# Tag Info

## Hot answers tagged energy-conservation

3

Here is the elementary particle table from which all others are built up , the standard model of particle physics. Which shows the conserved quantum numbers that characterize the particles (columns and rows have quantum numbers assigned too) plus the measured masses. The quantum numbers have to "annihilate" to have an annihilation event, i.e. they should ...

2

This chapter is on the conservation of energy. Ideally the machine he shows will work. If we jump ahead in the book to where we know about conservation of energy, we see that energy gained by one side is lost by the other. But in practice, some energy is lost to friction. The little extra weight is needed to overcome friction. You could also overcome ...

2

I suspect this is an example of the spinning-egg problem, in which a prolate spheroid (such as an egg) spun on a table about one of its "short" axes will tend to "stand up" so that it's spinning about its long axis. A few explanations have been proposed for this phenomenon, most notably: H. K. Moffatt & Y. Shimomura, "Spinning eggs — a paradox ...

2

It is better to think of the equation $E =mc^2$ as a true equality rather than a conversion. Mass is energy. If one has that mindset, then it is intuitive that energy has a gravitational field. A hot cup of tea weighs more than a cold one.

2

There are three types of matter/energy we consider when calculating how the universe expands: Matter - both normal matter and dark matter Radiation Dark energy We measure the expansion of the universe using a scale factor that we normally denote by $a$. The scale factor increases with time as the universe expands, and if we look backwards in time we see ...

2

Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current. If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the ...

2

You're confusing yourself. A black hole state has nothing to do with the lack of energy in a system. A star doesn't become a black hole because it can't fuse stuff anymore (i.e. the end of its energy), it becomes one because of what happens to the star after its fusion process ends**. Black holes form when a high amount of energy is concentrated at one ...

2

Half of the energy is lost to the battery's internal resistance (or other resistances in the circuit).if you try to consider an ideal battery with 0 internal resistance, the notion of charging the capacitor breaks down.since the capacitor and the battery are connected by a (0 resistance) wire, their voltages are the same the instant they are connected, no ...

2

Total energy is conserved. Let $E$ stand for total energy. Let $T$ stand for kinetic energy. Let $U% stand for potential energy.$E = T + U$If$T$increases and$E$is constant (which it is in a closed system due to conservation of energy), then$U$must decrease. Simply put, the kinetic comes from the potential energy. Potential energy comes from ... 2 The energy is not entirely from from the Big Bang since a lot of the material had to come from supernovas. This would mean that some of the energy came from the supernovas separating the dust that makes up the objects. Yes, you can harvest energy from the falling objects, if they fall onto the targets. 2 A significant source of power loss in a transformer is the induced eddy currents in the core. Just as the varying magnetic field induces current in the secondary coil, it can also induce currents in the core itself. These currents do nothing but dissipate energy, and so are to be avoided. To reduce eddy currents, you either build your transformer out of a ... 1 An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision. 1 When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is: $$W=\int F \,dx$$ where$F$is a force applied. It must be equal to the total kinetic energy of you and the block (if ... 1 here is what I think he is trying to explain. You have two machines A and B. A is reversible B is not necessarily reversible. Both these machines are placed side by side. let both machines A and B be initially horizontal. (Left , right pans of machine A will hold 3 unit and 1 unit mass respectively. Similarly, the left and right pans of machine B will hold ... 1 Energy, in the context of the universe and the models used in describing it , is in the stress energy tensor, which contains the energy/mass transformations. The accepted at the moment model for the universe is the Big Bang model, as summarized in this plot , is a good fit to the available observations using all known physics to date. Known physics is ... 1 A magnetic field is determined by the current and a changing electric field. And it has energy just for existing. It takes energy to make the magnetic field, for instance to increase the current, and you get energy back when magnetic fields decrease in strength. For a common inductor the magnetic field and associated stored energy are due solely to the ... 1 I suspect most of the loss is simply resistive heating in the coils and possibly some heating due to hysteresis in the iron core rather than coupling of the magnetic field to any external power leakage 1 The kinetic energy you calculated is in joules, not kilojoules. So x is 1.0m. You can calculate the peak acceleration using the elastic force: F=kx=ma; a=kx/m 1 This is a very, very good question, and the answer is not nearly as obvious as one might think. In fact, this system was the subject of a Nature article published in 2000, and of several other journal articles since then. To summarise, relating to the original question: Dissipation of the energy comes from rolling friction and from slippage between the ... 1 So, if we knew for sure that this was a harmonic potential ($V \propto x^2$) we could solve the actual equation itself. But I think we can establish the "oscillation" part without recourse to the differential equations. Let's look at this purely in terms of conservation of energy. First, conservation of energy tells us that the particle will not slow down, ... 1 Since there is no reference config for Potential Energy specifically mentioned, you can consider that the minimum potential (the lowest point on the graph) is zero, and hence the KE of the particle would be maximum at that point. For one half of that potential energy curve, this would be analogous to a very simple case of throwing a ball upwards. The ball ... 1 The block's$x$-velocity right before it hits the floor is not$v_b$, but$v_b \cos \alpha + v_I$. The$\cos \alpha$is because it's not moving horizontally relative to the wedge, but at an angle$\alpha$; the$+ v_I$term is because it's moving relative to the wedge, and momentum is only conserved in a frame that is at rest with respect to the table. ... 1 Someone already posted the exact mistake you made, so I won't repeat it here. I do, however, have some general advice for tackling this type of problem. As I think you realize, both of these answers should give you the same result if applied correctly. Since they both should be right, either the two of the answers you found are actually the same answer ... 1 The spinning coin is not a collision. Elastic Collisions involve the linear motion of two or more objects and the transfer or transformation of their kinetic energy. The situation described involves rotational motion, the contact forces between the table and coin, and angular velocity. I don't see a connection between the condition imposed of materials that ... 1 The whole energy-concept is a reformulation of Newtons laws. Starting from$\vec{F}=m.\vec{a}$, you could wonder about the effect of a force during a displacement. You call the concept 'work' and do the math $$W=\int_{\vec{x}_1}^{\vec{x}_2} \vec{F} d\vec{x} = \ldots = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$ To save yourself some work you define $$T = ... 1 For the sake of simplicity assume the binaries' motion to be circular. Then you can use the formalism of the circular restricted 3-body problem (CR3BP) to model the motion of the test particle m_3. Your Lagrangian will be time-independent and the conserved quantity (Jacobi constant) can be evaluated at infinity to give an equation for conditions of escape, ... 1 (After possibly introducing more variables) then OP is essentially considering an autonomous system of n coupled 1st order ODEs$$\tag{1} \frac{d\vec{z}(t)}{dt}~=\vec{f}(\vec{z}(t)), \qquad f: U \to \mathbb{R}^n , \qquad U\subseteq \mathbb{R}^n, $$i.e. without explicit time dependence, so that the system (1) possesses time translation symmetry. OP is ... 1 I think I can remember the derivation for a conservative force field in classical mechanics, wich is a somewhat stronger assumption than pure time-translation invariance. Let \vec{F} be a conservative force field, that is$$ \nabla \times \vec{F} = 0 $$or alternatively$$ \phi := -\int_\gamma \vec{F} \cdot d\vec{a}$$does not depend on the path$\gamma$... 1 The formula for kinetic energy is$\frac12mv^2$. If your initial velocity is$v_i$and your final velocity is$v_f$, then your initial kinetic energy is$KE_i = \frac12 m v_i^2$and your final kinetic energy is$KE_f = \frac12 mv_f^2$. The difference is$\Delta KE = KE_f - KE_i = \frac12 m(v_f^2 - v_i^2)\$ It appears you're thinking that you can define ...

1

My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal? The answer is obviously "YES." It must be so. I would refer at this point these words; Nature is relentless and unchangeable, and it is indifferent as to whether its hidden reasons and actions are understandable to man or not.- ...

Only top voted, non community-wiki answers of a minimum length are eligible