New answers tagged

1

You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...


0

Well, if you think about electric power, which includes current (notion of flux), then you'll end with the conclusion that if there's no flow, there is no power.


4

Are all heaters (same wattage, electric to thermal, no geothermal or other extra energy source) exactly as efficient as each other? No. Let's focus just on electrically powered heaters. If you have a heater that basically consists of a resistor with a current passing through it, you have 100% efficiency of electrical energy to heat energy conversion. ...


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 ...


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, ...


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 ...


0

Firstly, given a differentiable Lagrangian $L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }. $$ Secondly, make the assumption that $$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} $$ ...


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 ...


0

The initial energy to effectively fuel this expansion came from the Big Bang. Energy is not then continuously required to fuel further expansion. The temperature of the universe may decrease due to expansion, but the rate of expansion is accelerating due to currently unobserved dark energy, according to common theory. If the repulsive forces from the dark ...


0

The Heisenberg uncertainty principle is a basic foundation stone of quantum mechanics, and is derivable from the commutator relations of the quantum mechanical operators describing the pair of variables participating in the HUP. You are discussing the energy time uncertainty, . For an individual particle, it describes a locus in the time versus energy ...


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

A photon that escapes from black hole's neighborhood does work on the black hole. The photon causes the black hole not to be in the photon's gravity well after the photon has escaped. In other words the photon increases the potential energy of the black hole. The following paragraph may not be science, I just want to say something sane, as opposed to ...


0

This is just ordinary potential energy from first semester physics -- when the photon is close to the black hole, it's deep in the potential well. As it goes away from the black hole, it picks up gravitational potential energy, so therefore, it must lose kinetic energy. For a photon, the kinetic energy is given by the Planck formula $E = hf$, so the photon ...


0

Since due gravitational time dilation something takes forever to fall inside a black hole from the perspective of an outside observer, there is nothing in it yet that could come out. The light which is emitted from outside the horizon does reach an outside observer, but since it hasn't yet fallen in, it technically doesn't escape from inside the black ...


0

Simple-Every physical or chemical process always aims at decreasing the energy and increasing the entropy(randomness of things).Lets take some examples When an object falls on the ground then it would break and result in a large number of pieces completely scattered(randomness) thereby increasing entropy.The pieces would never come together to make the ...


0

If we assumed an "ideal fan", that had no forces slowing it at all (itself impossible), then you're right that if left alone, it would keep spinning forever. However, if you extracted energy from the fan - by whatever method - then that would cause it to slow down and stop, as the energy that you were "producing" would be the kinetic energy of the fan.


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 ...


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: ...


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: $$ 2T = -U \tag{1} $$ ...


0

This is true for central potential problems. Because in that case, both potential energy and kinetic energy are in (1/r) terms. Also, Potential energy is negative and its magnitude is twice that of Kinetic energy . Thus we conclude that only the mathematics of this problem allows it to be expressed as the formula u proposed and it is general for any central ...


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. ...


0

Work done is $\vec F \cdot \Delta \vec x$. If $\vec F$ and $\Delta \vec x$ are in the same direction then the work done is positive. If $\vec F$ and $\Delta \vec x$ are in opposite directions then the work done is negative. Consider a spring fixed at one end as a system and an external force $\vec F$ stretching the spring. If the external force $\vec ...


0

Potential energy of a body is its capacity to do work by virtue of its position in a conservative force field.So if the body is free to move it will do so in such a way as to reduce its potential energy and the reduction in the potential energy will be equal to the positive work done by the body which could result in the raising of a weight or displacement ...


0

It is just a convention. If we would use the opposite convention, we would get for conservative forces $$\vec F=+\vec\nabla U.$$ You can easily see (think in the one dimensional case) that the particle would move to points of maximal potential energy. This only sounds strange because we are used to the opposite. The mechanical energy principle would retain ...


0

The decrease or increase in potential energy is converted as mechanical work either to increase or decrease it's kinetic energy (we are considering here conservative systems). Suppose you have a system at rest at some height above the ground level, say, a ball placed on the top of a mountain. The work done in order to bring the ball of mass m to a height h ...


0

It's a convention. The real reason is so that we can have: $$\begin{align} \Delta {\rm KE} &= W_{\rm ext}\\ \Delta {\rm KE} &= W_{\rm con} + W_{\rm noncon}\\ \Delta{\rm KE} &= -\Delta {\rm PE} + W_{\rm noncon}\\ \Delta{\rm KE} + \Delta {\rm PE} &= W_{\rm noncon}\\ \Delta {\rm E} &= W_{\rm noncon} \end{align}$$ Which doesn't work ...


0

It would be better to say that potential energy is the amount of work that a system can do. Say you have a system consisting of two masses - a brick and the Earth. As the brick moves down U decreases, the force pulling the brick and the Earth together acts downwards on the brick and it can do some positive work. On the other hand, if brick moves up, U ...


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 ...


-1

This cant be fully derived but a part of it can be. In an elastic collision kinetic energy is conserved, so 1/2*m1*u1^2 + 1/2*m2*u2^2 = 1/2*m1*v1^2 + 1/2*m2*v2^2 m1*u1^2 + *m2*u2^2 = *m1*v1^2 + *m2*v2^2 m1*u1^2 - *m2*v1^2 = *m2*v2^2 - *m2*u2^2 m1(u1+v1)(u1-v1)=m2(v2+u2)(v2-u2) ---------- equation 1 NOW, Accordinbg to conservation of linear ...


0

The answer is to do with what happens when you stop moving the object upwards. You say that the upwards velocity is 'low enough that it does not leave your hand': the only value of velocity for which that is true is $0$. What this means is that, if at time $t_0$ the object is being moved upwards with velocity $v_0$ and is at $h_0$, and you suddenly stop ...


0

If you lift a book, then you are doing work on it. You are actually spending work on raising it's kinetic energy and on lifting it up. $$U=K+W$$ When you reach some height and stop, then all the work done is what determines your height. If you move your hand faster in order to speed up the book to a larger kinetic energy, then you've spent more work on ...


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 + ...


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 ...


0

I guess they can be seen as $$m_{_w}*g-T=m_{_w}*a;$$ $$T=m_{_c}*a$$ $$a=({m_{_w}*g})/(m_{_w}+m_{_c})$$ $$V={\sqrt (2*d*a)}$$ As @CuriousOne has mentioned it depends on the rolling resistance and it seems like 8 mts is no big deal for this.


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 ...


0

The idea of a massless spring is sometimes useful as is a spring whose mass is much, much less than the other masses within the system under consideration. In this case a massless spring implies that there will be an infinite acceleration when the stretching force is released so better to consider a spring with some mass. First assume that there are no ...


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

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

$W=\Delta K_\text{system}$ 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.


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 ...


0

The definition of potential requires you to work out either the work done by an external force on the charge or the work done by the field but not both together. If you consider both forces together you can use the work-energy theorem and say that as the net force on the charge is zero then the net work done on the charge is zero and hence the change in ...


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 ...


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 ...



Top 50 recent answers are included