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0

$\nabla T=0 $ is not a conservation law. You are not considering the energy of the gravitational field in this way. If you do the calculation you find something like $\partial_{\nu} (\sqrt{-g} T^{\mu \nu}) \neq 0 $, so you can't define a conserved charge like $P^{\mu}= \int \sqrt{-g} d^4x T^{\mu0} $. To account for the gravitational scalar field you ...


2

Energy conservation stems from Noether's theorem applied to time (i.e., time-invariance leads to energy conservation, similarly to how spatial-invariance leads to momentum conservation). Since the universe is expanding (and accelerating at that), the state of the universe today is different than it was yesterday and will be tomorrow, hence energy ...


1

What we like to call the energy, i.e., the total matter/energy content of space-time, might not be conserved. However, there is a lot of reason to suspect that fundamentally the universe is some big quantum system, and that space-time and particles and fields are emergent from this underlying idea. In that case, we expect there to be a Hamiltonian $H$ and ...


3

The torque on electricity generators is continuously adjusted to keep them running at a constant speed (e.g. 60Hz in the US and 50Hz in the UK). When you turn on some electrical item the current it consumes places a greater load on some electricity generator somewhere and this reduces the speed. To counter this, at the generating station more torque is ...


1

The difference is only in the properties of the material of a body. You can see in this video If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored in the lattices If it is not elastic the body will stay ...


2

Power generation and supply management is not easy and it is to their credit that most of the time power companies supply people with AC at the same voltage no matter what the demand for power is. So when we turn on appliances we do not see the voltage drop as a result. Or more realistically when everyone gets home from work and starts cooking/ boiling ...


0

I do not see any other possibility than doing the integral along the path. Do your telemetry data include the horizontal position? because if does you can calculate an approximation to the integral. Of course, there is no guarantee that the error will be larger than the effect you want to measure. You should have to do some pre-tests (using situations that ...


2

Step 1 The mass is dropped ... The rotational energy of the torus does not change. Going to stop you right there. Although total energy is indeed conserved, rotational energy does not need to be conserved. Remember that the potential energy of dropped objects is generally converted to heat when they go thud on the ground. Angular momentum is however ...


0

No, the elasticity of the collision depends on a couple of factors. 1) the properties of the materials. The carpet is easily deformable, and the fibers are not elastic, if one removes the ball that fell they return to their original form after some time, or, even don't return. So, the kinetic energy of the falling ball does the work of deforming the fibers. ...


5

Actually, if the mass comes to rest relative to the torus after landing, the energy of the system goes down. Let $I$ be the moment of inertia of the torus, $r$ be the radius of the mass from the axis of rotation before it is dropped, $R$ be its radius after it lands, $\omega_1$ be the angular velocity before the mass is dropped, and $\omega_2$ be the ...


1

Moving the mass inside the rotating object takes energy - when it is "lowered" against the artificial gravity it does work against the force resisting this motion so the final kinetic energy of the system with "lowered" mass is not the same as before. The equation: $$E=\frac12I\omega^2=\frac{L^2}{2I}$$ is the clearest way to see why this is so - when ...


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In step 1 you lower the mass and this generates energy. Let's say you store this energy is a spring, and for the sake of argument let's say the energy stored is 1J. The energy has to come from somewhere, and of course it comes from the rotational energy of the torus so the rotational energy of the torus is now 99J. In step 2 you slow the torus, perhaps by ...


3

Step 1: the rotational energy $E=1/2 \omega L$ does increase because $\omega$ increases (bacuse I decreases) and L is a constant ( $L=\omega I$). This is wrong again in step 3: The rotational energy of the torus does not change So the potential energies of the lifted objects is not the same in both situations, because they will also convert rotational ...


1

What is the difference that leads to conservation of kinetic energy in elastic collision ? The difference is only in the properties of the material of a body. If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored ...


2

The simple answer is that in an elastic collision (for objects >> in mass than typical molecules) energy moves from kinetic to potential then back to kinetic as long as the "elastic limits" of the materials are not exceeded. In other words, as long as they act like springs. In non-elastic collision the energy goes mostly from kinetic of the colliding masses ...


3

When one says that "kinetic energy is conserved in an elastic collision" that means that the total kinetic energy of the system of particles involved in the collision doesn't change. It does not mean that the kinetic energy of each particle is unchanged. For a two particle system, the kinetic energy of each will change, but the sum won't. Also, your ...


1

This is literally what I get when I Google "non-conservative force". From the very first link (to Wikipedia), here's a more detailed explanation: A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Equivalently, if a particle travels in a closed loop, the ...


0

The energy is "stored" in the configuration of the electromagnetic field. When the positions of the magnets are changed the configuration of the electromagnetic field changes accordingly. If the magnets are forced into a new position, then energy is transferred to the EM field. If the magnets spontaneously change position and acquire some kinetic energy, ...


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The question is like asking "when a body falls on the ground from a height, work is done and energy is used. but where does this energy come from?" The answer is, that force-fields (magnetic, electric, gravitational) provide a means for suitable materials to jump from a higher energy state to a lower energy state. When a magnet is present in the vicinity ...


0

When two magnetic object attract or repel it is because at that certain distance the electron movements oppose one another. When the magnets move toward eachother their potential energy from their attractuion is lost, resulting in equilibrium. The work that moves the magnets is equal to the potential energy of the magnets while seperated before they move.


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Exerting force does not necessarily lead to energy changes.


5

From the geometry, you can state that in order to move the screw by 1 unit of distance, you have to move the end of the handle by $10\times 20\times 2\pi$ units of distance. Let's call the unit of distance $[L]$ - in this case, an inch might be a good unit but we don't have to be explicit about that. Conservation of energy says that work done on the system ...


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Work is calculated as force times distance. $$W = Fd$$ The purpose of a simple machine like a screw jack is to lessen the force required. However, the work needed is still the same, so the distance over which you exert the force has to increase. Halving the force requires doubling the distance. In this problem, you want to lift 2000 lbs a distance of 1 ...


-1

As someone who has spent some time playing with gyros, I am very familiar (though on a non-mathematical level) with these objects. I have six battery powered high precision Gyros found here: http://www.gyroscope.com/d.asp?product=SUPER2 I have been intensely fascinated with the way they behave when spinning at maximum speed. I feel hesitant to comment ...


0

The are no means to hold kinetic energy in MASSLESS spring. So, the potential energy of extension, will instantly converted to infinite speed of contracting spring. If spring is absolutely rigid itself, it will contract to equivalent minimal length and will start to extending back. Once it extends to initial length, it will start to contract back, and so ...


0

The magnetic field is created by the current. Therefore, the energy of the magnetic field is supplied by the current. A magnetic field does no work on charged particles. This is because, it exerts a force $\vec{F} = q(\vec{v}\times\vec{B})$ on it, $q$ being the charge, $\vec{v}$ being the velocity of the charge and $\vec{B}$ being the magnetic field. ...


1

Frankly, no one really knows! In the day of Ampere, Faraday, and Maxwell they thought that the so-called ether passed the energy from one "ether particle" to its neighbor-much like a fire brigade passing a bucket of water along a line from hand to hand. Nowadays, there is no real certitude about precisely how the energy propagates through space. We can ...


0

If you observe the situation from ground the normal reaction due to incline would be doing no work, as it is perpendicular to the velocity all the time but from frame which is moving with the velocity work would be non zero, and in fact it is negative all the time (i.e extracting energy from the block system)and that's why at the bottom of the incline block ...


4

Rather than beating your student over the head with facts, try to approach the problem the way scientists did in the first place, by following the scientific method. Your student should come up with a hypothesis, and use known theory to make a prediction (calculate the momentum transfer in some idealized model), and then build a model to test the prediction. ...


0

Here is a version of Poynting's theorem, appropriate for media with $\epsilon_r=1$, $\mu_r=1$ (for simplicity). $$\nabla \cdot {\bf S} + \frac{1}{2}\frac{\partial}{\partial t} \left( \epsilon_0 E^2 + \frac{B^2}{\mu_0}\right) + {\bf E} \cdot {\bf J} = 0,$$ where ${\bf S}$ is the Poynting vector (${\bf E} \times {\bf B}/\mu_0$ or ${\bf E} \times {\bf H}$). ...


2

However, isn't any closed loop on a PV diagram reversible? The arrows can simply be drawn in the reverse way to create a refrigerator. If any closed loop is reversible then why does the specific Carnot engine (a specific loop) have the highest efficiency? This was exactly the question I asked myself ten years ago :-) The problem is that often students ...


0

In order for the condition as stated in the footnote (which you don't understand), you need to change the figure slightly to reflect the condition as follows: In the figure, the curve V and the line E meet at point A, and at infinity.


0

The comment in the book is wrong,, there is no reason why V should be a maximum. Just think of the case of a ball going upwards. V is a straight line and zero speed is reached at the turning point in finite time.


1

The other answers are great. I decided to plot it, however, because it's nice visualizing these things. Since your biggest doubt is about kinetic energy, be sure to pay attention to the last graph. SYSTEM. Motorcycle going to the left, truck going to the right, bound by an elastic rope ten meters long ($k=100\frac{\mathrm N}{\mathrm m}$). Masses and speeds ...


1

I suspect you are meant to treat the collisions between spheres 1 and 2 and between spheres 2 and 3 as separate collisions. Solve 1 and 2 first. There will always be a solution where 1 continues with unchanged speed - reject this solution, it would mean that 1 passed through 2 without colliding at all. Then, take your solution for the velocity of 2 and solve ...


1

In the standard homogeneous cosmological models the total energy in an expanding volume is zero. This is true for positive, negative or zero curvature and it must take into account the gravitational energy (which is negative), dark energy, matter and heat. Since the gravitational energy is negative the heat can be positive and increasing as you go back ...


1

In Zero-Energy Model, negative energy associated with Gravity counterbalances positive energy associated with matter, photons, etc. So, No, Big Bang wasn't cold. You are just looking at partial picture (you just ignored Gravity). This is what Zero-Energy Model says: With traditional Big Bang model (which doesn't contain Inflation), the universe started out ...


2

Yes. The photons play a vital role in balancing out the momentum and kinetic energy.



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