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6

Let me explain @ACuriousMind 's answer with some verbiage. The short, regrettably oracular, answer is that the Fabri-Picasso theorem does not hold in a finite superconductor, since translational invariance fails at its boundaries. Really, I do appreciate this is aggressively obscure: will strive to explain. First of all, if you have a chunk of warm ...


5

A physical constant is a physical quantity that is generally believed to be both universal in nature and constant in time. Well-known examples include the speed of light (c), elementary charge (e), and Planck's constant (h). (More about physical constants here.) A scalar (in the context of physics) is a physical quantity that can be described by a number ...


3

The forces acting on the fluid are changing its momentum. There are forces acting on the stationary portions of the control volume boundary, and there are also forces being applied to force fluid into, and acting to prevent it from flowing out of the control volume. The total rate of change of momentum in the control volume at any instant of time is equal ...


3

For generic initial conditions, the answer is Yes, due to Lagrange equations $$ \frac{dp_i}{dt}~\approx~ \frac{\partial L}{\partial q^i}, \qquad p_i~:=~\frac{\partial L}{\partial \dot{q}^i}. $$ [Here the $\approx$ symbol means equality modulo eom.]


2

@ChesterMiller: answer is good, but I would just simplify it. What you say is that the total momentum is the momentum of the contents of the control volume plus the sum of that which traverses the boundaries. If you take the change in momentum per unit time, you have momentum flux, which equals force.


2

It's unclear what you mean by "nuclear engine", but the main similar notion is a nuclear thermal rocket. Although it derives its energy from nuclear reactions, it uses this to heat gas (usually hydrogen) to very high velocity for propulsion. There is still matter being expelled. More common in space travel is the radioactive thermal generator, which uses ...


2

A conserved quantity is a quantity whose value remains the same over time. An invariant, or scalar quantity is a quantity whose value is the same in all reference frames. These two properties are completely independent. Energy is conserved but not invariant. Mass (i.e. $E^2 - c^2 \mathbf{p}^2$) is invariant but not conserved. Charge is both, and lots of ...


2

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of ...


1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for


1

You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...


1

When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ...


1

When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ...


1

I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ...


1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: $v_1, v_2, ...



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