# Tag Info

4

In an alpha decay no electrons are created or destroyed. There is a small correction needed for the Coulomb term when the alpha escapes without carrying two electrons with it, but that is at chemical, not nuclear energy scales and is (usually1) sorted out by chemical means in fairly short time scales. So, no you do not figure the mass of any electrons into ...

3

Edge effects. After the electron leaves the capacitor, the electric field winds up slowing it back down. Let's assume the capacitor is infinitely-massive and that the acceleration of the electron is small enough that we can ignore radiation. Then if you were to idealize the electric field of the capacitor, treating it as a uniform field between the plates ...

2

The enthalpy of combustion of hydrogen is 286kJ/mol i.e. 286kJ per 18g of water produced. So if the original mass of water in your comet is M (in kg), and you manage to split it completely into hydrogen and oxygen, then the energy released when you react the hydrogen and oxygen is given by:  E = 2.86 \times 10^5 \frac{M}{0.018} \approx 1.6 \times 10^7 M ...

2

When we're calculating the energy stored in a capacitor we normally assume it is isolated i.e. there are no other charges nearby to affect it. This makes the calculation nice and simple: the energy is proportional to $Q^2$ and the energy is stored in the electric field around the capacitor. However in your question you are introducing another charge, your ...

2

The energy is of course coming from the electric field of the capacitor. The energy of any capacitor is always stored in it's electric field. If an electron is initially positioned very far away and then moves close to the capacitor, it's being pulled by the field and that means energy is being transferred. The electric field get's a little weaker - loosing ...

2

Yes. Not only can it, but the values are tabulated for unstable isotopes. Compare the data provided for $^{206}\mathrm{Pb}$ (stable) with the provided for $^{210}\mathrm{Pb}$ (unstable). Look in the section headed "Decay properties". Moreover, note that there is a separate $Q$ provided for each decay mode (but not for each channel). In this instance, ...

2

First, of course there's no perfect mirror. But let's assume there was one. Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever. But what if it does lose energy ...

2

New version The problem in your demonstration is when you write down $\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case $||d\vec{r}||\neq dr$ because $dr<0$ when you go from $\infty$ to $r$ and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line. Old version The demonstration on ...

2

When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward. Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in ...

2

The question you have formulated is not an easy one to answer (correctly). But the question you've formulated isn't quite the question that I see. The good news is that the text of the question you've posted implies a much simpler question; it's just asking for the energy change. You can probably assume that the acceleration is dominated by the circular ...

1

Potential energy is a property of the system, not any one object. Thus there should only be one copy of the typical $1/r$ potential energy between two charges (plus an analogous gravitational term if that can't be neglected). The easiest way to see this is to start from "infinite" separation. Instead of pushing the two charges together, hold one fixed and ...

1

Actually, the Liouville theorem is more general - it is valid even if the distribution function depends on time, and even if the Hamiltonian depends on time. http://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) -> phase space volume preservation but no energy conservation: any Hamiltonian which depends on time, but you already know that. For ...

1

Liouville's theorem not only depends on the form of Hamilton's equations but also on the fact that $\partial\rho/\partial t = 0$, where $\rho$ is the statistical distribution function of the system. This is strictly true only for closed systems and is approximately true for quasi-closed systems when not observed for too long a time. Energy of a system is ...

1

$\mathbf{r}$ is a position vector and $\mathbf{s}$ is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed.

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Just to be clear, the potential energy of a particle of charge $q_2$ at a distance $r$ from a source of potential (supposidely at zero) of charge $q_1$ is the work that an external operator has to provide to bring the particle from infinity to $r$ at constant velocity. This reads then: $\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$ As people have said, the ...

1

Capacitor is losing energy, potential has changed as field is created even by this charge which is moving under the influence of force between capacitor plates . Take the point charge's potential , and then assume distance between capapcitor plate is d, now as -ve charge approaches +ve plate, it decreases the potential of the +ve capacitor plate more than ...

1

Ashish already essentially said this, but using only the equations of motion we can show conservation of total momentum with the following calculation: \begin{align} \frac{d}{dt}(m_1\dot{\vec x_1} +m_2\dot{\vec x_2}) &= m_1\ddot{\vec x_1} +m_2\ddot{\vec x_2} \\ &= m_1\left(- G m_2 \frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|^3}\right) + ...

1

OK, the first part: net force $\vec{F}=m_1 \ddot{\vec{x}}_1+m_2 \ddot{\vec{x}}_2=\vec{0}$, so by newton's second law total momentum is conserved(or force is rate of change of momentum, if total force is zero, total momentum doesn't change), and the second part: there is no sort of non conservative force acting, hence mechanical energy is conserved. Non ...

1

The answer is the energy goes into the gravitational field. If you take the simplest case of a spatially flat homogeneous cosmology with no cosmological constant then the equation for energy in an expanding volume $V(t) = a(t)^3$ is $E = Mc^2 + \frac{P}{a} - \frac{3a}{\kappa} (\frac{da}{dt})^2 = 0$ $M$ is the fixed mass of cold matter in the volume, ...

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