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1

First things first: waves that have already reached close vicinity to the beach DO displace water towards the shoreline - just notice how the water moves back and forth at the point it's ankle-deep. This is related to the phenomenon by which they lose their wave form and get a crest. There are many forces acting on a surfer, but two of them are the ...


1

In principle, the gravitational potential energy should be included into total internal energy, but in practice, most often it is not. I know of two reasons. because for systems that are discussed in thermodynamics, it is believed that gravitational energy is negligible compared to electromagnetic potential energy of the constituting particles; because it ...


1

Your separation into potential energy of the system as a whole due to external force fields and energy contained within the system known as internal energy seems a bit arbitrary. Still, if you want to split the PE up this way gravitational interactions within the system would have to go into internal energy. Take the Solar System as an example. Everything ...


1

The gravitational potential for particle 1 is $V_1(r_{12}) = -Gm_2/r_{12}$ and for particle 2 it is $V_2(r_{12}) = -Gm_1/r_{12}$. $m_2$ is in $V_1$ and vice versa because the gravitational potential energy is the potential energy in a gravitational field per unit mass, and therefore only depends on the mass that is generating that gravitational field. When ...


1

I think you are confused between potential difference and energy. The energy used up can vary, if the current flowing through the resistor varies. First of all, a battery is 'not' a constant energy source. It's a constant potential source. Secondly, if more current flows, more energy is dissipated, even though the per capita energy is constant. Now, to ...


2

Happy birthday. Once upon a time, there was a king and he had a sister who also liked coffee... Or let me omit this portion of the answer. A battery is motivating the electrons to buy a round trip ticket around the circuit by the voltage $V$. The voltage is nothing else than the energy $E$ per unit charge $Q$, i.e. $V=E/Q$; the unit 1 volt is nothing else ...


2

Shouldn't the energy gain be greater than this formula describes since the energy from the electric field is applied for so long? The electron gains energy and accelerates until it encounters a collision. This is a statistical process and there's a distribution for the energy loss for many electrons. Then, it can accelerate again from that point on, ...


4

Consider for a moment, a cell that is not connected to a circuit, i.e., there is no path for current external to the cell. The chemical reactions inside the cell remove electrons from the cathode and add electrons to the anode. Thus, as the chemical reactions proceed, an electric field builds between the anode and cathode due to the differing charge ...


1

There are many fundamental concepts that both you and Trimok have misunderstood. First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that ...


1

I know which book you are referring to. It is the book "Finite elements in Engineering" by Chandrupatla and Balegundu.I also have the same question. We learnt in Physics that the Work done by the force is stored as Potential Energy. There was no mention of Work Potential.


4

Let $\mathbf x(t)$ be the path of a particle. Let $\mathbf F(t)$ be a force acting on the particle as a function of time, then the work done by the force from time $t_a$ to time $t_b$ is \begin{align} W(t_b, t_a) = \int_{t_a}^{t_b} \mathbf F(t)\cdot \frac{d\mathbf x}{dt}(t)\, dt. \end{align} where the center dot denotes dot product; \begin{align} ...


0

The definition of work, it is done on vectors, let's say $$ \vec F = F_x\hat i+F_y\hat j+F_x\hat k, $$ that would be the force and the displacement vector it is $$ \vec \ell=\ell_x\hat i+\ell_y \hat j + \ell_z\hat k $$ so you have the 3D. Then the definition is $$W=\int \vec F \cdot d\vec\ell$$ also it would be good to check anyways the dot product and the ...


0

Okay, first of all, your question is unclear. 'Potential energy of a bar' has no meaning whatsoever. PE is of a system of interacting objects or particles. We sometimes just say that the PE of a mass $m$ at a height $h$ from the ground is $mgh$. But strictly speaking, this is actually the PE of the system: Mass-$m$ + Earth. So there you go. If you are ...


3

Integrating each mass slice and just taking the center of mass yield the same result. That is because: $$PE = g \int y \, {\rm d}m = m g y_{cm}$$ from the definition of the center of mass $$y_{cm} = \tfrac{\int y \,{\rm d}m}{m} $$


2

The formula: $$ \Delta U = mgh $$ is an approximation that applies when the distance $h$ is small enough that changes in $g$ can be ignored. As you say, the expression for $U$ is: $$ U= -\frac{G M m}{r} $$ So the change when moving a distance $h$ upwards is: $$ \Delta U = \frac{GMm}{r} - \frac{GMm}{r + h} $$ We rearrange this to get: $$\begin{align} ...


2

Your first potential energy arises from the approximation that the graviational field is approximately constant for "small heights" , i.e. $$\frac{GMm}{r^2} \approx mg$$ The full law leads to your second formula, the approximation to the first. For heights above the earth, it is justified, as we can see by taylor expanding $\frac{1}{r^2}$ around the ...


2

While the total displacement as shown in your figure is zero, this does not mean that the work is zero! Work is force scalar displacement, $W = \vec F\cdot \vec s$. Diving the path $A \to B \to A$ into infinitesimal steps we are led to $$dW = \vec F \cdot d\vec s$$ and for the total work, adding up the contribution $$W = \int \vec F\cdot d\vec s.$$ I think ...


-4

The maths is all impressive but entirely redundant. It is not zero BY DEFINITION.


5

Well, we can do a simple counter-example. Let $$ \vec{F}(\vec{x}) = F_0 \cdot \varrho(\vec x) $$ where $\varrho$ is the function that rotates vectors by 90° counter-clockwise (in matrix form $(\begin{smallmatrix}0 & -1\\1 & 0\end{smallmatrix})$ if you prefer that). Clearly, for the closed path $$ \vec{\gamma}\colon\quad [0, 2\pi]\ \to\ ...


2

Let us take Newton's law for a particle in special relativity $$F^\mu = \frac{d p^\mu}{d \tau} = \frac{d m_0}{d\tau} u^\mu + m_0\frac{du^\mu}{d\tau}$$ where $F^\mu = - \partial^\mu V$, $p^\mu = m_0 u^\mu = m_0 \gamma (c,\vec{v})$ and in our lab frame $V(x,t) = V(x)$. Taking $m=m_0 \gamma$ and a transformation from the proper time of the particle to our lab ...


5

For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is $$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$ which is a line integral. If ...


5

It isn't possible to measure potential energy because it has a (global) gauge symmetry. It's like trying to measure the height of a mountain - this could be the height above sea level, the height relative to the deepest sea trench, the height relative to the centre of the earth and so on. Any measurement can only measure the change in potential energy, and ...


3

Tabulated values of nuclear mass defects do in fact include an arbitrary offset. Usually the standard is that one mole of carbon-12 weighs exactly twelve grams, so that a bare proton or neutron has a positive mass defect, while a tightly-bound nucleus like iron or nickel has a negative mass defect. In computing the $Q$-values of decays it is only the ...


3

If you use relativity (which the use of $E = mc^2$ implies), we cannot choose the potential arbitrarily, because the relation between energy and mass makes absolute values of the energy measurable through the gravitational forces exerted by stored energy. EDIT: Since you asked, I will explain it in somewhat more detail: Let $V : \mathbb{R}^4 \rightarrow ...


1

I) Yes, it appears that the sentence [...] the $y$-axis vertically downwards [...] in Ref. 1 p. 81 should have been [...] the $y$-axis vertically upwards [...] II) Let us also mention that Ref. 1 p. 29 eq. (17.9) introduces a function $U$ to be minus the potential energy, however, this $U$ seems unrelated to above. References: C. Lanczos, The ...


0

I think that in a scattering process it is always understood that we are actually talking about asymptotic states for which mutual interactions are totally negligible (which is the reason why we treat asymptpotic states as free particles). Perhaps, a situation closer to what you have in mind is represented by pair production via a long lived bound states or ...


2

It depends what you mean by "electrostatic energy". When we are talking of pair production we are talking of physics at the quantum mechanics framework. FEYNMAN DIAGRAMS for pair production by a gamma ray (left) or an electron (right). These represent the processes in the preceding sketch. Lets take the simplest diagram on the left: a photon ...


3

The pair production is only possible due to relativistic quantum physics and one needs to describe all these processes by the so-called "quantum field theory" or its generalization (well, string theory is the only example). In quantum field theory, electromagnetic fields are, just like all other fields, quantized. All of the configurations of the ...



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