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4

This is anwered in Gravity on the International Space Station - General Relativity perspective, where we learn that time dilation in the ISS with respect to Earth equator is 1.00000000028655. So after 17 years for us, the astronauts would come back younger by about 0.15 seconds than if they have stayed on the ground. Note that a full GR treatment is ...


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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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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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Yes. Consider rearranging the equation as $$ \nabla (p+\rho E) = 0 $$ This implies that all the partial derivatives of $(p+\rho E)$ vanish at every point, and thus $(p+\rho E)$ is a constant.


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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 ...


3

From Kirchhoff's second law, the sum of all the voltages around a loop is equal to zero. That is, the sum of the voltages across the three elements of your circuit, R, L and C, must be equal to the time varying voltage from the source: $$V_R+V_L+V_C = V(t)$$ As $V_R=RI$, $V_L=L\frac{dI}{dt}$ and $V_C=\frac{Q}{C}$, we get your equation, which is correct: ...


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So, we have series LCR circuit. $V$ is a constant voltage source. $L$, $C$, and $R$ represents the inductance, capacitance and resistance in the circuit respectively. A current $I$ flows through the circuit. Now, the current through each component is the same. So, the potential difference between each component added up together gives the emf $V$. ...


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Observe that by chain rule, we have that $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=\frac{dv}{dx}\cdot v=v\frac{dv}{dx}$$


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It's not true. For example take $$ A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \qquad\qquad\qquad B = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} $$ you have $A^2=0$ so that $[A^2,B]=0$, but $$ [A,B] = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. $$ You must add some condition, for example if you know also $$ [A,[A,B]]=0 $$ ...


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(Source of image: Mohsin Khan, http://cslearners.blogspot.com/2009/08/equation-of-motion.html) Here they are! All the formulas. Sorry to say, you cannot find anything if you have only acceleration and distance. Think like this, Say you have an object that has an acceleration of 2 m/s^2. and if i say that if travels a distance of 2 meters. It can travel ...


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If you draw a velocity-time graph you will see that you do not have enough information to find the initial and final velocity. The gradient of the graph is fixed because it is the acceleration. The distance is the area under the velocity-time graph. As you will see from the graph you can draw an infinite number of trapeziums (or triangles) which satisfy ...


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The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer. It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you ...


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Let's recall why eigenvectors of Hermitian operators with different eigenvalues are orthogonal. If $A\vert\lambda_i\rangle=\lambda_i\vert \lambda_i\rangle$, then we have: $$\langle \lambda_1\vert \lambda_2\rangle=\frac{1}{\lambda_1}\langle\lambda_1\vert A\vert\lambda_2\rangle=\frac{1}{\lambda_2}\langle\lambda_1\vert A\vert \lambda_2\rangle,$$ so if ...


2

Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...


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When the wheel comes in contact with the belt friction will act as there will be relative motion between the belt and the point of contact. Now friction will tend to act on the belt opposite to the velocity of the belt until slipping ceases. To find the work done by the external agency lets consider the energy changes : 1) The K.E of the wheel increases. 2) ...


2

Hamiltonian H is indeed self-adjoint and its eigenvectors must be orthogonal. The reason why a subset of the "eigenbasis" appears to be non-orthogonal has to do with the particular structure assumed for the eigenvectors and with their degeneracy. The very short answer is that their two components carry the wrong normalization. But it is more interesting to ...


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This is a surprisingly difficult question to answer properly, but if you read the report from the Ricardo group (thanks @pentane for the reference!) it includes the following table: They do a lot of analysis to show that if you reduce the weight of a car, you can get away with a smaller (more efficient) engine; but I am assuming that you just want the ...


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$\nabla \cdot \vec E(r) = \dfrac {1}{r^2} \dfrac {d(r^2 E)}{dr} \ne \dfrac{dE}{dr}$ in spherical coordinates.


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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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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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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 ...


1

If the Fourier transform of $f(x)$ is $\tilde{f}(k)$, then the Fourier transform of $df/dx$ is $ik\tilde{f}(k)$. Proof: $$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$ This explains the momentum factors, so we've reduced the task to showing $$\Gamma(x, y, z) \sim ...


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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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Hint : A key to the solution is what is meant by the complex wave 3-vector $\:\mathbf{k}\:$. This vector is not any complex 3-vector in $\: \mathbb{C}^{3}\:$ $$ \mathbf{k} \ne \left(k_{1}, k_{2}, k_{3} \right) \in \mathbb{C}^{3}, \:\:\text{that is with} \:\: k_{\rho} \in \mathbb{C} \tag{a-01} $$ but $$ \mathbf{k}=\left(k_{1}, k_{2}, k_{3} \right) ...


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The question isn't entirely clear, but I suspect that you're being asked to prove that the given parametric equation for $R(\theta)$ and $t(\theta)$ satisfies the Friedman equation. If so, you shouldn't try to get rid of $\theta$ entirely. Instead, show that the left-hand side (as a function of $\theta$) is equal to the right-hand side (as a function of ...


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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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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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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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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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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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