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Potential energy stored in a body is relative. We have to first choose the potential at a finite point or infinity. In the case given above, we take potential energy to be 0 at the centre of the Earth. So according to the relation, $PE = mgh$ where $h$ is the height from the centre of the Earth. Generally, we take height from the surface of the Earth and ...

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Technically "potential difference" is the difference in electrical potential, i.e. $\Delta V$, not the difference in electrical potential energy, $\Delta U$. Potential difference ($\Delta V$) is also called voltage, in certain contexts. However, many people and sources are sloppy about their terminology, and they will say just "potential" when they really ...

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As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts ...

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In simple terms the internal energy can be thought of as the sum of the kinetic energy and the potential energy of the molecules. The kinetic energy of the molecules depends on the temperature - a higher temperature means that the molecules have more kinetic energy. The potential energy of the molecules depends on the bonds (interactions) between them - ...

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If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \... 1 Image object$m$is at some point$a$and you were to supply a force opposite to the gravitational force caused by$M$. This force is equal to$F_{stop}=-F_g$so that$m$hovers completely still at point$a$. M ------------------- a ↑ m Obviously this force can't do any work because the object doesn't move and ... 1 Saying that work done is +ve or -ve is a mathematical convention used for calculating energy transfers. It is +ve when it is done by us on the system, and -ve when it is done by the system on us. Positive work is done in pushing against a force to reach a configuration - eg pushing a car uphill. Negative work is done if a force pulls in the direction we ... 1 The object would experience a net negative force, and be moved a negative displacement. Potential is defined in terms of the work done by an external force. The object has a negative force acting on it due to the gravitational attraction so the external force acting on the object must be in the positive direction to have a net zero force on the object. It ... 1 The work done by an external agency in bringing the body to a point is needed. Note that the process must be quasistatic, otherwise kinetic energy terms will be needed. Now gravity will tend to attract a mass. Thus to keep the process quasistatic, one must oppose this gravitational attraction. In other words, F is directed oposite to the gravitational ... 1 Suppose you move the body down at constant speed, as small as you like, then the net force on the weight will be zero, that is, the force upwards you make will be exactly the same as the weight,$mg$. This upwards force makes negative work, as the displacement is opposite to its direction, and this work is exactly equal to the loss of potential energy,$mgh\$....

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When you write "energy given out" or "energy gained" you are expressing a choice of sign for the energy transfer that you are specifying, and each of them is different. Notice that in your scenario A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative. Employing a sign convention means always stating changes of ...

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there is an electric field inside the wire, and there is a loss of potential energy, or voltage as they move but this drop in voltage is usually negligible (thought not in some applications) and we only consider that the drop in voltage comes only from the circuits elements o loads. This idealization often fails not with the wires, but within the battery ...

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You have added the negative sign in front of your integral and then put in the cos(180) as well. Pick one. Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.

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