New answers tagged

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A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$). The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in ...


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Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. $$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ...


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Comments to the question (v4): Note that the formula (3) implies that the vector density ${\cal X}^{\mu}(\phi(x),\partial\phi(x),x)$ depends on spacetime derivatives $\partial\phi(x)$ of the field $\phi(x)$. This implies that the new Lagrangian density $${\cal L} ~\to~ {\cal L} + d_{\mu} {\cal X}^{\mu}\tag{A}$$ will depend on higher spacetime derivatives ...


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As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get $$\ddot x=0$$ $$\ddot y = -g$$ Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...


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In figure A the satellite is orbiting in a circular orbit. By the satellite increasing speed it would turn the orbit into a more egg shaped orbit as shown in B. This would rather increase the time for orbit revolution rather than decreasing it When the satellite is approaching B.1 it's velocity will begin to slow down as earths gravity begins to pull it. ...


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First of all we need the equation: \begin{equation}\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT} \end{equation} We have: $$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} ...


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I think this question does not actually address the photoelectric effect because this would be strongly dependent on the frequency of the light source. To me this seems to be a question regarding the relationship between intensity, power and work. While this is technically not how an electron is emitted in real life, one could calculate the power incident ...


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You first need to know that how the current flows through a capacitor. There are two parallel plates with a dielectric between them such as air etc. One plate is connected to the positive terminal of the battery while other is connected to the negative terminal of the battery. Charges from the negative terminal will accumulate on the plate connected to it. ...


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Anna, you are more or less right. It is pressure from matter at higher densities that can stop the gravitational collapse. It depends on the state of the matter, and in a simplistic description the equation of state. As it collapses as a hot gas after it exhausts it nuclear fuel, and maybe after a supernova explosion, it'll collapse. The first point at ...


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The definition of force is $F = m a = m \frac{dv}{dt}$. The average force is $<F> = \frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt$. Write $F= m \frac{d v}{d t}$ to find $<F> = \frac{1}{\Delta t} \times$ momentum change.


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If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration. In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.


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It depends on whether the current is carried by a conductor or is in free space (an electron beam). In the case of an electron beam, the current will appear to have reversed in direction if you travel faster than the charge carriers, even without relativistic effects. This web page does the transformation roughly like you have attempted, using a charge ...


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Q1. But how this can be explained "theoretically"? I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'. The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole. Naturally your probe charge ...


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If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion. Hence no work is done moving the charge.


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Indeed, $f$ is a symmetric form, since $\omega$ and $\omega '$ are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes $\omega$, the dimension of the space is $6$. A basis is given ...


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Easy way Let me first state the straight-forward way to do this computation. $$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$ First equality follows from compatibility, second equality uses ...


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The answer should be d). For Simple harmonic motion, the period (and frequency) are independent of the amplitude and the initial phase of the motion. Note: the period of oscillation IS the inverse of the frequency


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I think you are asking about how to transform various vector quantities between points attached to a rigid body. Here are a rundown of the rules of transformation between an arbitrary point A (located at $r_A$) and the center of mass C (located at $r_C$). $$\begin{align} v_C & = v_A + \omega \times (r_C - r_A) & & \text{linear velocity at C} \\ ...


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You're right that the Feynman propagator for a spinor field is indeed $(i \gamma^\mu D_\mu - m)^{-1}$. The tricky part is interpreting exactly what the "inverse" means. It doesn't just mean that you invert the gamma matrices (although you do do that). The derivative operator is also being inverted. That is, if we define a function $G(y, x) := (i ...


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Normalization Factor Let us define a generalized Gaussian probability density function (PDF) as: $$ f_{s}\left( x \right) = A_{o} \ e^{^{\displaystyle - \frac{ (x - x_{o})^{2} }{ 2 \sigma^{2} } }} \tag{0} $$ where $A_{o}$ is the normalization constant, $x$ is the argument, and $s$ denotes the set of distributions (e.g., particle species), $x_{o}$ is the ...


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If by "solve" you mean to get an answer that is independent of θ, the answer is... Nope. You need θ, it doesn't cancel out. However, here's a slightly similar problem that you CAN solve: You've got a similar hillside, and a sled at the top of it. The height of the hillside is h, the friction coefficient is μ, and the angle is θ. You let the sled slide ...


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Put a coordinate system on the center of mass and place each leg i at $$\vec{r}_i = \pmatrix{x_i, & y_i, & z_i}$$ where $z_i = z_{c}+\theta_x y_i - \theta_y x_i$ describes the vertical deflection of the point, given the center of mass vertical position $z_c$ and the two tilt angles $\theta_x$ and $\theta_y$. Add vertical loads the each point ...


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Here is a two slit calculator / simulator I designed in Excel. https://www.dropbox.com/s/kkopfv4xbratc9r/Alsept%20%202%20Slit%20Calculator.xlsx?dl=0 Download,save, open, enable editing The green cells are where you input distance, wavelength, slit width and separation.The yellow cells give you the fringe pattern spacing. The graph represents the right half ...


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Calculate the Fraunhofer diffraction integral of your slits and take the absolute square to get intensities.


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The vector sum of the forces will be zero: $$\Sigma \vec{F} = 0$$ If you can assume that the bottom of the block is planer and the weight acts perpendicular to that plane, (i.e., the lengths of the legs are identical) things are much simpler. The weight acts downward and the forces from the legs act upward. The sum of the torques (moments) about any point ...


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A lot of these problems are best done by first redrawing the circuit so that it is in a more accessible form. The correct answer is $\frac 8 3 \;\mu$F.


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Comment and Hint: The following figure may help, as long as you maintain all the connections this is the same circuit (will let you identify the capacitors). I get $8/3 \mu F$


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You can solve this strictly with adding capacitance in parallel and elastance in series. The 2uF is in parallel with the others. The two 1uF on the left are in parallel and the 1uF on the right is in series with the two on the left.


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With two lamps in series you will get double overall resistance, thus of Ohm's law overall current is I = U / (2R) that means overall current gets half of one-lamp case with I = U / R. As you said, this current is the same in the both lamps and voltage on each of the lamps gets half of the overall. In parallel you have two lamps attached to one the same ...


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If the pipe is cylindrical, there is no equilibrium height. All the air that goes in at the bottom must come out at the top, so the force balance will not depend on the height of the ball. In a ball flow meter (rotameter), the pipe is slightly conical, so that the gap between the ball and the pipe increases as the ball rises. If you want a very coarse ...


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If the strings have length d=0.08 m, then the distance from the center to each mass is $r=d/\sqrt{2}=0.056$ m, which you have noted yourself. The centripetal force is $F_c=m\omega^2r=0.678$ N. But the centripetal force vector is the sum of the tensile force vectors of the two strings, each of which is at 45 deg. Therefore, you have to divide $F_c$ by $\sqrt ...


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The answer is that it depends on the field on the outside (boundary conditions) and the dielectric constant of the insulator. For a imaginary insulating sphere of vacuum in a vacuum, it should be obvious that the sphere does not affect the electric field at all. Inside a dielectric, the field will be weaker than on the outside. For a dielectric sphere in a ...


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The component that ensures the current is zero just after the switch is closed is the inductor. Inductors do not like changes in current, since a change in current means the magnetic field linking the inductor is changing and this generates a back emf that opposes the change. If you replace the inductor with a piece of wire you would have an RC circuit and ...


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Setting up the differential equation $$L \frac {dI}{dt} + RI + \frac {Q}{C} = V$$ will not necessarily answer your question, "What and how can I conclude about the current in this circuit just after switch is closed." If you look at the methods of solving the differential equation somewhere on the way to the solution initial conditions are needed, one of ...


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Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? As there are no external forces the linear momentum of the system will be constant of motion and the constant can be taken as zero as well. If m1 and m2 are the masses at any time described by position ...


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To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses $x_1$ and $x_2$. The spring force (tension is positive) is $$ F = k (x_2-x_1 ) $$ and the two equations of motion $$\begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} $$ All this is trivial. ...


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It looks like a dubious exercise question, unless the pages preceding the exercise set up a framework of what kind of things you're allowed to assume here. Both the entrance and exit (speaker/ear) represent discontinuities in the acoustic impedance of the tube. The impededance of the speaker (will it absorb reflected acoustic energy or not?) is not described ...


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Assuming the spring is "ideal" (massless) you actually have 2 masses. You can describe your problem as the motion of the center of mass, and either of the masses. And if no external force is exerted on your system, you are only left with the motion of 1 mass relative to the center of the mass of the system. Let's say your masses are m1, m2, and the spring ...


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The speed of sound, which is effectively a measure of how fast particles are moving on average in a gas, is given by: $$ C_{s}^{2} = \frac{\partial P}{\partial \rho} \tag{1} $$ where $P$ is the pressure and $\rho$ is the mass density. For an ideal gas, one can use an adiabatic equation of state such that Equation 1 goes to: $$ C_{s}^{2} = \frac{ \gamma \ P ...


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The claims aren't true in complete generality. For example, $R^{2n}$ is a Kähler manifold and any vector $k$ is a Killing vector. But the Lie derivatives of $J$ and $\omega$ don't vanish for a general $k$. However, this example was special because it was insufficiently curved. For generic and curved enough Kähler manifold, the objects $J,\omega$ may be ...


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The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$ y = v_0t - \frac{1}{2}at^2 $$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...


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Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).


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For each wavelength you have been given (exactly) two values $\theta$, so $$\begin{align}n\lambda &= d\sin\theta_1\\ (n+1)\lambda &= d\sin\theta_2\end{align}$$ subtracting these two equations, we get $$\lambda = d\left(\sin\theta_2-\sin\theta_1\right)$$ You should be able to figure it out from there...


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From your diagram you have $x = y \tan \theta$. Taking the time derivative, and using the speed of the plane is $\dot{x} = v$ (taken as constant) you have $v = y \, \sec^2 \theta \, \dot{\theta}$ assuming that the height of the plane, $y$, is constant. So, for constant height (and taking the speed constant) you have $\dot{\theta} = C \cos^2 \theta$ where ...


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I'm too lazy to do the detailed math, but it's clear that, for constant velocity and constant rate of climb, it is possible to distinguish between a level path and a rising or falling path. Take point A as the intersection point of a level path and a rising path. At some time the aircraft both occupied A. Point B is any aircraft location on the level path ...


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Looking around, the root mean square speed of air at $20$ C is about $500 m/s$, and given that you have $\langle v^2 \rangle \propto \, T$ so that $v_{rms}(T) = \sqrt{\langle v^2\rangle}$ varies with $\sqrt{T}$ then have $$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$ and $$v_{rms}(25) = v_{rms}(20)\times ...


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The four vector $p$ is given as $(E,{\bf p})$ so that $$(p' -p)^2 = (E' - E)^2 - ({\bf p'} - {\bf p})^2 = (E'^2 - {\bf p'}^2) + (E^2 - {\bf p}^2)- 2E'E + 2 {\bf p'} \cdot {\bf p}$$ Then put $E'E \approx (m + \frac{{\bf p'}^2}{2m})(m + \frac{{\bf p}^2}{2m}) \approx 2m^2 + {\bf p'}^2 + {\bf p}^2$ (ignoring the term $\frac{{\bf p'}^2{\bf p}^2}{4 m^2}$). Using ...


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If the connection you are using is torsion-free (as the Levi-Civita one), you can systematically replace the coordinate derivative for the covariant one. So (please, pay attention to signs and positions of indexes, since I could use a convention different from yours) $${\cal L}_\xi \nabla_a K^b = \xi^c\nabla_c \nabla_a K^b - (\nabla_c\xi^b) \nabla_a K^c + ...


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It might be useful to know the formula for Lie derivative of metric itself. It is $\mathcal{L}_Xg_{ab}=\nabla_a X_b+\nabla_b X_a$. Then you might notice that covariant derivative can be split into symmetric and atisymmetric part: $$\nabla _a K_b = \nabla _{(a} K_{b)}+\nabla _{[a} K_{b]}$$ The first part is half of lie derivative of metric with respect to ...


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A slightly different and, perhaps, more simple proof follows. $\def\lie{\mathit{£}}$ For $K^a$ a Killing vector, we have (Kostant formula), $$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$ Using this, we may prove that the covariant and the Lie derivatives commute: $$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary $X^a$. We have: ...



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