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1

I would not write $dS=0$ for a cyclic process but rather its closed contour integral $\oint dS = 0$, then you also have $\oint \frac {\delta Q}{T} \le 0$ and from $T>0$ you will get that in a cyclic process you must have somewhere, sometime, at some stage of the cycle $\delta Q < 0$, that is some amount of heat must be rejected to complete the cycle. ...


1

There is no such thing as negative energy, except in the mathematical sense of representing a deficit of (positive) real energy. It's just like having money (real, positive), where having it in a financial sense is called "long", and if you owe money, then you are short (of a few bucks), but the actual money is always real and positive.


0

Since the block are on a frictionless surface, they don't interact with the Earth, so it doesn't make much sense to talk about the "block-earth" system. Within the context of this block-block system, the mechanical energy is always conserved. In terms of equations, at the point of collision we have $m_{1}v_{10}=m_{1}v_{1}+m_{2}v_{2}$ ...


0

Let us assume as you move (velocity $v$) you will see the relative kinetic energy of the Universe increases, therefore energy of $KE_0$ we measure is different to the kinetic energy in perspective of the Universe of us $KE_1$, and since the mass of universe is greater than you we can assume the kinetic energy will also be greater, therefore we will model ...


1

Your answer seems to be good. I solved it in the same way. However there are two points in which i have no clarity. The first thing i do not have clear is the conservation of energy in the problem. We use the energy conservation, but how do we guarantee that the normal force is conservative? And, still, in the conservation equations we do not consider its ...


1

The other answers are great. I decided to plot it, however, because it's nice visualizing these things. Since your biggest doubt is about kinetic energy, be sure to pay attention to the last graph. SYSTEM. Motorcycle going to the left, truck going to the right, bound by an elastic rope ten meters long ($k=100\frac{\mathrm N}{\mathrm m}$). Masses and speeds ...


2

I think the Hamiltonian is not necessarily the energy for the following reason: you can demonstrate that the Lagrangian may be deduced from the D'alembert principle which is linked to the concept of force, etc. but it may be also deduced from the Hamilton's principle which is a pure mathematical concept applied to physics (a certain quantity has to be an ...


1

The energy density in an electric field $E$ is proportional to $E^2$. See this article on Wikipedia for a proof if you want the gory details. That's why when you're calculating the total energy you integrate $E^2$, as you say in your question. This also explains why there is no negative energy associated with the negative half of the wave cycle, because the ...


0

We neglect NOTHING, when we calculate the energy. You have to distinguish between the AMPLITUDE of the signal, given by your formula s(t), and the ENERGY of the signal. By the way, negative energies do exist, e.g. the electrons in an atom have negative energy - to escape from the nuclear attraction and become free to run away from the atom, one has to supply ...


0

The negative energy doesn't exist. The negative symbol indicates that the direction of the signal is opposite. The signal may be voltage signal or current signal. Suppose the $R=1\: \mathrm{\Omega}$, as the energy is $\frac{V^2}{R}t$ or $I^2Rt$. Give $dt$ and $R=1$. The differential of the energy is $V^2dt$ or $I^2dt$. Instead both V and I with $s(t)$,the ...


0

I believe your question is perhaps ill posed. I confess, I don't understand what you mean by "intensity decreases much faster" at r=1 inch vs r=1 meter. The equation shown governs the intensity at a given distance: $$ I=\frac{S}{4\pi r^2}. $$ All this says is, for a sound source of intensity S, the intensity is inversely proportional to the distance ...


0

It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy. If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components: Kinetic energy of car: $\frac12 m v^2$ Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$ ...


2

I think EnergyNumbers's Answer is an excellent one, but could leave some people a bit mystified by what exactly a "direct ray" is and what exactly its relevance is. The essential idea here is that a Fresnel lens is an imaging machine: it puts a curvature on a low aberration wavefront so that that wavefront converges. Its working depends on there being ...


0

They are not generally used with the photovoltaic cells as they concentrate the sun light and heat to a temperature so high that many of the cells get, for lack of a better word, fried.


0

This is not a full answer, it is very partial but it gives the flavor of why we take the square of the signal, and why do we integrate over time. The origin of these things is in the FLUX of a signal. When on the path of a signal we place a detector, what impinges on it, is the flux (defined as quantity of incoming energy/unit surface in the unit time). So, ...


1

The intensity of your signal at every moment is given indeed by |x(t)|^2, s.t. the total intensity of the signal during the time is the integration of |x(t)|^2 over time. About the connection with the frequency, see below. You can describe the signal by its behavior in time x(t). But you can decompose it in sine and cosine functions, i.e. (1) $x(t)$ = ...


1

If $x(t)$ is the current through a resistor $R$, then the voltage is $Rx(t)$ and then the instantaneous power dissipated is $Rx(t)\bar{x}(t)$ and the dissipated energy over all times is $E_s=\int R\vert {x(t)}\vert ^2 dt$. Because of Parseval's theorem you can write mean square in time as the mean square of the frequency spectrum, ie., Fourier transform ...


1

Our current understanding says that the entirety of spacetime, matter, and energy was compressed to a singularity. Our current understanding also says that it is impossible to know anything about what happened on the other side of the singularity. Therefore, there is no answer to your question that does not involve pure speculation or theology.


0

You always have to satisfy the momentum equations, which is only the linear momentum equation for this one dimensional case: $$m_1v(t_1)-m_1v(t_0) = \int_{t_0}^{t_1} F dt$$ Assume the collision is completely elastic and all is conservative, so no plastic deformation, drag or any kind of damping. Then the only force which acts is the gravity: $$ F = ...


1

Kinetic energy is the energy associated with motion. Therefore, we can no more measure kinetic energy free of a reference frame than we can say something is moving free of a reference frame. It must always be moving with respect to something. Energy is conserved because the total energy of a system is the same before and after a process in the same frame of ...


0

The lighter one will travel further. (unless it is ridiculously light like 0.1g or something like that) Gravity means that the higher the initial velocity the further it will travel. Think about how gravity effects motion. Imagine throwing a tennis ball at 10 m/s and a softball/baseball/hockeyball (4 times as heavy) at 5 m/s - which would go further?


5

The exact quantities of kinetic energy (like momentum) depend on your choice of a reference frame. Don't get too worried though; regardless of your choice of a reference frame, you will find that energy (and likewise momentum) is conserved within the reference frame. Therefore, two observers may not agree on the kinetic energy or momentum of an object, but ...


18

Because in the frame of reference that is co-rotating, the object doesn't move, and therefore it has no kinetic energy in that frame, which is the frame in which most problems involving objects on earth are looked at. Note that kinetic energy is evidently not a frame-invariant quantity, but it is not required to be.


1

But can time also be measured by just changes in energy, and in the absence of matter? If there are no massive particles present, then general relativity has a symmetry called conformal invariance. In the presence of this symmetry, there is no way to measure time. According to evolutionary universe theories there was a point in the evolution where ...


1

Only if matter and energy were not intertwined. How can you exhibit and measure energy without the matter it works upon? Energy is just a property used to explain matter. See summary http://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/matter-and-energy-a-false-dichotomy/ Thinking outside the box: Does the idea ...


0

"Stange" harmonics cannot appear. Two 50Hz sinus waves always produce another 50Hz sinus wave : $$cos(\omega t)+cos(\omega t+\phi)=2cos(\omega t+\frac{\phi}{2})cos(\frac{\phi}{2}) $$ About synchronization, you should remember that the voltage in the network is fixed, and thus only the current changes. I don't see any problem with combining two different ...


0

Are quantum fluctuations one way to transform energy to matter? Yes, at least in theory. This is how Hawking Radiation is predicted to work. In this case the gravitational energy of a black hole 'boosts' the energy of a quantum fluctuation to create an actual particle/antiparticle pair, one of which gets sucked into the black hole and one of which ...


2

As well as the other answers, in particular, Anna V's comprehensive Answer, I would like to capture Ben Crowell's comment for permanence in this dicusssion: There's nothing special about nuclear reactions. Chemical reactions also result in a change in mass due to the energy released. It's just that the energy scale for chemical reactions is about $10^6$ ...


2

Wood burning is an example of converting mass into energy. Another is a seed which takes energy from the sun (and water, air etc.) and converts it into matter.


0

One way to derive the existence and expression for entropy is to consider a system that undergoes arbitrary cyclic process $\gamma$ in which its temperature is changing and is the same as temperature of a heat reservoir it is connected to. The cyclic process can be decomposed into a set of infinitesimal Carnot cycles, where the change of state $s$ during ...


6

First, the concept of matter and energy are entagled since Einstein. Most people have an intuition on the word energy as some untouchable fuild with destructive power,mainly due to hollywood. But energy and mass have a much more specific, mathematical meaning in physics. So instead, let me rephrase the question as "does it makes sense to talk about time in ...


0

They have both frequency and wavelength (or wave vector $\mathbf{k}$), but the frequency and the wavelength are not related as in a real photon. Instead, they are independent integration variables of a Fourier integral representing a near field. For example, you can represent a "moving" Coulomb field $V(\mathbf{r}(t))$ as a Fourier integral over $\omega$ and ...


0

Virtual photons are not on the mass shell, that means that their mass can be different than zero in the calculations using the Feynman diagrams. They are not observable in any sense other than as the result of the calculations that need them. They appear because in an interaction quantum numbers are conserved, and these photons are used to carry the quantum ...


0

Energy is the ability to do Work is a usual definition in textbooks and schools. But work is not the ability of energy (if i can use such a term for a second). This is made more explicit in thermodynamics. For mechanics and Newton's laws. Work is associated to a specific force acting on a body and only in the direction of that force (body moving in a ...


4

Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract: The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a ...


2

Prahar is correct that generally we have an energy contribution of ${1 \over 2} kT$ per degree of freedom in a system - so that atoms in a gas of atoms (e.g. Helium) will have an average energy of ${3 \over 2} kT$. Often people talk about thermal energy being '$kT$' because of the exponential expression in $N_i = N_0 {g_i \over g_0} e^{-{E_i \over kT}}$ ...


0

It is Minkowski spacetime. Both $\bar{p}=\left(\begin{array}{c}E\\{\vec p}\end{array}\right)$ and $\bar{x}=\left(\begin{array}{c}t\\ \vec x\end{array}\right)$ are invariant 4-vectors in spacetime. A 4-vector is composed of a time component and a spacial component (with 3 sub-components related to ordinary 3-space). By invariant we mean that their magnitudes ...


0

There are already a few answers that explain the mathematics behind it, but since you've said that your "math knowledge is quite limited" I'll try and break it down into simpler terms. You're already familiar with a 3D vector dot product, and it seems your confusion arises from dot products of a four-vector. Now what they didn't teach you when they taught ...


1

I had prepared this answer for a question that was made duplicate, so here it comes, because I found an instructive MIT video. (the second link) This answer is for electromagnetic waves mainly Have a look at this video to get an intuition how interference appears photon by photon in a two slit experiment. It comes because the probability distribution for ...


0

In Minkowski spacetime you assign four coordinates to your events: $x=(x^0,x^1,x^2,x^3)$ - in this notation $x^0 =ct$ and the other three coordinates are the spatial ones. Suppose these are the coordinates of a point in spacetime reached by a light ray which started at the origin, then you have that: $(x^0)^2=c^2 t^2 = \sum_{i=1}^3 (x^i)^2 \Rightarrow s^2 = ...


1

That's the definition of the dot product in Minkowski space-time. To be clear, any space-time is endowed with a metric. Standard ${\mathbb R}^3$ that you may be familiar with has a metric $\delta_{ij} = \text{diag}(1,1,1)$, $i=1,2,3$. Given two vector $\vec{v} = (v^1,v^2,v^3)$ or $v^i$ for short and similarly $w^i$, the dot product is defined as $$ \vec{v} ...


1

The thermal energy of a system is $$ E = f \frac{1}{2} k T $$ where $f$ is the number of degrees of freedom of the theory - which is roughly speaking the number of dimensions it is allowed to move in. For instance, if you are talking about an atom in 3 space dimensions, then the atom can move along the 3 axes and hence $f=3\implies E = \frac{3}{2} kT$. If ...


1

The probability of atomic excitation by photon absorption is given in terms of the "cross- section" for single photon absorption. This quantity shows a sharp peak if the photon energy and the atomic energy level difference match (the coupling between the atom and electromagnetic field causes the atomic states to broaden, the natural line width). Thus ...


0

Pistons do not convert linear to rotational motion. You are thinking of crankshaft. Having said that, in principle it is 100% efficient. Whether something better exists in the real world is debatable. For example, the piston could carry a magnet which moves through a coil which then uses the electricity generated to turn an electric motor. However, I doubt ...


0

Consider (at least perturbative) contribution of the additional effective "potential" in the radial equation when $l\ne 0$, analyze its sign and judge correspondingly. The physical explanation is simple - it is an additional kinetic energy of a rotational motion.


0

The idea behind this exercise is, to show you that everything CAN act like a wave in quantum mechanics, see slit experiments. You will have to use the same formula for all three entities. As stated before, the de Broglie wavelength is the way to go when confronted with such tasks. For the non-relativistic case, the de Broglie formula states: $$ \lambda = ...


1

Here is a hint - de Broglie wavelength for the neutron and electron - does that help?


2

All the Energy involved here comes from the boy. This is no different than jumping: the force involved in doing so propels you upwards and the Earth downwards, but no Energy came from Earth. In your example no Energy can be spontaneously produced by the block. The fact that Newton's third law expresses the action and reaction forces appearing together does ...


1

All such things as 'translational energy' or 'vibrational' are more of extrapolating concepts. Yes, there is rough approximation of $C_v$, but that works only for gases with low density. When density in your material increases, the pressure which you 'expect' is defined with not only 'impulse' part, but also 'force' part of pressure. $P = nkT + ...


0

Yes translational energy levels are very close together. These can be calculated with particle in a box type quantum mechanics for molecules in the gas phase, which are free to move about. In ice the molecules are held together next to each other and cannot move about - they can vibrate about the positions that they are held in. So their average position ...



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