# Tag Info

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You can try easily what is the good answer by calculating the difference in energy between the two participating levels. $E_{upper} - E_{lower} = -Rhc [\frac {1}{(n+1)^2} - \frac {1} {n^2}]$ $= Rhc [\frac {2n + 1}{n^4 + 2n^3 +n^2}]$ So, its obvious that the smaller n, the bigger $E_{upper} - E_{lower}$.

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It seems to me that a plane wave description of the field in the dielectric medium can be valid only at distances(-y direction) large compared to the wavelength of the wave. Here I am assuming that the xz plane itself is the upper boundary of the dielectric. The modified variation of the field below the dielectric boundary must be used to evaluate the fields ...

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$F1 - F2$ is the same as turning $F2$ around 'head to tail' and you get $-\Delta F$ or a vector, which is the same magnitude, but the opposite direction to $\Delta F$. To get $\Delta F$ you need to reverse $F1$ as in the diagram I have modified from your diagram below

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That's because you want $F_1 + \Delta F = F_2$ by additivity of vectors (for a more rigorous approach, see the formalization of affine spaces). Hence, $\Delta F = F_2 - F_1$ PS: I couldn't comment because of my low reputation, so I made an answer for so little

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I'm going to assume you have some familiarity with linear algebra, as the math becomes much less tedious than trying to do three-dimensional trigonometry with the $x$, $y$, $z$-coordinates directly. You are looking for a function describing the line of the horizon. Since it is a circle and thus a one-dimensional object, I'm going to call it $\vec{h}(t)$, ...

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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 ...

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The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...

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I am not going to solve this completely for you, but note just a few things: The charge of an electron is negative In a simple harmonic oscillator, the force must be in a direction opposite to the displacement You had earlier set $x=A\sin\omega t$ so it seems to me that $\frac{x}{A\sin\omega t}=1$ I think that if you ponder the above you will see where ...

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Put the arrows after one another. Then draw a new arrow to the point they reach. The net force vector is the new arrow. The direction its' angle. The maginute is its' lenght (just as the magnetude of the two original forces were the lenghts of each).

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I'll leave the problem for you to solve, but here's a hint. Remember that forces are vectors, and "net" means "sum of components". You can certainly use angles when summing vectors - just be careful with the signs so that you cancel what chould get canceled and add what should get added.

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UPDATE (Solution): The center of mass of $B_1$ has a displacement of $$\vec x_1 = \vec x_{1,1} + \vec x_{1,2}$$ One due to RWS and the other due to its contact with the other body. Differentiation (twice) leads to $$\vec a_1 = \vec a_{1,1} + \vec a_{1,2} \xrightarrow{algebraic} a_1 = a_{1,2} - a_{1,1}$$. But $$T = m a_1$$ and $$\sum \tau = I ... 0 You want to integrate with respect to t. You have \ddot x = g(1-\frac xR), which is a harmonic oscillator equation. With the initial conditions, it won't oscillate, but you can use the usual solution technique. Your choice of x as depth instead of radius complicates things-I suggest using radius r measured from the center so \ddot r=-gr/R Find ... 0 The answer to the above question is the definition of work i.e. how we define it. When we apply force on a body, it is displaced with respect to its position. Work is said to be on the body. The definition of work says, "Work is the product of the component of force in the direction of displacement or vice versa". So the case I stated above has it's force ... 1 Consider the following results: From the definition of scalar product of four vectors,$$ \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} ...

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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 ...

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The answers are really good, but I have (as a former long-time student who has not forgotten the practicality of things) a more practical approach without considering currents or voltages, just the topology of the circuit itself: When trying to reduce a circuit, the algorithm is as follows: Any number (> 1) of resistors on the same wire. That is, no ...

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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 ...

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In the context of a canonical transformation (CT) $$z^I~=~(q^i,p_i)~\longrightarrow ~(Q^j,P_j)~=~Z^J~=~f^J(z,t),$$ the matrix $$\textbf{M}^J{}_I~:=~\frac{\partial Z^J}{\partial z^I}$$ is the Jacobian matrix of the CT. Here the indices $$i,j~\in~\{1,\ldots, n\} \quad\text{and}\quad I,J~\in~\{1,\ldots, 2n\}.$$ If the CCR reads $$... 0 I think what I do is similar to what lionelbrits has posted. I calculate the voltage around a square loop anenna two ways: first, as the rate of change of magnetic flux from the peak of the magnetic wave passing through; and second, as the maximum difference of the voltages along the two vertical legs as calculated by integrating the e-vector. If you assume ... 2 The center of mass of an object is the point where the first moment of mass is zero. Put differently, when you support the object at that point, it will be balanced. Assume that point is x_0, then$$\int_0^\ell (x-x_0) \lambda(x) dx = 0$$Substitute \lambda and some simple manipulation will give you an expression for x_0. Let us know how far you ... 3 The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2: If you make the loops narrow enough, i.e., their widths are dx, then$$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}\oint_2\!\vec{B}\cdot ...

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The differential and integral forms of Maxwell's equations are truly equivalent; they are essentially the same set of equations. One can convert between the two using two mathematical theorems: Divergence Theorem (Wikipeda - Divergence Theorem) Stokes' Theorem (Wikipedia - Stokes Theorem) The divergence theorem states that the flux over a closed surface ...

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This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?) $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with ...

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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 $\rho_a=1 ... 1 From the definition of Centre of Mass, the entire mass of the system is assumed to acts at the centre of mass. Consider a frame in which the position vector of$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 ... 0 This may not require an iterative method I guess. Write down the conservation of mass, conservation of momentum and conservation of energy (if there is heat exchange or some energy transfer which is not your case) equations considering a control volume with control surfaces on the two holes. This would help you to solve the problem 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 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 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, ... 0 I have been wondering about a similar problem for over a year and have yet to find an answer. A vertical rod (or a rod set at an arbitrary angle with the ground) has its bottom in contact with the ground but the contact is NOT frictionless. The bottom of the rod is not 'pinned" or "hinged", it is just in contact (momentarily) with friction. Does the bottom ... 0 Sorry everyone for the stupidity, I never converted degrees celsius to kelvin. Sorry!$T=100+273.15T=373.15\hspace{1mm}K$Therefore,$L=T(S_2 - S_1)L=373.15(1.76-0.31)L\approx540\hspace{1mm}cal/g$1 In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation$r \to \lambda r$, the geometry is only preserved by$R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ... 5 If the car starts out going in a straight line, it will drop a little bit in the time it takes to cross the gap. If the drop is larger than the height of the chassis above the ground, the car will crash into the opposite wall. When the drop is less than that (small gap, high speed) and the wheels are able to absorb the shock, it is possible that the car will ... 3 The answer depends on many factors, but here are the basic bits of physics that play: The power and wavelength of the laser The reflectivity of the surface (function of wavelength of the laser) The size of the focal spot The thickness of the sheet The thermal conductivity of the sheet The reflectivity of the copper is a particularly important one. If you ... 3 When you push off, as you go forwards, the boat moves backwards. Momentum is conserved, so that your mass$\times$velocity equals the boat's mass$\times$the boat's velocity. If you do the sums you will see that if you and the boat have comparable masses then the boat moves backwards quite quickly. Because you exert the force for your leap over a small, but ... 2 First, the obvious explanation for the sign is that if$J$has a minus sign in (1), then there should be a minus sign in (4). For some reason your$G$turned into$\phi_i$. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, ... 1 Well it all depends on what assumptions you are able and willing to make. For a start you can work out how much of the laser light is reflected. The reflected power will be something like (for normal incidence) $$\frac{P_{r}}{P_i} = \frac{(\eta_m - \eta_{vac})^{2}}{(\eta_{vac} + \eta_m)^2},$$ where$\eta_{vac}=377$Ohms and$\eta_m$is the impedance of the ... 0 There are so many variables here i think its nearly impossible to give an answer like how thick is the lazer, how far away is the lazer how reflective is the surface of the metal ......... the more variables there are the harder something is to predict if you wanted an answer it would be much more practical to get one by doing the experinment, probably not ... 1 I think we are supposed to assume that the buoyancy force of the balloon is equal to the weight force of the balloon, ladder, and climber. If this is the case, the system is in equilibrium with its environment, with no net forces to the environment. You could look at it as a center of mass problem. We assume the ladder has negligible mass. When the climber ... -1 You'll need to first calculate the Poynting vector or the momentum density of the fields. The angular momentum density is L= r x p. Once you have that, you can find torque easily. 0 To answer where did the g - a come from is that form the period formula T = 2Pi * sqrt(l/g-a) when the elevator is going downwards it will be g-a but if its going upwards against the gravitational force it will be g+a. 0 what you need to do is replace the springs with equal and opposite forces on both ends and do a Free Body Diagram. The answer comes straight out of this. 0 Hard to answer this without just suggesting an answer. Consider the forces on the mass. There is gravity that is downwards, but I guess the mass is not moving so think about the forces on the mass - you could also think about the forces the point where the two springs joint together. Hope this is helpful. 1 If pulleys are not weightless and maybe rope is not weightless. Weight$Q_1=m_1 g$accelerates whole setup. Tension$T_1$is responsible for acclereating two pulleys, rope and mass$m_2$,$T_2$accelerates one pulley less to the same acceleration, so that it is lower. By the same reason$T_3$is lower than$T_2$. 2 The first one is correct,$L_{new}=2L$. In your second expression$L=\frac{v^2}{2a}$, when changing$a$you must also find a new end speed$v$, $$L_{new}=\frac{v_{new}^2}{2a_{new}}=\frac{1}{2}\frac{v_{new}^2}{2a}\neq\frac{1}{2}\frac{v^2}{2a}=\frac{1}{2}L$$ A note from reading the comments to the question: I am assuming the objects fall during the same ... 2 The error in the OP's question comes from the second Kronecker delta, which is not correct (furthermore, the sum over$k$is not dealt with...). After using the first Kronecker delta$\delta_{k\,k'}$, one has to use the identity $$\frac{1}{N}\sum_{k=0}^{N-1}\exp{[\frac{2\pi i k(j'-j)}{N}]}=\delta_{j'\,j},$$ which directly gives that$F^\dagger F =1$. One ... 0 I don't believe it's true that$\langle k | k' \rangle = \delta_{k',k}$. If I recall correctly, the whole point of the$\frac{1}{\sqrt{N}}$factor is to normalize the momentum eigenstates, so we should have$\langle k | k' \rangle = N\delta_{k',k}$. 2 All your thinking is very good and correct. BUT you are missing a point in the question :) Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True. BUT let's read the question: A door is hinged at one end and is free to rotate about ... 0 The torque equation is $$\tau = (r \times F) = rF \sin \theta$$$r$is a vector from the point from which torque is measured to the point where force is applied.$\times$is the cross product$\theta\$ is the angle between the position vector and force. F is the amount of force directed perpendicularly to the position of the particle. Any force directed ...

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