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2

Say the time evolution for Hamiltonian $H$ is given by $U(t) = \exp(-iHt)$ and the corresponding evolution on the support of $P$ is $$PU(t)P = P\exp(-iHt)P = \exp(-iH_\text{eff}t) \equiv U_\text{eff}(t)$$ assuming $H_\text{eff}$ exists. The desired identity follows from $$ \lim_{\eta \rightarrow 0} \int_0^\infty dt \; U(t) e^{i (\epsilon + i \eta) t} = ...


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Consider the angle between the two vectors as $\theta$ and the following rules $$\begin{align} \vec{a}\cdot\vec{b} &= \|\vec{a}\| \|\vec{b}\| \cos\theta & \|\vec{a}\times\vec{b}\| & = \|\vec{a}\| \|\vec{b}\| \sin\theta \end{align} $$ Now to construct the parallel vector use the direction of $\vec{b}$ and the adjacent side of the triangle ...


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Lets to this step by step and take care of the signs! Let $q_1=3$µC, $q_2=5$µC and $q_3=-8$µC. The formula for the force, acting on particle one due two the presence of particle two, is given by $$\vec{F}_{12}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{{| \vec{r}_{21}|}^2} {\hat{r}}_{21},$$ where $\hat{r}_{21}$ is the unit vector pointing from charge two to ...


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At the risk of sounding like a broken record... it is a good idea to draw a diagram for all but the very simplest problems (and even then): You can immediately see that the normal force is made up of two components: $F_c \sin\alpha$ and $F_g\cos\alpha$. The friction results from the combination of both of these. In your approach, you ignore the normal ...


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Yes. One way to see this: if $S'$ and $S$ have coordinates $x'$ and $x$, then by the usual rule we know that $S'$ observes a distance of $\Delta x = x-x'$ between them. Differentiating on both sides, we get $\Delta v = v - v'$, $\Delta a = a - a'$, and so on. In other words, velocity, acceleration, and all higher derivatives behave like you think they ...


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The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...


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I'm going to use Einstein summation notation throughout. You're almost there. You just need to use $$ {\bf B} = \nabla \times {\bf A} $$ or, equivalently, $$ \begin{eqnarray} B_k &=& \frac{\partial}{\partial r_i} A_j \epsilon_{i j k} \\ &=& \frac{1}{2}\left(\frac{\partial}{\partial r_i} A_j \epsilon_{i j k} + \frac{\partial}{\partial r_j} ...


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yep, think of $ \xi $ as a unit vector and replace all instances of it with $ \epsilon \xi$ where $ \epsilon $ is some small number. Then you will see that those two terms are second order in $ \epsilon $.


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When you calculate the "parallel" vector, you should not use the dot product of $a\cdot b$ but instead the normalized dot product $$\frac{a\cdot b}{|b|}$$ times the unit vector $b$. The projection of $a$ onto $b$ should always be independent of the length of $b$.


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The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is ...


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Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates $(t,x)$, and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one. Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of ...


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The Laplace equation $\nabla^2 \psi = 0$ is a linear differential equation. Now note that if $\phi$ is real, then so is $\nabla^2 \phi$. Moreover, by the linearity of the equation, if $\phi$ is real, then $i\phi$ is pure imaginary, and so is $\nabla^2(i\phi) = i \nabla^2(\phi)$. Okay, back to your situation. Let's say the solution is $\phi_1 + i\phi_2$ for ...


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First you half fill the pipette with hot water. So at that point it's filled half with hot water in the lower part, half filled with air at room temperature in the upper part. Then you block the top end of the pipette with the tip of your finger and turn the pipette upside down, so the hot water starts flowing down to the blocked side. However, the presence ...


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for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


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First you need to define the orientation of the cube relative to the axis you want to measure. Typically a 3×3 rotation matrix $E$ does the job transforming local coordinates along the principal axes to the world coordinates. The use the transformation $E I_{body} E^\intercal$ Example: A single rotation $\theta$ about the world $z$ axis is $$E = ...


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The ball was in flight for four seconds: we can safely say that the ball reached maximum height at $t = 2$. (The gravitational pull is constant and there are no other forces acting, so the flight path is symmetrical). The ball was stationary at $t=2$ so its speed is $=0$ So now use the formula $v= u + at$, where $a$= acceleration, $t$= time, $u$= initial ...


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Molecules are not that large. The density of air at 1 Atm and 20°C is $2.5\times 10^{25}$ m$^{-3}$. So, the average spacing, using your teacher's method, is about $(2.5\times 10^{25})^{-1/3} = 3.4$ nm. The radius of molecular nitrogen is 0.2 nm. So, the diameter is about 0.4 nm. The means that only about $(0.4/3.4)^3 = 0.0016$ of the volume of air is ...


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How to get from $\left(\frac{-\hbar^2}{2m_e} \Delta_{r_e} + \frac{-\hbar^2}{2M_P} \Delta_{r_p} +V(r) \right)\Psi(\vec r_e,\vec r_p) = E \Psi(\vec r_e,\vec r_p)$ to: $\left(\frac{-\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} + \frac{-\hbar^2}{2\mu} \Delta_{r} +V(r) \right)\Psi(\vec r,\vec R) = E \Psi(\vec r,\vec R)$ with: $\vec R=\frac{m_e \overrightarrow{r_e} + ...


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Maybe a little story will help. In a country far, far away, and a long long time ago, the king died in battle. His generals sent messengers to the castle. One messenger rode the swiftest horse in the country called Lightning; it was so fast, it arrived almost as soon as it left, as it was able to cover 300 km in one millisecond. The second messenger rode a ...


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So you have that $\vec{B} = \frac{a}{r^2}\begin{bmatrix}0 & z & -y\end{bmatrix}$ thus the magnitude is $B = \frac{a}{r^2}\sqrt{z^2+(-y)^2},$ where $a$ is unknown. Can you write that as a function of $r?$ Can you investigate what happens as $r$ goes to zero? Are magnetic fields continuous in empty space (a vacuum)? If so, try the next five: What ...


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The magnitude of the B-field is $a/r$ and circulates around the axis. By symmetry, you understand that the magnitude is zero on-axis. But if $a$ is anything but zero, your expression gives an infinite B-field magnitude. Therefore $a$ must be zero and therefore the B-field is also zero everywhere else inside the pipe. The result also follows from Ampere's ...


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Any Bipolar junction transistor (BJT) constitutes : a heavily doped emitter a lightly doped and small size base a moderately doped collector So according to your question, the emitter should have been doped to a concentration of 10^23/cm^3, the base doped to 10^16/cm^3 and the collector doped to a concentration of 10^18/cm^3. Now your assumption about ...


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Yes, you're correct. Since $S'$ is an non-inertial/accelerated frame of reference, all objects within this frame is acted upon by a pseudo force that is proportional to the mass of the object and whose direction is opposite to the direction of acceleration of $S'$. The fact that $S$ is accelerating(with respect to a rest frame on ground) with $a=5 m/s^2$ ...



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