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14

The simplest formula for the centrifugal acceleration is $$a = r\omega^2$$ Here, $r$ is the radius which is 0.25 meters in your case. $\omega$ is the angular velocity which is $2\pi$ times the frequency $f$. Your $f$ is 1500 revolutions per minute which is $1500/60=25$ revolutions per second. In the SI units, we have $$a = 0.25\times 4\pi^2 \times 25^2 = ... 4 Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation$$\frac{dx}{ds}= X(x(s))\tag{1}$$with a generic initial condition z at s=0 in the manifold M (Minkowski spacetime in your example). X is your vector field on ... 3 Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle \theta of a light ray in the ... 3 The ladder operators satisfy: \bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1>\bf{a}$$|n>=\sqrt{n}|n-1> Taking into account <n|m>=\delta_{n,m} , you get the answer. 3 Take a trace of Einstein equations (trace of g_{\mu \nu} is D), you obtain$$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}$$3 You meant the angular velocity vector \vec{\omega}, I think. But adding the velocity vector \vec{v} to the angular velocity vector \vec{\omega} would be like adding apples to oranges. Look even at the dimensions of the scalars of these vectors, for velocity that is \mathrm{m/s}, for angular momentum it is \mathrm{s^{-1}} (angles have no dimension ... 2 If you write the current as a function of time, I(t), then the root mean square current is:$$ I_\text{RMS}^2 = \frac{1}{\tau}\int_0^\tau I^2(t)dt $$where \tau is the period of the waveform. In this case I is always \pm 2 so I^2 is always 4 and the integral becomes:$$ I_\text{RMS}^2 = \frac{1}{\tau}\int_0^\tau 4dt = \frac{1}{\tau} 4\tau = 4 ...

2

I'm guessing you're misinterpreting this diagram: as meaning $\vec \omega$ is "along the plane of motion". That's not what the curved arrow is meant to denote, it is meant to denote the rotation around $\vec \omega$. The angular velocity $\vec \omega$ itself, since $\vec \omega\propto\vec r \times \vec v$, is always orthogonal to $\vec r$ and $\vec p$ and ...

2

It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only. What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to $$\gamma^0 S^\dagger \gamma^0 = S^{-1}$$ because $( \gamma^0 )^2 = 1$. Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} ... 2 It gets easier if you use the result from part 1. Then you also don't have to deal with the$\mathcal O(\omega^2)$(see my answer to your other question). In your calculation, you transformed$\bar\psi$and$\psi$, but not$\gamma^\lambda$. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: ... 1 The confusion lies with the definition of the transition probability. The transition amplitude between an$H_0$eigenstate$H_0 \left|m\right> = E_m \left|m\right>$and another eigenstate$H_0 \left|n\right> = E_n \left|n\right>$due to a perturbation$V$after a time$tis given by \begin{align} \left< n \right| U(t) \left| m \right> ... 1 The key is this: how is the force applied to the rider? Gravity pulls directly down on the passenger's mass, but what keeps the passenger from heading towards the center of the Earth? When you're in the passenger seat of a sports car, and the driver floors it, you feel pushed back into the seat. But what's actually happening is that the seat is pushing you ... 1 The rider feels "forced down" because the object to which they are attached is accelerating upwards. Because the acceleration is opposite to gravity, the normal force,\mathbf{F}_{N}$, being exerted on the rider must increase in magnitude (relative to the "at rest" magnitude on a horizontal track) in order to produce a net upwards acceleration. Thus, it ... 1 First of all, there are a few problems with your question:$J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b$is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of$\epsilon$is a bit weird, too. What you mean is $$J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ... 1 The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both m and the spring is characterized by constant k the angular frequency of ... 1 The flask become most stable when its centre of gravity is at the smallest height. If you start pouring water, you will notice that the effective centre of gravity gets down to a lower postion. As you keep on filling, it would be at the lowest height for some level of water and rises again, afterwards. You will have to find that point of minimum height. Just ... 1 Hint: Use$$m\ddot{x}=-kx-x^3 \\\ddot{x}=v\frac{dv}{dx} \\-\frac{kx^2}{2}-\frac{ax^4}{4}=\frac{m}{2}\left(\frac{dx}{dt}\right)^2$$It will reduce to a form$$\frac{dx}{dt}=ix\sqrt{c^2+x^2}$$This is a standard integral, and can be solved, then use$$U=-\int f(x) dx \\T=\frac{1}{2}m\dot{x}^2$$Total energy E=T+U\; . 1 Goutham is quite correct in some ways but overlooks something. Look at the diagram below: We known the centre of gravity (COG) of the empty cylinder is z_1=10\:\mathrm{cm} and the mass of the empty cylinder is 100\:\mathrm{g}. If we fill the cylinder up with water to height 2z_2 then the COG of the water is z_2 and the mass of water is (assuming ... 1 If you accelerate your car with constant acceleration AND we then assume that the friction decelerates the car with constant (negative) acceleration, then you simply consider each situation for itself:$$x_{acc}=x_{0,acc}+v_{0,acc}t_{acc}+\frac12a_{acc}t_{acc}^2=\frac12a_{acc}t_{acc}^2$$... 1 Inserting the expansion$$ \psi=\int\frac{d^3p}{(2\pi)^32\omega_p}(a_pe^{-ipx}+b_p^\dagger e^{ipx}) $$into the expression for the Hamiltonian$$ H=\int d^3x(\dot{\psi}^\dagger\dot{\psi}+\nabla\psi^\dagger\cdot\nabla\psi+m^2\psi^\dagger\psi) $$we get$$ H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 1 By using work energy theorem it can be solved. The velocity of the 2kg object till it reaches the 6kg object is given by $$\sqrt{2*g*5}=9.90m/s^2$$ apply the conservation of momentum for plastic impact. $$m_1u_1+m_2u_2=m_1m_2V$$ $$2*9.90+6*0=8*V$$ $$V=2.475m/s^2$$ work energy theorem $$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*x$$ $$x=1.801472656m$$ ... 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 ...

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