# Tag Info

1

They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

0

I understand what it takes to hover with wings but how would an animal or a person (like Superman) hover without wings? I know that the definition of flying is falling and missing the Earth, but how do you hover over the ground when you don't have wings and you have gravity working against you. As you can probably tell I just watched Man of Steel recently ...

5

To augment what tpg2114 wrote, work is usually defined as: $$W = fd$$ Where $W$ is the total work done, $f$ is the force, and $d$ is the distance (actually it's the $\Delta d$ caused by the force). As long as $d = 0$ no work is being done. For example, if you're standing on the surface of the Earth, energy is not being consumed to keep you from falling ...

2

Since $f = -\frac{dE}{dx}$ you don't need "energy" per se, you just need a gradient of energy between the two states. On the other hand, force is independent of work because it's possible to have forces that do no work.

2

I would think that a hardened hole saw approach would be most efficient. Bore a short distance into the rock with a self-augering hole saw the diameter of the desired tunnel and then impacting the center of the core to break out the center, remove the broken material and then take the next bore. Where is the sense of pulverizing the center material to dust ...

1

As I cannot post any comments I have to post this as an answer although the essential points were already given: In classical mechanics energy itself was no meaning. Only energy differences have a physical interpretation. Thus in the classical case energy is only defined up to an arbitrary constant. So any fixed state's energy can be set to zero (but of ...

6

As weird as it sounds, the answer is "yes." Take, for instance, a satellite in gravitational orbit around some heavy body. It's energy is given by $$H=\frac{p^2}{2m}-\frac{GMm}{r}$$ Clearly, there are solutions to this equation which have $0$ energy (look at a slowly moving particle that's really far away), but those solutions necessarily involve a ...

0

No. According to the energy-momentum relation, the magnitudes of the energy and momentum of a particle of rest mass $m$ are related through the equation $E^2=p^2+m^2$. Obviously $0\leq p^2\leq p^2+m^2=E^2$, so if the energy is zero we have $0\leq p^2\leq0$, or $p=0$.

1

It's up to you whether or not to include elastic potential energy. You'll get the same answer as long as you're careful about what you define to be inside and outside of your system. The basic idea to use here is the work-energy theorem when only mechanical energy is of concern: $$W_\text{net,external}=\Delta K_\text{total}+\Delta U_\text{mechanical}.$$ ...

0

Yes, a particle can have potential energy in one dimension as potential energy depends upon the configuration of the body. We can have configuration in one dimension as in hook's law and hence potential energy.

2

Sure, a particle can have potential energy in one dimension. Just look at Hooke's Law or the gravitational force. Both of those are conservative forces in one dimension ($x$ and $r$, respectively) that have a corresponding potential energy. In higher dimensions, nothing has to change, but it is possible to have potential energies which depend on the value ...

3

Most gamma rays from $pp$ collisions come from neutral pions ($p+p\to p+p+\pi^0$), you'd first have to do some relativistic momentum & energy conservation to determine the energy of the neutral pion. It's easiest if you consider the two subsequent reactions: $$p+p\to p+\Delta^+ \\ \Delta^+\to p+\pi^{0}$$ (it's up to you to figure out the kinematics). ...

0

As far as I know, your AC is kicking in automatically when you select defogger, so the system is cooling the air just as if you were using your air-conditioner though the vents.

0

I'm not quite sure what your goal is. If the user specifies both $p$ and $q$, you end up with one value for the energy, rather than a plot. For a proper plot you'd have to allow for a range of values $p,q\in\mathbb{Z}$. This will of course result in a 3d graph. You may cut that down to 2d, like in the plot you posted, but you'd have to give some relation ...

1

On your first question: absolutely, energy gravitates (or induces curvature in spacetime) the same way that mass gravitates. If you read general relativity, you will learn that it is in fact the Stress-Energy Tensor that is the source of gravitational interaction (or equivalently spacetime curvature). Energy can be localized very easily; a parallel-plate ...

0

I thought, if it was a possibility, then electron would constantly need to lose energy, whenever excited, at last, it would collapse into the nucleus. You seem to have forgotten that when the electron is excited, it gets energy, which is then released when it emits it. So it wouldn't collapse because energy absorption and emission are balanced.

1

Here in case of electron, it has already emitted absorbed energy as quanta. So, is it that electron losses some energy other than the energy absorbed from the source, to come down to ground state. I thought, if it was a possibility, then electron would constantly need to lose energy, whenever excited, at last, it would collapse into the nucleus. But, this ...

1

Based on some of your comments, I think what might be tripping you up is the first statement you started with: From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels. and ...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that ...

2

You are right that in this theoretical problem the breaking force to achieve constant velocity is the same for all velocities. You either stipulate that the bike has the desired speed as a initial condition, or that it coasts with no breaking applied until the desired velocity is reached. In the real world, there will always be some friction or drag forces ...

1

Draw a force diagram and determine the braking force required to counteract the (portion of) gravitational force pulling the bike down the slope. If the net force is zero, delta-v will be zero. I think the problem you were given assumes the bike starts out with velocity v . If it doesn't then you'll need to derive a braking force whose magnitude varies ...

1

Reading your comment in reply to mine, I understood what you wanted to ask. This is where so many people are confused. Since we start middle school chemistry, we are taught about electron orbits which are like concentric circles. Everyone innately assumes that the 1st orbit is inside the 2nd orbit, which is inside the 3rd and so on. It isn't like that! ...

0

Even if an electron gives up energy (and the quantum rules require certain energy states in a bound system), the nuclear repulsive forces (not electromagnetic) would keep it from getting "near" to the nucleus itself. Take a look at the kind of speeds particles are accelerated to in CERN or LINAC to allow them to collide with a nucleus.

0

No, that isn't right. Think about the gravitational field of a sphere. Equi-potential surfaces form concentric spheres about the original sphere. An object in the gravitational field of a sphere will follow the field lines (or lines of action of force) - which in this case are radially inward. Now if you imagine a field with two spheres, they will be ...

0

Because then why would they tell us the distance between the centers? Your mistake is that you have considered $U_f$ as the gravitational potential energy (up to sign, which you may want to be careful with) of the object on moon's surface due to the moon, and $U_i$ as the gravitational potential energy on Earth's surface due to the Earth. But what ...

2

Look up something called the Carnot efficiency. That is the theoretical limit of how effecient any heat engine can be at converting heat power to some other form. This maximum possible efficiency is    Carnot efficiency = Tdiff / Thot = (Thot - Tcold) / Thot By simple 8th grade algebra, you can see that you get a higher value by decreasing ...

0

As you guessed in the end of your question: We already use the energy released or absorbed in phase transitions to transport energy. For example, when we put ice on a drink, and also in refrigerators and air conditioners we remove thermal energy from a region that is already cold and release it where it is already hot. The problem is that these processes do ...

0

One common way that this happens is through spontaneous parametric down-conversion. From Wiki: an important process in quantum optics, used especially as a source of entangled photon pairs, and of single photons. A nonlinear crystal is used to split photons into pairs of photons that, in accordance with the law of conservation of energy, have combined ...

1

If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation: $$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$

0

On the internet there is plenty of talk of how the continuity equation applies to conservation of charge, fluid dynamics, and so forth, but I can't find any mention of how it applies to the conservation of energy. Why? Is it because it is problematic to talk about energy current density (j)? The continuity equation is fine for energy, and sometimes ...

1

You're almost right. But... 1138 kilowatts power output will give you 1138 kilowatt-hours in, well... one hour, not 1 second. Just leave out the $\times 3600$ It's better to avoid weird units(like kilowatt-hours as much as possible, so another longer way is this: $1138 \text{ kilowatts}$ mean $1,138,000\text{ joules/sec}$ So in one year you'll get ...

0

First of all, your measurement unit is wrong. Energy is measured in, say, kWh, not kW/h. Second, you should not multiply by 3600, as the final result is in kilowatt-hours.

0

The fundamental reason why energy is conserved, is invariance of the physical laws by a time translation $t \to t + t_0$, in a Lagrangian formulation. This is a particular case of the Noether theorem, which states, that if a Lagrangian has a continuous symmetry, there is a corresponding conserved quantity. Now, if we look in detail, conservation of a ...

0

Usually, when physicists talk about energy being conserved, they mean Energy being a Noether charge on the fundamental level, c.f. wikipedia. As a very general result, one can derive that the time derivative of Energy is zero $$\frac{d}{dt} E = 0.$$ This result is only true if the Lagrangian description of your system does not explicitly depend on time ...

7

No. Newtonian gravity is a conservative vector field (or conservative force) which mean that energy that you extract from the field has to be put in first. This is technically stated as the work done around any closed looped is equal to zero. For example, you raise your pet cat up 1 meter (you do work against gravity) you let go and gravity does the same ...

0

If we assume that mirrors will leak some energy, then is it possible to put objects such as a photo multiplier tube (in combination with a mirror) and adjust it in such a way that only the amount of energy lost by reflection of the mirror is recovered and sent back to the other mirror. This cannot be done even as a thought experiment. Photomultipliers ...

0

Though it is not common to use Fresnel lenses for electricity generation but 100 MW power plant is nearing completion in Rajasthan state of India using linear Fresnel lens technology. So to say that this technology is not feasible for large scale use is not correct and time may come if that above mentioned power generation goes smoothly, the scene may change ...

12

"Total energy of the Earth" is somewhat of an odd concept, but there's no reason we can't really entertain it. It brings up some genuinely difficult questions. The right way to approach this is to define the system correctly and then identify forms of energy content and flows. Things to "count" in the Earth's energy: Heat content Nuclear energy ...

6

The heat generated from the Earth's core is about 4x10^13 W while the Sun provides about 1.7x10^17W so although the Earth's core is slowly cooling this has very little effect on the Earth's temperature. The Earth is in equilibrium between the energy received from the sun and the energy it emits into space. If the amount received changes, then temperature of ...

2

I'd say depends on the definition of 'total energy' - see Alan's answer above. The total thermal energy is not in equilibrium, it is increasing: global warming, that is an imbalance of around 0.5 W/m^2, corresponding to a total imbalance of 2.5x10^14 watts (if I did the multiplication correctly...)

1

It's impossible like anna v. said, but let's entertain the possibility for a second. Assume you have a star-trek like device, which is capable of transforming every atom in your body into energy. However, before this process is even started, the device would first have to register and store every bit of information about every atom, their exact configuration ...

4

Consider the time derivative of the Hamiltonian $$\frac{dH(q,p,t)}{dt}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p}+\frac{\partial H}{\partial t}=-\dot{p}\dot{q}+\dot{q}\dot{p}+\frac{\partial H}{\partial t}$$ From this you see that the Hamiltonian is conserved if it does not depend on time,$t$, explicitly. $H$ may or may not be ...

1

No, you can't do that. You can write $\lvert x\rvert = \sqrt{x^2}$, and you can then go to $\langle\lvert x\rvert\rangle = \langle\sqrt{x^2}\rangle$, but it's not valid to say $\langle\sqrt{x^2}\rangle = \sqrt{\langle x^2\rangle}$. You can pick almost any wavefunction, $\psi(x) = \pi^{-1/4}e^{-x^2/2}$ for instance, and show that the two are not equal. In ...

1

Let's say you have a catapult that can be turned into a trebuchet by attaching a sling to the end of the projecting rod. And let's say that it always launches the object with its projecting rod moving at some particular angular velocity instantaneously before it is stopped. Adding the sling has some benefits in terms of launching something at an enemy. The ...

1

Both operate on the same principle: the velocity of a point on a rigid body undergoing rotation is proportional to its distance from the pivot. The sling is simply an ingenious way to extend the distance of the the projectile from the pivot without extending the rigid arm. As the trebuchet arm moves in an arc, the sling exerts a centripetal force on the ...

-1

Resolving (Decreasing) energy gradients to lower realized potential states is what drives every process on many levels including evolution. www.intothecool.com Here is a super cool paper that shows the process across universal, biological and socio-economic domains. http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=185965 if ...

3

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...

0

A system that is in thermal contact with it's environment will tend towards both a lower energy state and a higher entropy state. Basically, the energy of the system + environment is fixed, but energy will flow between the two until they are in a state of maximum entropy. It might be more informative to ask why systems tend towards increased entropy. What ...

0

Well, chemical reactions almost always require heat (energy) to take a place, and almost always release heat upon reaction, so by that logic state when elements is unable to keep reacting is a state with insufficient energy or, in other words, lowest energy state (or we probably should say "lower energy state" then one that required for reactions)

0

I'll try to explain with the help of an classical example. Take the situations in the picture above. What you're interested in are the first to cases. The unstable state of equilibrium is such a state that when you slightly displace the ball, it departs from the original position. Being at the top of the hill, it has an excess of potential energy (may it ...

1

Roughly: Becouse $F=-\overrightarrow\nabla U$, with $U$ some potential energy (coud be an effective potential energy). Then, if you aren't in a minimum of potetial, your system isn't in equilibrium. Edit: Can you see that 1. and 2. are stables equilibrium?. I chemistry your effective potential energy is some function called Gibbs free energy.

Top 50 recent answers are included