# Tag Info

16

Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = $6.4 \times 10^6~\text{m}$ mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$ Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$ This gives a value for ...

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 $$a = 0.25\times 4\pi^2 \times 25^2 = ... 10 Hey! The question keeps getting edited! Make up your mind! You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not? What if Earth had the mass of Jupiter? More ... 4 Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation$$\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 ... 4 The Hamiltonian H(\theta,p_\theta) needs to be formulated in terms of the coordinate \theta and its canonically conjugate momentum p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ... 4 Actually you can go to the orbit of Jupiter with a \approx 2500 tonnes rocket and a 3 tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at 12.6~\text{km}/\text{s} needs just an increase in speed v = 29.5~\text{km}/{\text{s}}.v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := ...

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

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 ...

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 $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

2

you have to notice that this motion is accelerated, so if you define velocity as $d/t$ you will get the wrong result. For uniformly accelerated motion (which is your case, the acceleration is contant: g), you have to use the following relationship: $y(t)=y_0+v_0t+\frac{1}{2}at^2$ When you release the ball, $y_0=H$, $v_0=0$ and $a=-g$ so you get $... 2 As you correctly note, you need to prove$t_B > t_A$is preserved by Lorentz transformations. It's not clear to me whether your final answer demonstrates this (since the sign of$x_b - x_a$isn't totally obvious), but I think you're on the right track. I might have responded with the following argument. The invariant interval $$ds^2 = -dt^2 + dx^2 ... 2 If you write the current as a function of time, I(t), then the root mean square current is:$$ 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 $t$ is 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

We'll give some details based on Andrew's comments. We are trying to solve $$\frac{ \partial \Delta \rho(t) }{ \partial t } = -i\mathcal L_0 \Delta \rho(t) -\{A, \rho_0(t) \} \, \mathcal F(t). \qquad (1)$$ Homogeneous solution To develop some familiarity with the Liouville operator, let us first consider the special case of $A = 0$. Then we have a ...

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

"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

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 ...

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