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90

What you cannot see by drawing the picture is the velocity of the individual points of the string. Even if the string is flat at the moment of "cancellation", the string is still moving in that instant. It doesn't stop moving just because it looked flat for one instant. Your "extra" or "hidden" energy here is plain old kinetic energy. Mathematically, the ...


48

ACuriousMind's excellent description was missing a picture. Here it is: This clearly shows that for the wave moving to the right, the front is moving up and the rear is moving down. For the opposite wave traveling to the left, the front (now on the left) is moving down and the rear is moving up. Summing them, you get a straight line with significant ...


25

Just to complement the other excellent answers, here's an animation showing what two wave pulses with opposite amplitude passing through each other actually look like: You can clearly see that, at the instant when the string is momentarily flat, it's not stationary but rather moving quite rapidly, and thus will not stay flat for long. (Obviously, the ...


6

If there is friction (air resistance), that friction will extract energy from the spinning fan, thus slowing it to a stop. If you extract energy in any other way, you are also applying friction to the fan, again slowing it to a stop. In other words, yes you can extract energy from the fan, but no more than the rotational energy that the fan possesses: ...


4

Your gut feel is correct. Both are exactly the same. Look at the acceleration in both scenarios. 35 mph to 0 in the time it takes for the cars to fold up and stop. Everybody gets this wrong. Good question.


4

Let the electromagnetic field has $u$ as its energy density (amount of energy per unit volume in the field) and let $\bf S$ represents the energy flux- the amount of energy per unit time flowing across a unit area perpendicular to the flow). Now electromagnetic field can interact with matter and do work on them; so this energy interaction must be considered ...


3

Poynting's theorem is the work-energy theorem in electrodynamics. The equation tells us that the total power (or energy) carried by an electromagnetic wave is equal to the decrease in the energy stores in the field (first term) minus the energy radiated out from the filed (second term). The radiated energy will never come back. It's gone. The radiated energy ...


3

Minkowski spacetime has the symmetries of the Poincaré group, which include the four spacetime translations. Noether's theorem then says that there are four conserved quantities, $p_0, p_1, p_2, p_3$, associated with these four symmetries. Typically $p_0$ is denoted by $E$. The structure of the Poincare group implies that these four quantities are related ...


3

Well spotted. The average intensity of the whole fringe system must be $1+4=5\;\mu$W. However there will be places where the intensity is a maximum, $1+4+2\sqrt{1\times4} = 9\;\mu$W and places where the intensity is a minimum, $1+4-2\sqrt{1\times4} = 1\;\mu$W. Your chosen position is one which is nearer a maximum than a minimum. Later Here is a ...


3

This probably more a philosophical than a physical discussion. Let's take a simple everyday example: The air molecules in the room where you are sitting are fairly evenly distributed through the room. Because the molecules are subject to random motion, it's perfectly POSSIBLE to have all molecules bunch up in one half of the room and that there is a perfect ...


3

This question appears to be a pseudo-duplicate on the Skeptics exchange, as pointed out by @CraigGidney. The highlights of the comments here and answer there appear to be that: 1) Yes, one could potentially accrue some electricity from soil. 2) No, it would not (ever) be sufficient to charge an iPhone, let alone 3 times. 3) In the comments here, "there ...


3

You've discovered the virial theorem. The virial theorem tells us that for a bound system where the potential energy $V$ is given by an equation: $$ V(r) \propto r^{-n} $$ The average kinetic energy $T$ and average potential energy $U$ are related by: $$ 2T = -nU $$ For the electrostatic force $V(r) \propto r^{-1}$ so $n = 1$ and: $$ 2T = -U \tag{1} $$ ...


2

The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole. If you want to see gravitational waves, you need to allow the metric to evolve dynamically, ...


2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...


2

Since $sin(\theta) = \frac{h_{i}}{D+d} = \frac{h_{f}}{D}$ and $U_{g} = mgh$, it follows that there should be only one factor of $sin(\theta)$ in equations 4 and 5 corresponding to the initial and final potential energy of the object. If you make those corrections, you should be arriving at correct value of velocity for part (a) of the problem.


2

Your understanding is actually perfectly correct. In the general case we have indeed something like $E_m = K(r) + V(r)$ for a central force say. Now, as you rightly point out, the valid positions are those that satisfy $K(r) \geq 0$. This implies that $E_m - V(r) \geq 0$ If $V(r)$ is an ever increasing function of $r$ that has an asymptote at $V_0$, then ...


2

For example in simple circuit with lamp. I understand that the energy is spent for heating and lighting. But how exactly it happens? The MIT Teal project has an excellent short video which can be used as an animated diagram of the (Classical physics) process inside a resistor such as a tungsten filament. In the video below, imagine that the vertical ...


1

For two particles to influence each other you need some sort of interaction. For (macroscopic) mass this is clearly Coulomb-interaction. Two atoms can not be at the same place, because their cores repell each other. If you look at smaller scales, strong and weak interaction might add their part. Photons have no charge, no color-charge and don't interact ...


1

$W=\Delta K_\text{system}$ and $W_\text{external}=\Delta K_\text{system}+\Delta V$ are consistent with each other iff $\Delta V=-W_\text{internal}$. The latter is the definition of the potential energy for conservative forces. The cited equation is thus valid iff all internal forces are conservative.


1

So is this genuinely the creation of energy via the casimir effect, if so it seems extraordinary and a violation of the principle of conservation of energy. You gave no link, but I discovered the following by googling: The experiment is based on one of the most counterintuitive, yet, one of the most important principles in quantum mechanics: that ...


1

I always wondered the same thing, my guesses are all kinetic energy in tires turns into KE in engine and then lost through heat, as you may notice engine reving up when you down shift , the engine isn't getting any energy from fuel it must from the tires, so its the opposite the tires move the engine.


1

I'll try to show here that energy conservation (the energy continuity equation), as well as the interpretation of the Poynting vector as the energy flux, are consequences of Maxwell's equations. Poynting's theorem: $$\int_V\left(\vec{E}\cdot\vec{J}\right)\,\mathrm dV = -\dfrac{\partial}{\partial t}\int_V\dfrac{1}{2}\left(\epsilon_0 E^2 + ...


1

Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is: $$dE=dK+dU=0,$$ that is $$mvdv=-\frac{K}{x^2}dx,$$ where $K\equiv Gm_1m_2$. Integrating from $(x_i,v_i)$ to $(x,v)$ we get $$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$ This is the correct ...



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