New answers tagged

0

Since it is absolutely clear that the problem is symmetrical, you could introduce a fixed support at the center point and it would make no difference. Each mass moves 2m so that is $\Delta x$. But you must realize then that the appropriate length when you are considering Hooke's law is not 80m but 40m. The proportional change in length is $4/80 = 2/40$


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The way I know it one just sums ver all available positions and momenta of all particles to get the partition function. This can be achieved by instead computing the phase space volume and give it some prefactors which you probably know. The phase space volume is just the integral over all allowed positions and momenta, so we basically just replaced the sum ...


0

There is no $f(k)$. Frequency is a function of wavenumber: $$ \omega(k) = \sqrt{gh\tanh{hk}}$$ From that: $$\frac{d\omega}{dk} =\frac{g(\tanh{hk}+hk{\rm sech}^2hk)} {2\sqrt{gk\tanh{hk}}}=v_{\rm group}(k)$$ is the group velocity (vs. wave number).


1

Charles' law is for a constant pressure process, not a constant volume process. In any case, you are dealing with an adiabatic (Q=0) expansion which is neither constant pressure nor constant volume. That said, the book answer is correct if one can assume that the adiabatic process is reversible. It should have stated so, because without that assumption you ...


0

Consider a frictionless table, with the spring providing the only force on each mass. Considering the two masses as a system there is no external force on the system in the horizontal direction; therefore, the center of mass of the system does not change, and both masses each move the same distance in opposite directions, 2 cm for your example. If the string ...


1

Charles' "law" must only hold for fixed volumes. Starting from the ideal gas law: $$PV=NkT$$, $$\frac{P}{T}=\frac{Nk}{V}$$ so the ratio $P/T$ is only fixed if $V$ is held fixed. Here, you have a "frictionless piston" that is allowed to move and change the volume of the cylinder. However, you do know that the cylinder is thermally ...


1

By setting $P_1/T_1 = P_2/T_2$, you are assuming that the volume of the gas is constant, c.f. the ideal gas law: $$\frac{P}{T} = \frac{N k_B}{V}$$ In this case, the pressure, volume, and temperature are changing, so one has only that $P_1 V_1/T_1 = P_2 V_2/T_2$. This could be re-expressed as $$\frac{T_2}{T_1} = \frac{P_2}{P_1}\frac{V_2}{V_1}$$ Since you ...


0

Amperage can be higher as long as the voltage matches. This is true for many types of sources and loads, but not for battery chargers. It is true when the source is a voltage source (fixed voltage, variable current depending on load). A battery charger is likely to actually work as a current source (fixed current, variable voltage depending on load) during ...


0

I don't think your electronic device will operate at all when connected to the charger while charging it. Battery is not feeding power to the device when charging and the device is powered by the charger output voltage. With half the voltage supply to your device needed if it is an electronic device and not some kind of electrical motor I don't think it will ...


0

You are working this problem backwards. Starting from the velocity $u$ and resolving it on $M$ in the vertical direction and adding the two sides up. This is incorrect since velocities do not add up like forces. While a body might have multiple forces acting on it, there is always only one motion state (velocity + rotation). So start with the velocity of $M$ ...


0

You took the components wrong Since Mass M is moving just vertically but there's also a horizontal displacement in the strings (As the strings are inextensible $\Delta$AB = $\Delta$AM Let the vertical displacement of M be $y$, So as $$\Delta AM = \frac{y}{\cos{\theta} }$$ On differentiating we get, $$V_p = \frac{V_M}{\cos \theta}$$


0

You have \begin{equation*} \mathbf{A}(\mathbf{r}) = \frac{\mu_{0}}{4\pi} \iint \frac{\mathbf{K}(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|} \text{d}^{2}S' \end{equation*} I'll only write the $x$ component. \begin{equation*} A_{x}(\mathbf{r}) = \frac{\mu_{0}}{4\pi} \iint \frac{-M\sin(\theta')\sin(\phi')} {|\mathbf{r} - \mathbf{r'}|} {R}^{2} \sin(\theta') \text{...


0

I agree with @John Rennie’s answer to this question, which is essentially “no” but I disagree with his calculated accelerations for the reasons given below. The gravitational acceleration of the Earth due to the Sun is $\frac{GM}{r^2}$ , agreed. The centripetal acceleration of the Earth due to its circular motion (orbit) around the Sun is $rω^2$ where $ω$ is ...


3

the "work" that done due to the tension force $~T$ is $$T\,\cos(\theta)\,\delta x=T\,\delta u\quad \Rightarrow\\ \delta x=\frac{\delta u}{\cos(\theta)}$$


0

The problem with your reasoning is that you assume for PQ that Q is not moving and the rod is purely rotating about Q. This is evident by you dividing by $\ell$ the velocitiy components. To get the answer you need to find the instant center of rotation of each rod. Rod $QR$ is pivoting about $R$ and the velocity of $Q$ is shown above with the blue arrow. ...


1

A problem like this is solved in the Landau & Livshitz quantum mechanics (Problem 4 in chapter 23, Linear oscillator, in my Russian edition). These cite Ph. Morse, 1929- so, if you are interested to go to the sources or get more information, then look for Morse potential. Wikipedia link above actually contains the reference to the original More's ...


4

You cannot use vector addition for velocity like that. Imagine two horses running in the same direction at the speed $v$ pulling the same wagon. Does the wagon move at speed $v$ or $2v$? The wagon moves at the speed $v$, i.e. you can apply velocity vector addition only to velocity components (vectors) of one body, and two horses are separate bodies. In your ...


3

Note that, deliberately, this is not a completed answer to the question asked. You are treating $\cos \theta$ as a constant. It isn't. What is constant (or almost constant) is the horizontal separation (call it $2X$) of the tops of the strings. Then if $y$ is the vertical distance of the point where the strings meet below the line joining the tops of the ...


0

It's a similar issue to that addressed here Help me in understanding use of vector in this problem The questioner wondered why the speed that the lower block slid along the table wasn't $5\cos53$ Instead it's $\frac{5}{\cos53}$


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When I read this part recently I got the same question, and I believe it is wrong too. Take a simple example that $\phi$ is a state of definite occupation with $p$ particles and $\psi$ is a state of definite occupation with $q$ particles, then switching the two amplitudes will cause an extra sign of $(-1)^{pq}$, whereas changing $\langle\xi|\psi\rangle$ to $\...


1

As in the comments, you may omit terms independent of $\varphi$ in the Lagrangian (they typically arise from changing the zero point of potential energy, etc.). You can also integrate by parts / remove total derivatives. Here, we note that the derivative of $\cos(\omega t - \varphi)$ means we can exchange $\dot{\varphi}$ for $\omega$. (NB this does not mean $...


0

These are so-called Fock-Darwin states, see, e.g., these notes. Solving for them yourself starts with choosing a suitable gauge for the magnetic field, so that one can separate the angular momentum and the radial part of teh Hamiltonian in the cylindrical coordinates. One then will need to look for special functions - Bessel or Laguerre polinomials.


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After the measurement the wave function collapses in the eigenstate of the Hamiltonian (more generally the operator whose value is measured), corresponding to the measured value, so...


1

If $F_y$ on the upper hinge is positive, then you have chosen the +y axis as down (assuming x and y are horizontal and vertical). With a right handed system, +z is into the page. The y component of force from the hinge acting on the slanted stick would give a clockwise torque about the lower end of the stick. That would be represented by a torque vector in ...


1

This is up my street because I’m a structural engineer. My first question to you is “is this a problem of jointed rods – a pin-jointed structure?” because if any of the three joints is rigid then the structure is not statically determinate i.e. you cannot solve it by statics alone but need to consider the elastic properties of the members. I assume that it ...


0

If the acceleration was constant then the calculated figure of $140m/s^2$ would be correct and the distance of 17.5m would also be correct. However, the uniform force required to cause that acceleration would be 115x140 = 16100N not 4000N. Furthermore, if the applied force was 4000N for a short period then it would need to be more than 16100N at another time....


1

The general calculation to this potential is far from trivial and its not known in the present literature, since no symmetry can be invoked. However, in the particular case of point lying above the plate along the symmetry axis passing through the center of the plate, at distance z, the exact solution follows: $$V(z)= \frac{\sigma}{\pi\epsilon_0}\left[a \...


0

By symmetry, given both balls are identical, if you select the x-axis such that $\theta_1 = \theta_2$, then $\theta_1' = \theta_2'$. The initial momentum in the x-direction is then given by, $2mv\cos\theta_1$. Similarly the final momentum in the x-direction is then given by $\frac{2}{3}mv\cos\theta_1'$. Clearly these can only be equal when $\cos\theta_1 = \...


0

Here is a more direct approach which is outlined in the solution of Problem 5.2 of Alar P. Lightman's book 'Problem book in relativity and gravitation'. I've slighted edited the book's presentation so that it looks similar with the accepted answer. We assume the magnitude of velocity $v$ is fixed, so the particles have the same magnitude of velocity but they ...


0

Consider the situation from the top where $T$ is the tension and $f=\mu_sN$ is the friction force. The net central force must equal the centripetal force which is $m\omega^2 R$.


0

Consider the free body diagram:


0

Gravity is an attractive force, so we tend to think of it as "pulling" rather than "pushing". But you're right that block B experiences a gravitational force from A. There is a second force. Block A doesn't fall through Block B, even though there's an attractive gravitational force – according to Newton's Second Law, it seems like it ...


2

Always draw a free body diagram. Your $F_{up}$ is the tension in the string, and your $T$ is the net force $=Ma$ acting upward. See the free body diagram below. Hope this helps.


0

I think the other answers are answering a more difficult question: this is about a simple pendulum so there is no need to go nonlinear. The questioner has been fooled by the ambiguity in our use of $\omega$. It is the 'angular' velocity of the wave, and for a mass on a spring we happily write $x=A \sin(\omega t)$ as equivalent to $x=A \sin(2 \pi t /T)$ and ...


-1

When the amplitude is really small ($\theta_0<< 1\,\mathrm{radians}$) the math required to find the period of oscillation is very basic. But when that condition isn't met, the math becomes much harder, as you can see here.


0

I think your confusion comes from the fact that you might not be familiar with the concept of the total derivative. It basically is the chain rule already. Let me make an example: Imagine you have a function $f(x,y)$. Right now $x,y$ are just some coordinates. But you could consider a specific trajectory described by $x(t),y(t)$ where $t$ is time and both ...


0

There is already an accepted answer, but the OP is asking whether one can get the same answer without adding a mass term, so let me show how this is done. An alternative to using mass regularization is to use regularization in the scaling dimension. Let us assume for a minute that we are not dealing with a free field, but with an interacting field in a ...


0

You are not wrong. In fact, the ball, which moves freely, doesn't always go from the point A to the point C. If you write the equality of two results for time $t$, then you will obtain the condition on $v$, $\theta$, $d$, that regulates that your ball will fall in the point C.


0

There are two ways to look at this problem that can be useful: 1. Using forces: The work $W$ done by moving something along curve C is defined as follows $$ W = \int_C \mathbf{F} \cdot\text{d}\mathbf{s} $$ where $\mathbf{F}$ is force and $\text{d}\mathbf{s}$ is element of the curve. If the force is constant then simpler formula applies: $$ W = Fs \cos \alpha ...


2

For an object in circular motion, we have the standard circular motion equation: Under your above assumptions, if $k$ is the absolute value of your proportionality constant: $$v_c = \sqrt{\frac{k}{r_c}}$$ $$v_c^2 r_c= k$$ An object on a parabolic trajectory, under the influence of an inverse-square centripetal force comes to rest at $r=\infty$. Under your ...


0

We do not answer homework problems on this site, but the following information may help you. Consider the cart as your system of interest. Due to the constraint of the earth, the acceleration of the cart is only in the horizontal direction. In an inertial frame (observer at rest on the ground) there are no pseudo-forces; the forces on the cart are gravity, ...


1

I don't have the book, so I'm not sure of the details, but they probably defined the tangent space to a manifold at a point as the set of all derivations (linear maps which eat smooth functions and output real numbers, and satisfying the Leibniz rule). So, saying $\dot{C}(0)$ is a tangent vector to $SO(3)$ at the identity, i.e $\dot{C}(0)\in T_I(SO(3))$ ...


0

It might depend on the total length of the vine. Let's say the vine only has length $y$ and you swing all the way to the top when you jump from height $h$. Then, when you start with more energy because you jump from height $2h$, there's no higher you can go when you're still holding on to the vine, so you'll also make it to height $y$ but with a greater ...


0

No, the spring constant of a simple spring doesn't change. Here is what is happening: With zero added mass, the spring is trying to collapse into a shorter length but cannot because the coils are interfering with each other. They are exerting forces, keeping the spring longer than it would be if zero force was acting on each coil. The addition of the 80 g ...


2

The term "orbital" comes from the word "orbit", found in the original semi classical Bohr model. It describes the quantum mechanical space distribution for measuring an electron at (x,y,z,t) for a given nucleus given by the probability distribution calculated with the wavefunction. Even for the simple hydrogen atom, the energy of the ...


0

The second answer you found is wrong. This is because the amplitude is not $mg/k$. When the mass falls a distance $L$, it has some speed $v=\sqrt{2gL}$. However, when it has fallen a distance of $h$, its speed is zero. You are assuming the oscillation occurs between these two points when in fact it does not. In other words, the amplitude of oscillations ...


0

There is no reason this motion cannot be modeled as a simple pendulum (given meaningful data). In part (a)(ii) the 2π/T is the angular frequency of oscillation which is not the same as the maximum angular velocity. And, the given width tells you nothing about the radius. The given solution is nonsense.


1

If the track is a segment of a circle, the given setup is exactly equivalent to a regular pendulum on a string (sometimes called a mathematical pendulum when we ignore the mass of the string). The motion of the mass follows the exact same path, and the involved forces are equivalent. The only difference is that for a mass on a string, the force is provided ...


0

As you used only verbiage in your question, I think the most heuristic (reliably understandable) answer's offered by Einstein-Cartan Theory, which was developed by them in 1929, after the fact that subatomic particles spin had been discovered. In the cosmological model using ECT, the formation of a black hole occurs in the gravitational collapse of a ...


0

this is the function that you want to linearized $$f={\frac {g\sin \left( \theta \right) }{l}}-\frac 12\,{\dot\phi }^{2}\,\sin(\theta)\,\cos(\theta) \tag 1$$ from the conservation of the energy $~E=E_0~$ you obtain $~\dot{\phi}^2~$ where \begin{align*} &E= \frac m2\left( \left( \sin \left( \theta \right) \right) ^{2}{\dot\phi }^{2}+{ \dot\theta }^{2} \...


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