# Tag Info

5

When two quantities of water ($m_1$ and $m_2$) at different temperatures (resp. $T_1$ and $T_2$) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature $T$ of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ...

3

Take a trace of Einstein equations (trace of $g_{\mu \nu}$ is $D$), you obtain $$R - \frac{D}{2} R + D \Lambda = 0$$ Or $$R=\frac{D \Lambda}{D/2-1}$$ Then substitute this expression for $R$ into full Einstein equations and you obtain trivially $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

3

Using Lagrange's equation $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{-m r^2}{l^2}F\left(r\right)$$ And plugging in $$F\left(r\right)=\frac{-k}{r^3}$$ We get $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{k m}{l^2 r}$$ Which simplifies to $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}\left(1-\frac{k ... 3 Yes. If \bf B \times V=C and {\bf B \cdot V}=\lambda, the BAC-CAB rule tells you: \bf B\times C=B\times(B\times V)=B(B\cdot V)-V(B\cdot B) So {\bf B\times C}={\bf B}\lambda-{\bf V}B^2 and$${\bf V}=\frac{{\bf B}\lambda-{\bf B\times C}}{B^2}2 We know from the fact that the wood block is floating that it is less dense than water. I will assume the coin is more dense than water. I will also assume that we are considering steady state after any waves have died away. I will also assume that the wooden block remains in the same orientation as before. I will also assume the block is cubiod. I will also ... 2 He uses that the action is dimensionless so that \begin{align} [ d^d x \left(\partial\varphi\right)^2] &= 0 \\ &=[d^d x]+2[\partial\varphi]\\ &= -d +2 + 2[\varphi] \end{align} using that [dx]=-1 and [\partial\varphi] = [\partial] + [\varphi] = 1+[\varphi]. This gives [\varphi] = (d-2)/2 1 Use the equation of Lorentz force to calculate the field vector. Quoting from this link, If a particle of charge q moves with velocity v in the presence of an electric field E and a magnetic field B, then it will experience a force\mathbf{F} = q\left[\mathbf{E} + (\mathbf{v} \times \mathbf{B})\right]$$In your case, there is no electric ... 1 Not always, but it in almost all cases you can. To find the conditions whereunder it can be done and the inversion formula, reason as below where I give a more "visual" analysis than the other answers. It's also worth mentioning that the consideration of your question leads to one conception of the skew field \mathbb{H} of quaternions (which I'll say more ... 1 since you are concerned about a long solenoid, this problem has a very simple solution. Suppose you have two identical long solenoids, each of them having magnetic field B at the ends. You join them end to end, such that their magnetic moments are in same direction. Thus, at the junction the magnetic field adds up to B+B=2B. But this junction is ... 1 So if the path is described parametrically with \vec{r}(q) = (x(q),y(q)) you need to define the tangent and normal vectors in order to describe the motion \ddot{q} and the reaction force F. The kinematic velocity vector is$$\vec{v}(q,\dot{q}) = (\dot{x},\dot{y}) = ( \frac{{\rm d}x}{{\rm d}q} \dot{q}, \frac{{\rm d}y}{{\rm d}q} \dot{q} ) = (x' ...

1

It is correct except for one sign, note that the work done by friction is negative (since you move the block in the opposite direction w.r.t. the friction force) and thus it is equal to $$-\mu mg d \cos \theta$$ with this solving for $v$ gives you 11.49 m/s.

1

As the ball swings downward, its gravitational potential energy is converted to kinetic energy. At the bottom of the swing, its velocity will all be in the horizontal direction. From this, you can calculate the velocity with which the ball strikes the block. In a perfectly elastic collision, both kinetic energy and momentum are conserved. From this you can ...

1

Remember Lenz's Law: as you change the flux through a coil, an e.m.f. is generated that opposes this change. Therefore, if I have two coils that are a certain distance apart, they will have a certain "shared" flux - flux due to $A$ appearing in coil $B$, for example. Now if we bring $A$ closer to $B$, we change the flux in $B$ due to $A$, and will get a ...

1

Let the wavefunction be $$y(x,t)= A\sin\left[\frac{2\pi}{\lambda}(x- vt)\right].$$ Now,$$\frac{\partial y}{\partial x}= \text{Rate of change of wavefunction when time is constant}\;,\\ \frac{\partial y}{\partial t}=\text{Rate of change of wavefunction when position is constant or transverse velocity}\; .$$ Now differentiate $y$ w.r.t. $x$ keeping $t$ ...

1

You should also match the derivatives at $x=0$ so that they took into account the $\delta$-function. If you take smooth well $V_\epsilon(x)$ and consider small region near zero $(-\epsilon,\epsilon)$ where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ...

1

"The frequency doesn't change" is only true when the core is perfectly linear. For a real transformer, there will be some nonlinear effects (saturation) meaning that the sinusoidal input waveform will create harmonics in the output - second harmonics and higher frequencies will appear. But if you ignore those, then the flux change will vary sinusoidally at ...

1

A short pendulum with a large rigid bob will run slower than expected, because the force required to rotate the bob is added to the force required to swing the bob, reducing the accelerations. As the pendulum length becomes "long" relative to the size of the bob, then this effect becomes "small" and you can better and better approximate the bob as a point.

1

You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ...

1

using $$a^{\dagger} |n\rangle = \sqrt{n+1}| n+1 \rangle$$ and $$a |n\rangle = \sqrt{n} |n-1\rangle,$$ apply these rules in order: $$a(a^{\dagger}|n\rangle) = a\sqrt{n+1}|n+1\rangle = \sqrt{n+1}a|n+1\rangle = (\sqrt{n+1})^2 |n\rangle = (n+1)|n\rangle$$

1

Newton's law of cooling is a corollary of Fourier's law of heat conduction: $$q=-\kappa \nabla T,$$ where $q$ is the heat flux, $\kappa$ the heat conductivity and $\nabla T$ the temperature gradient (in a single dimension $\nabla T=\frac{dT}{dx}$). In essence this law tells us that heat flows from hot to cold and that the heat flow is proportional to the ...

1

This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance $h$ below and moves down at a constant $6$ m/s. The bolt catches up in $3$ seconds. Same problem, but the equations are simpler...

1

The summation indices on $Z$ must match the summation indices on $\bar{E}$, because $Z$ is the normalization constant for the total probabilities of being in every state. The $n=0$ state has zero wavenumber and doesn't exist. You're right to discard the analogy with para hydrogen. In that case, the peak is caused by interaction of the nuclear spins of the ...

1

Just solve the second order differential equation obtained from using Newton's Laws i.e. $$F=-\frac{k}{r^3}$$ or $$m\frac{d^2r}{dt^2}=-\frac{k}{r^3}$$ If you solve this differential equation, then your equation for the path will be of the radius as a function of time. The equation will be a non-central conic. HINT (TO SOLVE THE DIFFERENTIAL EQUATION): ...

1

The issue here is whether or not you have a sum over $l$. If you want $r_l^2$ to mean any of $r_1^2, r_2^2, r_3^2$, then when you write $r_l^2 = r_l r_l$ you should not be summing over $l$. So $[r_l^2,L_i] = 2i\hbar \epsilon_{ijl}r_jr_l$ is correct as long as you sum over $j$ but not over $l$. On the other hand, if in $r_lr_l$ you sum over $l$ you get the ...

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