# Tag Info

6

Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...

3

The answer you cite is somewhat simplified to suit the level of the question. The vacuum does not consist of pairs of particles and anti-particles popping into existence and then disappearing again. When calculating the properties of the vacuum it's true that we use Feynman diagrams showing the creation of particle/antiparticle pairs, but these are virtual ...

3

Some energy would come from the part where "the water would be collected and sent back into the container", which you somewhat glossed over.

3

The Schrödinger equation cannot be derived from classical physics. There are various consistency checks and motivations, such as its consistency with conservation of energy, but it is not derived from those considerations. However, that the Schrödinger equation conserves energy is built in when one knows that the Hamiltonian is the energy operator since $$... 2 There are two possibilities: water (or whatever fluid you're using) is compressible or it's incompressible. If it is incompressible, when the red valve is closed the tank pressure drops to zero, and opening the yellow valve produces no effect on the generator. If it is compressible, its pressure remains constant when the red valve is closed, and it ... 2 In a static spacetime, there is (by definition) a timelike Killing vector field \xi^\mu, which implies that geodesics with four-velocity u^\nu have a conserved quantity \epsilon = -g_{\mu\nu}\xi^\mu u^\nu. For example, in Schwarzschild spacetime, this is$$\epsilon = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\text{,}$$where ... 2 Your first part was correct; but for the second part, you have to equate the energy of A plus the stored energy in the spring to the energy of B (because you start with no energy in the spring, and all the energy as kinetic energy in B). So the expression for the stored energy is E_\mathrm{spring}=\frac12 k x^2 = \frac12 m_b v_b^2 - \frac12 m_a v_a^2. ... 2 Considering the way how v^2=v_0 ^2-2g\Delta y is derived, It is derived through energy conservation, that is$$\frac{1}{2}mv^2-\frac{GMm}{r+\Delta y}=\frac{1}{2}mv_0^2-\frac{GMm}{r} \tag{1}v^2=v_0^2-2\frac{GM}{r}\bigg(1-\frac{1}{1+\frac{\Delta y}{r}}\bigg)$$Using Taylors series$$v^2=v_0^2-2GM\bigg(\frac{\Delta y}{r^2}-\frac{\Delta ...

2

When studying physics, it is often necessary to make simplifying assumptions in order to keep the math manageable. In the case of an elastic collision between two balls, the textbook example ends with the kinetic energy of the two balls being conserved through the collision. As you've noted, this ignores the energy lost to sound - it also ignores the ...

2

Quantum mechanics is not derived from classical mechanics or energy conservation, but there are "jumping off points" in classical mechanics that may serve to answer your question. If you study classical mechanics at a sufficiently advanced level you will discover the Hamiltonian formalism. The Hamiltonian for an isolated system with only conservative ...

2

All friction can be ignored. In that case Conservation of Energy applies: $$U+K=mgh$$ So that when the mass reaches the bottom of the ramp: $$K=\frac{mv^2}{2}=mgh$$ My question is: The mechanical energy of which system is conserved in the progress? The one formed by m and M, or the one formed by m, M and the Earth? The system here is mass, ...

2

In a collision momentum is conserved if there are no external forces. enefgy is also cionserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones when kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in ...

2

Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe. Conservation means that these quantities cannot spontaneously change. Let's ...

1

I came up with a solution seeing John Rennie's comment. The centripetal force, $\vec F= -F \hat r$ so infinitesimal work done by centripetal force, $$dW=\vec F.d \vec r= -F \hat r.d\vec r$$ but, $\hat r⊥d \vec r$ so $$dW=0$$ is this correct ?

1

There is no conflict here. Let the two charged particles ($M,Q$) be the system with no external forces acting. Momentum is conserved and so for all time $M_BV = M_A V_{Af} + M_BV_{Bf}$ Energy is also conserved and so for all time $\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$ The ...

1

It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...

1

Just think about how you might push a child's swing. You apply a push once every oscillation of the swing and thus build up the amplitude of the swing. This is a resonance condition whereas if you pushes the swing at a slightly lower frequency you would not be able to increase the amplitude of the swing as much. Once the swing is at a constant amplitude ...

1

Lift is approximately proportional to velocity squared of the aircraft, not the thrust. That is why runways are required. So the thrust is used the accelerate the aircraft to take-off velocity, which will produce enough lift to overcome gravity. Also, the fact that the thrust is less than the gravity in Antonov implies that it can't do this: ...

1

This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle. $$\frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx$$ For a constant force and mass $\int F\cdot dx = F (x_A - ... 1 How much sound colliding objects make depends entirely on the objects and the medium they are in. In the vacuum of space collisions don't make a sound, at all. In Earth's atmosphere the total energy of sound released by collisions that are caused by solid objects is very small compared to the energy of the objects. This is because of the large difference in ... 1 Let's make things a little more fun. How can we use the conservation of energy equation to derive Schrödinger equation in QM? Let's say we know the system's Hilbert space$\mathcal H$and we know how to define a Hamiltonian$H:\mathcal H \rightarrow \mathcal H$whose average value$\langle \psi | H | \psi\rangle\$ provides the average energy in state ...

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