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An isolated body that doesn't exchange any angular momentum with the outside universe will never stop spinning (by conservation of angular momentum). There is no way to absorb angular momentum within the body in internal degrees of freedom; the angular momentum must be transported away if you want to stop. For example: if something is not rigid, you could ...

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Conservation of angular momentum means that the sphere will continue to spin forever. In order to change the angular momentum you need to apply an external torque. Note that this treats the sphere as a rigid body. If you consider just a small part of the sphere there are forces acting on it in such a way that the sphere remains undeformed. On a microscopic ...

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This problem has a recursive flavor that we'll not try to avoid. Conservation of momentum tells us that $$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$ Imposing the boundary condition $v(0)=0$ we find $$v(n) = \frac{n}{n+p}v_0$$ as claimed. Let $a_n$ be the time at which the $n$th bullet strike occurs. We have $a_1=x_0/v_0$ and $$v_0 (a_n - T) = v_0 ... -2 The time taken between the N-1 collision and the N collision is T-T\frac{N-1}{p}=T\frac{p-1+N}{p} Edit: Reasoning: The difference in T is due to the N-1 collision and is given by: T\frac{N-1}{p} 0 The decay of a neutral \pi^0 to three photons would indeed violate charge conjugation. The charge conjugation argument goes as follows: The reaction$$\pi^0 \to 3 \gamma$$is mediated by electromagnetism. QED has a charge conjugation symmetry, so you should be able to apply a charge conjugation to both sides of the equation. Under charge conjugation, ... 0 In regards to the current question, for a single charge going in a loop, i = qf. f = 1/(2pi x sq root(LC)), w = 1/(sq root LC) = V/R. After the corresponding substitutions, i = qV/2piR 2 How can we apply angular momentum conservation when friction is present? Why not? If we have a closed system, momentum and angular momentum are conserved. In this case, the full system is disk A and disk B, and there are no external forces, so the system is closed. There are internal forces, namely in this case, friction, but that doesn't matter. You ... 3 Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will". If you take q|\vec v| instead of q\vec v, you get something related to current (generally current times a factor). Current is conserved at a junction. Regarding your equal-and-opposite situation, the closest ... 0 When a particle is deflected by gravity the gravitational field will also be modified by the particle. To form a conservation law for momentum you need to take into account the momentum in the gravitational field as well as the particle. This can be done e.g. using pseudo-tensor methods. This works but remember that momentum is a relative concept. Even in ... 0 The cart and the ramp have to be able to move at different speeds from one another, otherwise the cart could never move up the ramp or back down it again. So they can't both have the same velocity at all times. However, the centre of mass of the system always keeps moving with the same velocity in the x direction (both before and after the collision), and ... 0 With your current assumptions you do not have enough equations to solve this problem, since it is two dimensional, which gives you 4 unknown variables: u_{1}',v_{1}',u_{1}',v_{2}' where u and v are the speeds in the respectively x and y direction, the indexes 1 and 2 indicate which object it is and the apostrophe (') indicates that these ... 0 Here's a formula to help you find final velocities from initial velocities:$$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2} These formulas you get from combining momentum and energy equations. You have to apply both of the above formulas separately in 2 dimensions: $x$ and $y$. So you should get ...

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Vorticity can certainly be destroyed, this is the basis of the energy cascade in 3D turbulence where energy is channeled across wavenumber space from large to small scales all the way down to the Kolmogorov scale at which point it dissipates into heat. Of course in order to do that you need to be looking at the complete equations... this link should answer ...

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