New answers tagged

-1

Note- Assuming pulleys and strings are massless


1

So if I were reading this, I would interpret this as saying that $m_2$ lifts upwards right as $m_1$ is at the bottom of its swing, rather than right as it is released. When it is released, it has no centrifugal force to back up its weight in tensioning the string and thus lift up the other mass: that comes as it starts moving faster. Basics you should know ...


0

You should draw a free-body diagram for the swinging mass, $m_1$. Then you will see that the tension in the rope is a function of $\theta$ if the mass is released at some angle $\theta_0 > 0$. You should be very careful to note that the radial acceleration is not zero, but is a function of speed. You can calculate the speed of $m_1$ as a function of angle,...


0

Your setup is correct, as you are tensoring/entangling eigenstates of $L_z$ and such of $S_z$, and your expectation values are sound. (This is not quite the notation suggested in the referent question. There, you start with a two-spinor, but, implicitly, and confusingly, you tensor it with infinite-dimensional kets you never write down, which you then dot ...


0

What do you mean by "at the moment of dropping?" Is it when $m_1$ swings down, and then $m_2$ is released to move freely?


0

Assuming there is no force of friction from the rod, then your basic equation is F = ma. In this case: $$(m/20)[x – (20-x)]g = m\frac{{d^2}x}{{dt^2}}$$ Where, $m$, is the total mass and, $x$, is the length on the heavier side (which goes from $12$ to $20$ with, $t$, starting at zero.


-2

just think that when would the tension be maximum in string and choose appropriate $\theta$ to utilize that maximum tension wisely. You can ask more explanation but in no way I am going to provide you the solution to it EDIT- So as you have asked for more so let us start it step by step Now first of all if an object is swinging ignoring other effects(even ...


0

There are two errors in your reasoning. First, the initial acceleration of the rope is not $g$, and second the acceleration of the rope is not constant. Suppose the rope has a mass $\rho$ per metre. Then the total mass of the rope is $20 \rho$. But initially the weight of the rope on one side of the bar is $8\rho g$ and the weight of the rope on the other ...


1

So I made a little diagram to work out what is positive or not Where $+x$ is to the right, $+y$ is into the plane, and $+z$ is up. This is a right-hand coordinate system. The ball moves along $+x$ and also rotates about $+y$. Is friction, as shown, acting along $-x$ on the ball, and $+x$ on the ground? Or is the other way around? Friction opposes relative ...


2

Dirichlet is not enough for your operator to be self-adjoint, or even Hermitian. Indeed, be careful! "Hermitian" is probably not what you want. There is a distinction between mere hermiticity and self-adjointness. It is the latter condition that is needed for an operator to have a complete set of eigenvalues and hence be a QM observable. There is ...


2

Let's break down each term: $\phi^3$ means that each interaction vertex has three lines attached to it 2-point means the diagram has two external lines 1-loop means that any other lines need to be contracted to form a single loop Now for these conditions, as you can verify, there are two possible diagrams [source: physicspages.com, Lancaster]: But the ...


0

The following is a Hint: Canonical transform generated by $F_2(q,Q,t)$ give $$p=\frac{\partial F_2}{\partial q} \ \\ \ \ \ \\ P=-\frac{\partial F_2}{\partial Q}$$ Consider a legendre transform to $\Phi(q,p,t)$ from $F_2(q,Q,t)$ $$F_2(q,Q,t)=\Phi(q,p,t)+Qp$$ That's all. From here Take the total derivative of both hand sides and use the fact that $Q=Q(q,p,t)$ ...


0

Only for interest: How old are you? To convert units where is an easy method. Note the conversion factor for each unit. For example from this page https://www.convertunits.com/from/newton/to/OZ 1 N = 3,60 OZ (1) 1 mm = 0,04 in (2) Put in (1) and (2) $1\frac{OZ}{in} = \frac{1N}{3.60 OZ} * \frac{0.04 in}{mm} = \frac{0.04}{3.60}\frac{N}{mm}$


0

As the other commenter says, you can use the Euler-Lagrange equations, but you might find the presence of the square root a little awkward. With a reparameterization, and being careful with indices, one can manipulate the EL equations into the form $$ \frac{d}{d \tau} \Big(\frac{\partial L^2}{\partial \dot{x}^{\alpha}}\Big) - \frac{\partial L^2}{\partial x^{\...


0

You do not need to perform the integral; use the Euler-Lagrange equations. These only need the Lagrangian which is the integrand.


0

You have used the formulas correctly but did a silly mistake.


0

Here, Sin 30° = 12/AB => AB = 24 We will use work energy theorem to solve this question. Work done by all forces = Change in kinetic energy of the block. Work (mg) + Work (Friction) = ∆K.E. mgh - f(AB) = 1/2m(v^2 - v^2) 81012 - f*24 = 1/2 * 8 * (12^2 - 6^2) f = 22N Note : Here friction does negative work as direction of motion is opposite to friction ...


1

If the light takes longer to go from A to P than from B to P, then that introduces a lag in phase for the source A. Think about how the phase of a wave changes as it propagates through space. For every $\lambda$ a wave travels, its phase will advance by $360^\circ$. A is ahead of B in phase by $37^\circ$, but lags by $30^\circ$ due to the path difference. ...


-1

Assuming you don't care about wasting propellant, that's the way to go. If we take a frame with the $z-$axis in the direction of the velocity vector of your spaceship, you have to ensure that the sum of the thrust applied is zero along the $x$ and $y$ axis (otherwise your ship will not only decelerate but also that deviating from its trajectory). If this is ...


-1

It doesn't matter only about the amount of air in the balloon... think about it this way...think about some action that shifts the balance (like for example throwing a ball up into the air). When your doing that you sort of shift the balance because when the ball was on the ground it was balanced - all the forces balanced out. But when you throw it, the ...


-1

When solving a non-homogeneous second-order ordinary differential equation, there are two solutions, one known as the solution to the homogeneous part $x_h(t)$(which contains some arbitrary constants that can be set by the initial conditions), and one known as the particular solution $x_p(t)$ which -- in general -- does not. The general solution, as you have ...


0

In an inertial reference frame the groove $AB$ is rotating, so you cannot assume that the force exerted on the particle by the groove is perpendicular to $\vec{AB}$. The vector $\vec{AB}$ has a time dependency - it should more properly be written $\vec{AB}(t)$ - so the force exerted on the particle by the groove has a component along $\vec{AB}$. It is is ...


0

As seen from the inertial frame, due to the rotation you get zentrifugal force which perpendicular to the position of the mass ,vector R. The component of this force towards the incline A-B cause the mass to move to point B, but only if this component is greater then the sum of the inertia force of the mass and the component of the weight towards the incline....


1

I did the following proof (it might be useful but it's pretty large though): $$\hat n|n\rangle=n|n\rangle$$ $$[\hat n,\hat \phi]=i \ \ \ (\mathrm{given})$$ $$[\hat n,e^{i\hat \phi}]=-e^{i\hat \phi}$$ From the identity (see the proof here): $$[\hat n,e^{i\hat \phi}]|n\rangle=(\hat ne^{i\hat \phi}-e^{i\hat \phi} \hat n)|n\rangle=\hat ne^{i \hat \phi}|n\rangle-...


6

Let us compute $$\hat{n} \, e^{i \hat{\phi}} | n \rangle = \hat{n} \sum_{k=0}^\infty \frac{(i \hat{\phi})^k}{k!} | n \rangle = \sum_{k=0}^\infty \frac{i^k \big( [ \hat{n},\hat{\phi}^k] + \hat{\phi}^k \hat{n} \big)}{k!} | n \rangle = \sum_{k=0}^\infty \frac{i^k \big( k \, \hat{\phi}^{k-1} [\hat{n},\hat{\phi}] + \hat{\phi}^k \hat{n} \big)}{k!} | n \rangle = \\...


0

So I think that my problem was that I forgot to replace $r=t\text{sinh}(\sqrt{|k|}\chi)$ in my Minkowski metric ! But after replacing it, and replacing the derivative I kind of get the Robertson-Walker metric again, with $a(t)=t$ and with $S_k(r)=R\text{sinh}(\frac{\chi}{R})$ for an open universe. I don't know if this can help anyone but I just let the post ...


0

You can convert Newtons to ounces and then convert millimeters to inches and you will be describing the same amount of torque. This can be done with any system of measurements. Or you can type in " torque conversion table " in a search engine and you should find results such as; http://www.unitconversion.org/unit_converter/torque.html Ihope I have ...


0

The Hamiltonian is $$ H=-J\sum_i s_i s_{i+1}$$ Imagine starting from an all $+$ configuration and flipping all spins right of $s_k$ (including $s_k$) $$ H=-J(s_1s_2+\dots+s_{k-1} s_k + s_k s_{k+1}+\dots s_{N-1}s_N)$$ All of the terms will keep the same sign, except for $s_{k-1}s_k$, which will go from $+1$ to $-1$, hence while the mismatched link only ...


0

If you have an even number of charges, all of the same magnitude, and equally spaced along a line, the field will be zero at the center of the configuration. Otherwise, you have a problem which may have no answer.


1

The Fresnel equations give the relationships between the reflected/transmitted fields (or intensities) as a function of incidence angle.


0

Although this question can be likened to a homework type question there is an important idea embedded in it, in that the act of catching is an inelastic collision so kinetic energy will not be conserved. This means that although energy will be conserved, kinetic energy would not, so it would be extremely difficult, if not impossible(?), to solve this ...


1

The mistake is in (i). Angular momentum equation for a point mass would be $\vec{L} = m\vec{r}\times\vec{v}$. The magnitude is therefore $L = mrv\sin\theta$. Since you're in an elliptical orbit (and $\theta\neq 90^\circ$ thus $\sin\theta\neq 1$), the angle between $\vec{r}$ and $\vec{v}$ will keep changing. Therefore, your equations for angular momentum ...


3

The following may be useful to consider. The distance that car A travels during the accelerating portion is computed as $533.33$ ft. During that same amount of time, $13.33$ seconds, car B will have traveled $13.33 \times 60 = 799.8$ feet. Since car A and B are 6000 feet apart at $t=0$, then they will still be $6000-799.8-533.33=4666.87$ feet apart after $...


4

We do not post full solutions to homework questions on this website. With that said, here are a few hints. Hint 1. The object is being pulled up by three ropes. Each of those ropes will apply a force I'll call $T$. In this case, the three ropes will all exert the same tension on the object. I.e. the upwards force on the block is $3T$. As for the inclined ...


0

Taking average on both sides of the equation we get: $<H> = -Jdm<\sum \sigma_i> + \frac{JdNm^2}{2}$ By definition $m = \frac{<\sum \sigma_i>}{N}$. Substituting it in the above equation we get: $<H> = -\frac{JdNm^2}{2}$ This is the average energy.


1

When you have acceleration as a function of displacement you can use conservation laws to describe the velocity-position relationship. Do the energy after minus energy before = work done. Work is $W=\int F \,{\rm d}s$ and thus $$ \tfrac{1}{2} m v^2 - \tfrac{1}{2} m v_1^2 = \int \limits_{s_1}^s (m a)\,{\rm d}s $$ with $s_1 = 27\text{ [m]}$ and $v_1=27\text{ ...


1

HINT: $\langle \textbf{p}|\psi(t)\rangle := \psi(\textbf{p}, t)$ is the fourier-transform of $\langle \textbf{r}|\psi(t)\rangle := \psi(\textbf{r},t)$.


0

$$ a=f(s)=6\,s^{1/3}=\frac{d^2s}{dt^2}$$ solve this differential equation with the initial conditions $~s(t=2)=27\,,D(s)(t=2)=27~$ you obtain $s(t)$ $$s(t)=(t+1)^3$$ and $$v=\frac{ds}{dt}=3\,(t+1)^2$$ thus: the acceleration a $$a(t=4)=6\,s(t=4)^{1/3}=30~ \text{[m/s^2]}$$ $$v(t=4)=75 ~\text{[m/s]}$$


0

The work done during a short interval of time, $[t, t+\Delta t]$ is given by usual formula $$ \Delta W = \mathbf{F}(t)\cdot \Delta \mathbf{r}(t), $$ where $\mathbf{F}(t)$ and $\Delta \mathbf{r}(t)$ are the force and the displacement at the beginning of the interval. The total work is then approximately a sum over all the intervals, and this approximation ...


0

The 20 W is a maximum rating for the transformer and is of no concern unless it is exceeded. At 100% efficiency the power in equals the power out. In this case the power out (7.2 W) gets cut in half, so the current in (0.03 A) will also be cut in half.


0

So the first thing we need to do is find the "step-down ration". Due to how a transformer works, a small change in the secondary coil/output current will result in an even smaller change in the primary coil current. Ratio = $\frac{\text{Output Voltage}}{\text{Input Voltage}}$ = $\frac{12V}{240V}$ = $0.05A$ Use the ratio to calculate the change in ...


1

The possible trajectories of a particle subject to an inverse-cube force $F = - k/r^3$ can actually be derived exactly; they are known as Cotes's spirals. Depending on the relative values of the particle's angular momentum $\ell$, its mass $m$, and the constant of proportionality $k$ in the force law, they take on the form $$ r(\theta) = \begin{cases} (A \...


0

In case this helps anyone in the future. Defining Veff essentially turns this into a 1d problem with only r as the dynamical variable. If we want circular orbit you can either (do explicit calculation as Layla's answer does) or just see that Veff = const × 1/r^2 for inverse-cubic force law. Now, this function can be either monotonically increasing (const<...


-2

To find the electric potential energy of the initial configuration, bring the charges in from infinity one at a time. This will equal the total final kinetic energy. If the charges are released simultaneously (removing all external forces), then momentum will be conserved (giving two component equations). To take advantage of symmetry, I would choose an x ...


0

X-ray absorption is measured here by the total electron yield (TEY) method which is often assumed to be sufficiently limited to a surface region to be representative. At some of these strongly absorbing "white lines" that is not the case. Corrections need to be made when the absorption is so strong that most of the x-ray attenuation is within this ...


2

Take a look at the picture bellow (adapted from here https://guitar-auctions.co.uk/wp-content/uploads/2018/02/lot0126.jpg) where I give you the recursive idea to calculate the length $x_{n+1}$ from $x_n$ for the frets on a guitar. $x_0$ is simply the length of your string (in your case $x_0 = L = 1 $ m). The density of the string does not matter. Why? As you ...


2

The frequency spectrum is given by: $$f_n=\frac{n}{2L}\sqrt{\frac{T}{\rho}}$$ and: $$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho}}$$ where $T$ is the string tension. Alternatively, write: $$f_n=\frac{1}{2x(n)}\sqrt{\frac{T}{\rho}}$$ So that: $$\boxed{\frac{f_n}{f_1}=\frac{x(n)}{L}}$$ Where $x(1)=L$. Calculate the $x(n)$ from there.


-2

I think that none of the answers hold. As all spheres are connected with conducting wires, the voltages should be equal for all of them. So they are all in series with charges q each. The analogy with capacitors shall take this into account.


1

Since the pencil is in equilibrium, the frictional force is equal to the component of gravitational force along the slope. Take a component of that force along the perpendicular to the pencil, and balance the torques due to gravity, normal force, and friction. For the borderline case, the normal force must be acting at the edge of the pencil. You'll get the ...


2

From $U^\dagger U = 1$, you get $U^\dagger = U^{-1}$; then $UU^\dagger = UU^{-1}= 1$. You don't have to worry about anything; $\det U = 1$, this matrix is invertible.


Top 50 recent answers are included