# Tag Info

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You might get an order of magnitude estimate as follows. We make the rough assumption that everything ends up in its vessel as a monoatomic ideal gas - actually it will be a plasma, with a thermal energy per mole of $\frac{3}{2}\,R\,T_{final}$, where $T_{final}$ is the thermodynamic temperature of the plasma. Neglecting heats of vaporisation (we assume ...

8

I think the colloquial term for that type of plot is "spaghetti diagram" because you have a bunch of lines running across it. It's really the mass fraction as a function of interior mass. From our stellar structure equations, we have that $$\frac{dm}{dr}=4\pi r^2\rho,$$ which is derived from the mass-continuity equation, so you can relate the radius, $r$ ...

5

If the car starts out going in a straight line, it will drop a little bit in the time it takes to cross the gap. If the drop is larger than the height of the chassis above the ground, the car will crash into the opposite wall. When the drop is less than that (small gap, high speed) and the wheels are able to absorb the shock, it is possible that the car will ...

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1

You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...

1

The idea is that since the steel beam has conduction electrons that are free to move, the movement of the charges in a magnetic field causes a magnetic force to act. The magnetic force causes the electrons to accumulate at one part of the curved surface of the rod, thereby creating a potential difference. The charges keep accumulating till the potential ...

1

Consider the following results: From the definition of scalar product of four vectors, $$\tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$\tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$\tag{3} ... 1 In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation r \to \lambda r, the geometry is only preserved by R_S \to \lambda R_S. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ... 1 Since the surface is frictionless there is only vertical force. The torque is given by the normal force of the surface multiplied by the horizontal distance to the center of mass (c.o.m.). Now the normal force depends on the vertical acceleration of the c.o.m. - you know that the acceleration of the c.o.m. is a result of all the forces acting on the object, ... 1 Unlike in QFT where you can derive spin from more basic principles, in ordinary non-relativistic QM spin is essentially defined into existence as a group of operators S^i = (\hbar/2) \sigma^i that satisfy the algebra$$[\sigma^i, \sigma^j] = i \epsilon^{i j}_{\,\,k} \sigma^k.$$The dimensions in the Hilbert space on which the Pauli operators act are ... 1 Without providing a numerical answer to your question, to do so would help nobody: It would help to first draw a 'free-body-diagram', doing so will help you visualise the forces acting on the pole. You need to calculate the moment at the hand 1m from the end, this is your pivot, from the force exerted by the mass of the pole. Calculate the force required ... 1 From the definition of Centre of Mass, the entire mass of the system is assumed to acts at the centre of mass. Consider a frame in which the position vector of i^{th} particle is R_i and its mass is M_i. Total mass of the system is$$M=\sum_i M_i Let $R_{cm}$ be the location of the centre of mass measured in this frame. Then the placing the body ...

1

Assuming a direct hit - so traveling through about 50 km of atmosphere - at 0.1 c that would take about 2 ms if it didn't get slowed down too much by the atmosphere. What about drag force? Let's assume a radius $r$, density $\rho$, mass $m = \frac43 \pi r^3 \rho$. If it is a sphere, it experiences a drag force $F=\frac12 \rho_a v^2 C_d A$. Putting \$\rho_a=1 ...

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