attachment ...

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I guess I shouldn't try more than a one liner in a "comment", it looks like (enter) causes a new comment.... In any case, it is an interesting question. My reading was that the unexpectedly high abundance was for eliptical galaxies, which are usually the largest ones. So the significance may be higher than you assume. Supposedly the Milky Way (which is a ...

2

This was intended to be a comment, but is too long so I will post it as an answer. First of all, a disclaimer, I am a physict and all that I know about quantitative finance comes from self-learning, so please feel free of correcting me if I am mistaken (also in the physics stuff, of course!). I have been doing a little of research, and perhaps you are right ...

2

If drag is ignored, it won't fall. Note: There is no centrifugal force, only centripetal force. Without drag, both of them have the same moment of inertia, ... etc. such that they are balanced so they won't fall. However if drag is considered obviously the object with a larger surface area will be exerted on a larger force, causing a net force + net torque, ...

1

The continuity equation is not violated in either of the situations you describe above. The generic continuity equation for some scalar quantity (such as density) can be written as $\frac{\partial \psi}{\partial t} + \nabla . \psi\mathbf{u} = \sigma$, (1) where here $\psi$ is the some conseverd scalar quantity, $\mathbf{u}$ is the velocity ...

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Your first port of call might the relevant wiki page, Conservation of mass. Though I admit that the page is confusing because it includes the "relativistic mass" and "rest mass" terminology, e.g. "For the special type of mass called invariant mass$\ldots$" - I suggest it is rewritten. It was found that in classical physics, mass is conserved, which tallies ...

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I believe that what you are suggesting would violate causality. The original up-quarks are spacelike separated from the final ones, so they couldn't possibly interact. Your observation that this would contravene the conservation of mass is in fact a deep one. Particle interactions like the one you describe are usually considered in the context of quantum ...

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As far as the total weight being lifted obviously it's the same. But, there are two factors I can think of. 1) is the psychological factor of lifting what appears to be more mass. But, 2) is that the distribution of mass affects the moment of inertia. This means that it's harder to make corrections to the bar once it's rotating. Also, each half of the ...

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To start with, the downward force is the same in both cases, so you're right to say that might 'feel' heavier rather than actually being heavier (of course!). My guess at the reason for this illusion is that in example 2, the mass is distributed further from the centre of mass (of the bar+weights), giving a higher moment of inertia for the bar+weights. This ...

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I think that there is some confusion in your understanding of relativistic physics in the statement here: In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per E=Δmc2. When a neutron decays and you are left ...

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