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You can view it pictorally. For when the second velocity is very high, the LHS tends to the second velocity. However the LHS tends to the average of the two velocities.


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See this link for a list of how to compute tensors of various types. The one you are interested in is the following one: $$ T^{ab}_{\ \ \ \ \ ;c} = \partial_{c} T^{ab} + \Gamma^{a}_{\ cd} T^{db} + \Gamma^{b}_{\ cd} T^{a d} $$ You are contracting your indices in the sense that $$ T^{\mu\nu}_{\ \ \ \ \ ;\nu} = \partial_{\nu} T^{\mu\nu} + \Gamma^{\mu}_{\ \nu ...


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your equations seems all right to me. Note that for 26th car has negative acceleration time ti=15-(26-1)=-10 and therefore you should use the first equation $t_i=0$. But if you don't, this means you assume car was accelerating with the same acceleration for the infinite amount of time both to the past and to the future. In reality the 26th car was ...


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The $9.8m/s^2 + 5 m/s^2 = 14.8 m/s^2$ is the case. This is because to even stand still on the launchpad requires a force equivalent to an acceleration of $9.8m/s^2$ pushing up, to oppose the force of gravity, equivalent to $-9.8m/s^2$ pulling down. This is due to Newton's 3rd law of motion. 'For every action there is an equal and opposite reaction (for an ...


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You are confused because you think that since the two accelerations are in opposite direction, so for finding the acceleration by the given thrust in the absence of gravity you are subtracting one from another. But you must note that that isn't the case over here. Here is a thought experiment which might help you: Suppose that you are falling from a given ...


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By convention, the direction of an electric field is the direction of the force that a positive charge would experience if placed in the field. What does that tell you about the direction of the force that a negative charge would experience if placed in the field? Hope this helps.


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What you are missing is the fact that the electron has a negative charge, so the field and the force are in opposite directions.


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If the rocket were hovering in mid air, not moving, it would have to be delivering a thrust to counter gravity, 9.8 meters per second squared. To accelerate, it would have to deliver an ADDITIONAL thrust. So you add the thrust for actual acceleration to the thrust needed to overcome gravity.


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You generally assume the force is varying temporally fast, but spatially slow such that you can define a particle's position as a function of time as: $$ x(t) = x_{o}(t) + \delta x(t) $$ where the $x_{o}$ term is slowly varying term and the $\delta x$ term is for the quickly varying temporal term. From this we also assume that $\lvert \delta x \rvert$ $\ll$...


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At any point other than $x=0$ the potential does not exist. So the TISE is a free one. However about $x=0$ when trying to solve the equation there is a potential. We can use the fact that $$\int_{a}^{b}dxf(x)\delta(x) = f(0)$$ if $0\in(a,b)$ and zero otherwise (this is the definition of the delta-function, in fact), to have $$\frac{d}{dx}\psi(x)|_{\epsilon}-...


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Your assumption is right.Regarding how to treat the delta function, $\delta(x)$ is just $\delta(x-0)$, and the integral is just equal to $\psi(0)$ since the delta function is located on the $x=0$. To clarify more, the integral of $\delta(x)$ from $-\infty$ to $\infty$ is the same as $-\varepsilon$ to $\varepsilon$ since the delta function is zero beyond this ...


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As said in the comments to your question, the quantum numbers assigned to particles and the resonances in order to fit experimental data, will give the allowed reactions. It is simpler to think in terms of the possible Feynman diagrams for the specific decay one is looking at (similar to the one in the title) for example: Particles can decay via the ...


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Timaeus succinctly gives the fundamental Physics. But it's worth remembering Faraday's empirical rule: an emf is induced in a conductor when it cuts lines of magnetic flux. In your diagram lines of magnetic flux run from left to right. As the skeleton cube moves 'upwards', only EH, FG, AD, BC cut these lines.


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Imagine a bunch of rays travelling parallel to the principle axis and diverge after travelling through the lens. If you had to take the focal length to be positive you would have to trace it on the positive side which is not possible. When you trace the paths of the diverging rays to find which is the spot of convergence you end up at the focal point which ...


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If I'm not mistaken, you are referring to the second term in the following equation from page 5 of the paper K = $\frac{1}{2}M_{CM}v^{2}$ + K(rel. to CM) Well... writer commented (on page 1) further explanation would go on about fluid later but I can’t find anything relevant. I believe the further explanation alluded to by the writer is probably ...


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You should study the condition of buoyancy stability for these three cases: the center of gravity must be below the metacenter (http://www.cns.gatech.edu/~predrag/courses/PHYS-4421-13/Lautrup/buoyancy.pdf).


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Find the buoyant force related to the position of the top and bottom surface of the object. The acceleration may change.You need to find the viscous resistance as well and then take Newton's second law of motion and you will find the final result.


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Kinetic energy of a system is the ability of the system to create events based on the motion of the objects which make of the system.


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There are two points here, I believe: one is that acceleration is something that is defined for very short times. It tells us the rate of change in velocity, so we can think of it as "how much is $v(t+\delta t)$ different than $v(t)$ for very small $\delta t$?" When deriving the formula you are referring to, of $x=\frac{1}{2}at^2$, an assumption was made ...


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For this to make sense, we have to talk about $F$ as a 2-form, i.e. $\mathbf{F}=\frac{1}{2}F_{\mu\nu} \: dx^\mu \wedge dx^\nu$ Then $\star\mathbf{F}$ is its dual. In 4d space: $\star \left(dx^\mu\wedge dx^\nu\right) = \frac{1}{2}g^{\mu\kappa}g^{\nu\sigma}\epsilon_{\kappa\sigma\rho\phi} dx^\rho\wedge dx^\phi$ Thus: $\star \mathbf{F} = \frac{1}{4}F_{\mu\...


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to see it intuitively just charge both toU against earth the will have a charge of Q1 and Q2 then connect then, they still have U and now together Q1+Q2 . (the capacity of the wire is neglected, otherwise u would diminish a little.


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Hint: By spherical symmetry the electric field $E_r=-\frac{d\phi}{dr}$ should vanish at $r=0$, which leads to $c_1=-e_0$.


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You should follow the right sequence of statements about conversation of energy: Work-energy theorem $$A = K_2 - K_1$$ In the left part of this equation one can see the summary work done by all the forces applied to the body. There are different types of forces: Potential: $\qquad A_{potential} = \text{П}_1 - \text{П}_2 $ Dissipative:   &...


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The Question is more of a combinatorics question. "What are the number of interactions between 6 particles;Given that each interaction involves 2 particles?" The answer is of course $4C2$ which is 6. (The Mistake You had done is that You had not considered an interaction Between 2 Particles at a time) Now for the Physics Part. If the Particles are A,B C,D ...


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No, if you included the other ones you would be double-counting (notice your way has 12 terms instead of 6). An interaction involves two charges. So you have the A-B interaction, the A-C interaction, etc. The X-Y interaction Isn't the combination "X on Y" and "Y on X" interaction. It's just a single interaction between the two charges. Another way to look ...


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Picture the current as a number of electrons moving along the wire in a given time, and you'll realise it has to stay the same throughout the wire, otherwise you would have to be creating extra electrons at the point where the diameter of the wire increases. The doubled cross section is reflected in the fact that there are now twice as many moving electrons ...


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First, let's start by completing the square as suggested. The equation will now be of the form $$ \frac{d^2\Psi}{dz^2}+(\nu+\frac{1}{2}-\frac{1}{4}z^2)\Psi=0 $$ with boundary conditions $\Psi(z=L)=\Psi(z=\infty)=0$ and $$ z=\sqrt{\frac{2m\omega}{\hbar}}x+L \\ \nu=\frac{E}{\hbar\omega}+\frac{L^2}{4}-\frac{1}{2} \\ L=-\sqrt{\frac{2mg^2}{\hbar\omega^3}} $$ This ...


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What other forces might be acting on it (other than normal force by the loop and the gravitational force)? Actually, those are the only forces acting on the bead. However, keep in mind that there will be "inertial effects" coming into play here. For another example of this, think of the example of a bead on a rotating, horizontal, straight wire. In that ...


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If the carts are the same mass (including the wheels), and the wheels are rolling without friction, then the only difference is the extra force needed to increase the angular velocity of the fourth wheel.


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If $F$ denotes the QFT operation, then by definition $\sqrt N F|j\rangle=\sum_k \omega^{jk}|k\rangle$. By linearity, you therefore have $$F\left(\sum_{x=0}^{N/r-1}|xr\rangle\right)= \sum_{x=0}^{N/r-1}F|xr\rangle =\frac{1}{\sqrt N}\sum_{y=0}^{N-1}\left(\sum_{x=0}^{N/r-1}\omega_N^{xyr}\right)|y\rangle.$$ Now by assumption $r\mid N$, and thus $m\equiv N/r\in\...


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Presumably you're referring to the Heisenberg Uncertainty Principle, a part of quantum mechanics, which states that there exist pairs of so-called 'conjugate variables', like the position and momentum of a particle, which cannot be measured with infinite precision simultaneously. (More technically, it states that the uncertainty in position, $\Delta x$, ...


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The way to obtain the coefficients $c_i$ is straightforward. You have to remember that: $$\Psi = \sum\limits_{i}c_i\varphi_i = \sum\limits_{i}\langle\varphi_i\vert\Psi\rangle \varphi_i $$ So $c_i=\langle\varphi_i\vert\Psi\rangle$. Since you are in the position representation in 1D, the inner product reduces to: $$\langle\varphi_i\vert\Psi\rangle=\int\...


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It is just the projection on the ground state. \begin{equation} \mid \psi \rangle = \sum_n \langle n \mid \psi \rangle \mid n \rangle \end{equation} So $c_0 = \langle n \mid \psi \rangle $. In this specific example, this corresponds to a Fourier-series.


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The Gell-Mann matrices do not belong to the group ${\rm SU}(3)$. They are a vector-space basis for ${\rm Lie}\{{\rm SU}(3)\}$, the Lie algebra associated with the group. The centre $Z\subset {\rm SU}(3)$ is the set of unit-determinant unitary matrices that commute with all elements of ${\rm SU}(3)$. Schur's lemma tells us that they have to be multiples ...


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You can use the lagrangian approach to solve it. You know that it is fixed to the surface so you can use $x, y$ as the independent coordinates. So the lagrangian is: $$L = T-V =\frac{1}{2}m\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)-mgz$$ Where $T$ is kinetic energy and $V$ is potential energy. You now use the fact that $z$ can be expressed in terms of ...


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Let $A$ be a $n\times n$ matrix. Recall that $$A \text{adj}(A)=\det(A)\cdot1$$ where $1$ is the identity matrix, $\text{adj}(A)=C^T$ is the adjoint matrix, $C_{ij}=(-1)^{i+j}M_{ij}$ is the cofactor matrix, $M_{ij}$ is the minor of the $A_{ij}$. This is justified because $$\det{A}=\sum_j A_{ij}C_{ij}=\sum_j A_{ij}adj_{ji}=(A\text{adj}(A))_{ii}.$$ Now ...


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Poisson's equation is a general equation, which takes the following shape: $$\Delta \phi = f \text{, where $\phi$ and $f$ are real or complex-valued functions}$$ $\Delta$ is as we know the Laplacian operator, the divergence of the gradient: $\Delta = \nabla^2$. I believe you may have not taken into account the fact that $\vec{E} = -\nabla V$ if you have a ...


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Determinant of the metric, or any matrix can be expressed through Levi-Civita (relative) tensors, or through generalized Kroenecker deltas I have written about it here https://www.physicsforums.com/threads/i-cant-verify-a-relationship-between-cofactor-and-determinant.970419/post-6165630


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Note that both the answers are the same. $v_2=\frac{m}{M+m}v_p$. Put this in $h'=\frac{v_2^2}{2g}$ and get the first expression. Your teacher has just expressed the answer in the variables given in the question.


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I'm going to use a somewhat more complicated method than the one I hinted at. Given a charge q at position $z_0$ with $z_0<R$, our image charge is $-qR/z_0$ at $R^2/z_0$. We can generalize this. Let us have N charges Q from 0 to $R/2$ such that $2NQ/R=\lambda$, N>2. This means we have a total charge NQ divided by length R/2, giving our line charge ...


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For the simplest case if there is no frictional force between the wall and the ball the ball would move in the circular part solely due to the normal force acting on it and hence there would be no change in speed of the ball[Note that at any given instant the Normal force $N$ would be perpendicular to the direction of motion and hence the work $W_N$ done by ...


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Mmmm... I don't think there is a proof as such, since it's simply a choice of phase for the individual highest weight states $|j,j\rangle$ states in the decomposition . It would be like asking for a proof that the coefficients $\sqrt{j(j+1)-m(m-1)}$ in the action of the ladder operator $J_-$ are real. This latter reality is merely a choice of phase for ...


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Advice: don't rely on mechanically applying a formula. Think about what is actually happening in the problem instead. The average speed for a journey is total distance travelled divided by total journey time. So we need to find the total distance travelled and the total journey time. The total time for the journey is $2.5$ hours plus $30$ minutes plus $20$ ...


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You are failing into the trap that many students fall into. Moving in a certain direction does not mean acceleration is in that direction. Acceleration is a vector quantity that describes how the velocity vector (not the speed) changes. Therefore, there is nothing saying that the direction of motion must coincide with the acceleration (it usually does not). ...


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You might be misunderstanding the notation. $$ \delta \Phi = \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi} -\Phi +O(\alpha^2)= \frac{i}{2}\alpha_j\sigma^j \Phi, $$ so that $$ \delta \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} = \frac{i}{2} \begin{pmatrix} \alpha_3&\alpha_1-i\alpha_2\\ \alpha_1+i\alpha_2&-\alpha_3 ...


0

If you have a generic volume charge $\rho(x, y, z)$, each "small piece" of the charge object generates an electric field at a generic point $\vec{P}$ is given by $$ d\vec{E} =\frac{\rho(x, y, z) \hat{r}}{4 \pi \epsilon_0 r^2} $$ where $\hat{r} = \vec{r}/r$, $\vec{r}$ is the vector joining $(x, y, z)$ and $\vec{P}$ and $r$ is its module. This distance is ...


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By this argument, shouldn't the potential at $P$, due to $Q_1$ be equal to that on the surface of the second sphere (due to $Q_1$)? Yes, it should. However, in this type of problems is is usually assumed that the spheres are very far away from each other or, in other words, that $AB \gg r_2$. In this way the two spheres don't influence each other so that ...


-1

The work of a force that is constant (which is the case for all 4 forces acting on this block), is calculated like this: Work = Magnitude of the force x Displacement in the direction of the force. Work done by applied force: The magnitude is 200N and the displacement in it's direction is 12m. Hence W=200*12=2400J Work done by the normal: the displacement ...


0

Couple of issues here. First, assuming these “carts” have wheels we are dealing with rolling resistance (aka rolling friction) not kinetic friction (aka sliding friction). So the force opposing motion is $$F=C_{cc}N$$ Where $C_{cc}$ is the coefficient of rolling resistance and $N$ is the normal force =$mg$. Second, unlike the coefficient of kinetic ...


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another expression for the vertical displacement is given by: $$\frac{-(rsin\theta)^2}{2(-9.81)}$$ whatever your vertical distance may be, multiply it by $2(-9.81)$, and continue rearranging for r.


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