New answers tagged

1

You can trivially obtain the solutions to the shifted problem by replacing $x$ with $x-x_0$ in the original solutions. From there, it's a matter of elementary trigonometry (i.e. the angle addition formula) to express the result in terms of $\sin([\ldots]x)$ and $\cos([\ldots]x)$.


0

Let us assume that when you shorten the pendulum you are not doing any work on the bob.We know that that total energy of a pendulum is equal to the potential energy possessed by it at the highest point.Also the Maximum height reached by the pendulum is $l(1-cos\theta)$ where l is effective length and $\theta$ is the angular amplitude. So total mechanical ...


0

In Spherical coordinates, the shift in $\pi/2$ will correspond to a charge located at the plane $z=0$. Since the polar angle $\theta$ is measured from "north" to "south" and goes from $0$ to $\pi$, a shift of $\pi/2$ means it exactly in the middle of the 2 hemispheres. you can also see that "mathematically": $$\delta(\theta-\...


1

The torque generated by W2 is not constant. The higher the arm, the greater the torque. Imagine the extreme case of the arm being almost vertical. W2 must travel a much longer vertical distance to cause a the same angular rotation of the drum as when the arm is horizontal. Here's a more detailed diagram of the connection between W2 and the arm. There are two ...


1

Suppose we have an arbitrary state $|\phi\rangle$ in the subspace $\{|\phi_1\rangle,|\phi_2\rangle\}$. What this means is that we can write $|\phi\rangle$ as a superposition (linear combination) of these two states. For example, $|\phi\rangle = a|\phi_1\rangle + b|\phi_2\rangle$ for which $|a|^2 + |b|^2 = 1$. Now, we can find the eigenvalue of $|\phi\rangle$ ...


0

The potential in (b) is due to an effective dipole (polarization charge) from $q_1$, as considering the force from it on $q_2$, and in (c) the field internal to the dielectric due to its polarization is zero, and so we simply have $F_3 \lt F_1$, according to K.


-1

The question is a simple question. The only thing that spoiled the question is that it's incomplete. You omitted the distances between the point charges. Whether x,y or z. It's not enough to simply say it's a square without telling us the length of it's sides. Without knowing the position of the four charges with respect to each other there's no way to ...


1

For completeness sake and later reference i'll add the actual result here. The derivation requires besides the evaluation of the sum also a fair amount of hyperbolic trig function manipulation. Here it is, $$\begin{aligned} \rho(x,x') &\equiv \langle x | \hat \rho |x'\rangle \\[1.0em] &= \frac{\langle x|\exp(-\beta \hat H)| x'\rangle }{Z}\\[1.0em] &...


3

I assume you're looking for the components of the state in the angular momentum representation? The state itself is technically the same, but you want it in a different basis. We need to think about the actions of $L^2$ and $L_z$ in the Fock space. Consider $$L_z(a^{\dagger})^n |0\rangle = \frac{n}{2} (a^{\dagger})^n |0\rangle$$ $L_z=\frac{1}{2}(N_a-N_b)$ is ...


2

67.96875 pounds of weight vertically supported evenly on an area of 0.15625 square inches would equal a pressure of 435 pounds per square inch, neglecting any additional atmospheric pressure.


0

In relativity, a force can and does change the perpendicular components of the (3-) velocity of a particle. You can slow down a particle along one direction without applying a force along that direction. How do we know? Imagine a particle traveling very close tor the speed of light along the $x$-axis with momentum $\rho_x$. If we gave it some momentum in the ...


1

This is often a point of confusion for people new to the concept of work in the physical sense. If we strictly look at the motion of the baby, carrying it is equivalent to placing it in a stroller and pushing it across the room. You can see that the stroller is not doing any work to keep the baby at the vertical position it is in. Assuming the stroller ...


1

The definition of work made by a force is $$W=\int\vec{F}\cdot d\vec{r},$$ where $\vec{F}$ is the force and $d\vec{r}$ a differential displacement. In your case, the force to hold the baby is "upwards in the vertical" (opposing to gravity), and the displacement is "horizontal", i.e. parallel to the floor, so the dot product between the ...


1

While writing the question, I found the answer. The part of $\delta S_{\mathrm{CS}}$ that should be made to vanish is: $$\delta S_{\mathrm{CS}} = 3\theta^M{f_{LM}}^IC_{IJK}\int A^L \wedge F^J \wedge F^K = 0$$ The wedge product on the right is symmetric in $J, K, L$ (since $F$ is a $4$ form), so the previous equation constrains only the part of ${f_{LM}}^IC_{...


1

The OP is a good exercise in kinematics and dynamics. The solution to the first problem is accurate and clear since all velocities are measured relative to an absolute or the ground frame. Clearly, there are typographical errors in the solution to the second problem. Let us first correct the analysis. Interpreting the question (in reverse) from the solution ...


1

Assuming that the vertical reaction force $N$ is uniformly distributed over the area, you can integrate the torque $d\tau = r\cdot \mu dN$, given $dN/N = dA/A$. The area $dA$ is easier to write in polar coordinates $dA=r\,dr\,d\phi$. Your plan for the rest of the problem is correct. However, there is an alternative approach. From angular momentum ...


0

From the diagram, you can see that: $$V\cos( D )= 40m/s$$ Since $V$ is known, you can find $D$


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If the string (fixed at the top) passes through a hole in a movable mounted rod, and you move the rod downward (when the string is straight), then you have done no work on the pendulum and its maximum kinetic or potential energy do not change. The maximum height of swing will not change, but with a shorter string, the angle of swing will increase. If you ...


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The expression $$\left( 1 - \frac{r_s}{r}\right)\frac{dt}{d\tau} = \frac{E}{mc^2} = {\rm constant}$$ would apply to an inertial observer in the Schwarzschild metric. i.e. The $\tau$ here corresponds to the proper time experienced by an inertial (free-falling) observer. Your first expression $$ \left( 1-\frac{r_s}{r}\right)^{1/2}dt = d\tau$$ would apply in a ...


2

The fact that, $$ \phi_{1}=p_{x}+\frac{qB}{2c}y\approx0 $$ Does not imply that, $$ p_{x}=-\frac{qB}{2c}y $$ But only, $$ p_{x}\approx-\frac{qB}{2c}y $$ and conversely for $y$, $x$. That is, $p_x$ is a function of $y$ only on the constraint surface. Because the Dirac bracket is a derivative operator built by design to keep you on the constraint surface, while ...


-1

Electric Potential is a scalar quantity. On equatorial position of the dipole, the distance from the positive charge and the negative charge is equal and they have equal, but opposite magnitude. Therefore, on equatorial position net charge is zero, q=0. This implies that no work will be done in moving a charge along the equitorial line of the dipole. This ...


0

I suggest you that always solving it w.r.t ground. It is easier and also avoids confusions .you can directly conserve momentum and energy which is not possible if u solve using relative motion.


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I am a little late on this, the correct mathematical theorem (in the sanse implied by @Ryan Unger in the comments) states: Theorem: If $A:D(A)\subsetneq \mathcal{H}\rightarrow \mathcal{R}(A)\subset\mathcal{H}$ is a linear operator in a complex separable Hilbert space, then we have the following equivalence: $$ \forall \psi\in D(A), \langle \psi,A\psi\rangle ...


-1

This turns out to be a complex question. Given numbers, a numeric simulation provides a fairly accurate answer. Or, look up an analysis of two objects which are orbiting their center of mass. You want half the period for a flat orbit (with an eccentricity of zero).


0

Your sketch looks correct. I agree with your formulas. I put them into a spreadsheet and tried different inputs. With MKS units, a 30 degree angle, 1 revolution/sec, and g = 10 m/s/s, I got your answers. I found no combination of inputs that gives answers resembling the answers from your book.


3

Yes, several quantities vary with distance from the source. One useful quantity is the radiant flux (the amount of energy passing through a given area in a given amount of time -- basically how 'bright' it looks from your position). In fact, for a source that is spherically symmetric or far enough away that we can regard it as spherically symmetric, the ...


0

Intensity changes because energy of the wave is absorbed by the medium in which it propagates. Also the wavefront appear different. Far away from a source, waves have plane parallel wavefront, like light from the stars.


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A ball of mass $m$ moving with velocity $u$ in the negative x direction strikes the wall and rebounds with velocity $v$ which is in the positive x direction with $|v|<|u|$. This change in velocity is acquired in very little time, meaning the force applied by wall is very large so the quantity of interest is impulse. Initial velocity = $-u \hat{i}$ and ...


3

If you want to write $\exp\{-\beta \hat H\}$ in closed form in the position representation, you can use Mehler's formula: $$ \sum_{n=0}^\infty s^n \varphi_n(x)\varphi_n(y) =\\ \frac 1{\sqrt{\pi (1-s^2)}} \exp\left\{\frac{4xys -(x^2+y^2)(1+s^2)}{2(1-s^2)}\right\}, \quad 0\le |s|<1. $$ with $$ s= e^{-\beta(n+1/2)}. $$ Here $$ \varphi_n(x)\equiv \frac{1}{\...


0

Let the force(F) act on the body and displace it by 'S'. Then work done by this force equals change in Kinetic energy. Let initial velocity be u and final velocity be v. Now, Work done by F = $\Delta$ KE F.S = $\frac12$ mv$^2$ - $\frac12$ mu$^2$ ... i Also from Newton's second law : F = ma Substituting this result in i we get, ma.S = $\frac12$ mv$^2$ - $\...


0

To obtain die Equations of motion (@ Dale) you need those steps Step I obtain the kinetic energy $~T$ and the potential energy $~U$ The Lagrange is $~\mathcal L=T-U$ Step II The Hamiltonian is the total energy $\mathcal H=T+U~,\mathcal H=\mathcal H(\boldsymbol q~,\boldsymbol{\dot{q}})$ where $$\boldsymbol q=\begin{bmatrix} \theta_1\\ \theta_2 \end{bmatrix}...


1

$F=ma=m\frac{dv}{dt}$ by chain rule $m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dx}v=F$ Rearrangment yield $\int Fdx=m\int vdv$, this derivation implicitly imply force is constant then $F=ma$ Integration gives $max=m(\frac{v^2}{2}-\frac{u^2}{2})$, cancel common factor mass rearrange gives final result: $2ax=v^2-u^2$


2

Increase of kinetic energy is equal to the work done, which is given by the force multiplied by the distance moved in the direction of the force. In other words: (final kinetic energy) = (initial kinetic energy) + $F s$ which is $$ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + m a s $$ Now you can multiply by 2 and divide by $m$. The advantage of this method is ...


-2

$$v=u+at$$$$v^2=u^2+2uat+a^2t^2$$$$v^2=u^2+2a(ut+\frac{1}{2}at^2)$$$$v^2=u^2+2as$$ Quite starightforward.


0

I'm pretty confident that the problem means for us to consider the plane as a system not including the passenger. In this case, as other answers have pointed out, the passenger is 'external' to the plane system, and therefore any force the passenger exerts on the plane is an external force on the plane system. Now for the nuts and bolts of the force ...


0

Homework questions are not appreciated here but I'm giving you hint a nonetheless. Biot-savart law is a cross product. So it should already give a vector result. My guess is you are using a simplified version which assumes l and r are always perpendicular.


-1

Approach 1: Consider the plane and the man as a single system. Since there are no external forces acting on them, the center of mass shouldn't accelerate. But if the man starts running to the right and the plane doesn't react to this, the center of mass would go to the right as well. So in order for the center of mass to stay where it is the plane that ...


0

You have three main systems in your question - the plane, the passenger and the plane+passenger. The plane + passenger system can be considered isolated in the example, no external force is acting on it, so its center of mass remains at constant velocity. The plane itself has an external force acting on it by passengers feet when he/she tries to change his/...


0

Well here's the thing : If you consider the aeroplane as a separate body its velocity would change because friction applied by you on the plane would then be considered as an external force but if you consider yourself and the plane as a system then the net velocity of the centre of mass of the " you and plane" system would not change since the ...


0

What would happen if instead you imagine the plane to be no longer in air, but instead it is floating on a large, still bed of water? I use this example so as to have a situation where there is no friction. Now if you are inside the plane and then run forward will the plane begin to move backward? The answer is obviously no. You are correct in that Newton’s ...


1

Your problem is how to do the integral: \begin{align} I =\frac{1}{4\pi\epsilon_0}\int_{-a}^{a}\frac{\eta}{\sqrt{r^2+z'^2}}dz' \tag{1} \end{align} This is carried by change varaible to triangular function: \begin{align} z' = & r \tan\theta; \,\,\, \text{ therefore } \,\, dz'= r\sec^2\theta.\\ \sqrt{r^2 + z'^2} = & r \sec\theta\\ z' = & \,\,\,a \,\...


2

Starting from these two equations $$(e_++p_+)\frac{v_+^2}{1-v_+^2}+p_+=(e_-+p_-)\frac{v_-^2}{1-v_-^2}+p_-, \tag{1}$$ $$(e_++p_+)\frac{v_+}{1-v_+^2}=(e_-+p_-)\frac{v_-}{1-v_-^2}. \tag{2} $$ Show that $$v_+v_-=\frac{p_+-p_-}{e_+-e_-}, \tag{3}$$ $$\frac{v_+}{v_-}=\frac{e_-+p_+}{e_+ + p_-}.\tag{4}$$ Rewrite Eq.(1) as: \begin{align} & (e_++p_+)\frac{v_+^2}{1-...


1

Note that for questions similar to that in the OP and in general for dynamical modeling, executing an analysis which considers the geometry, kinematics and dynamics in that order is a good strategy. In the figure provided, if we denote the distance of the fixed pulley from the material particle model of $m_2$ by $l_2$ and similarly the distance of $m_1$ the ...


0

Its much better to use energy conservation for this type of sums.You probably didn't take care of the fact that acceleration is not positive but negative. Also motion might not be along a horizontal line but its certainly along a straight line.


0

The acceleration used in the equation $v^2=u^2+2as$ should be $-g\sin 30^\circ\approx-5m/s^2$ so that $v^2=10^2-2\cdot5\cdot4$ so that $v=\sqrt{60}=2\sqrt{15}\;m/s$. The kinematic equation can be recognized as the energy conservation equation in disguise.


0

if you take a closer look here you'll notice something really interesting. See, when you raise mass m2 by 'x' units both segments of the string around the pulley are loosened by 'x' units so m1 travels a total of '2x' units along the incline. So, X(m1)=2 * X(m2) (where X(m1) and X(m2) are displacements of m1 and m2 respectively.) Now if we differentiate ...


1

This would just be Hamiltonian mechanics. It is very similar to Lagrangian mechanics except instead of getting a few second order differential equations you get twice as many first order differential equations. Essentially you write $H(q,p,t)=E$ where $q$ is your generalized position (your angles) and $p$ is the conjugate momenta. Then $$\frac{dp}{dt}=-\frac{...


0

You are right about this one: $h^{\mu\nu}_{,\nu}=0$ corresponds to $k_{\nu}\epsilon^{\mu\nu}=0$. This is a familiar result for deriving the gravitational waves in the weak field limit. And about your question, $k_{\mu}\epsilon^{\mu}_{\nu}=0$ is simply obtained as $${\rm{if}}\,\,{k_\nu }{\varepsilon ^{\lambda \nu }} = 0\,\,{\rm{then}}\,{\rm{we}}\,{\rm{also}}\,...


0

We know from Poisson's equation that: $$ \nabla^{2} \phi = - \frac{ e }{ \varepsilon_{o} } \left( n_{o} - n_{e} \right) \tag{0} $$ where $$ is an electric potential, $e$ is the fundamental charge, $\varepsilon_{o}$ is the permittivity of free space, $n_{o}$ is a fixed number density of positively charged ions, and $n_{e}$ is the electron number density. From ...


0

For the first point - the ring actually acquires its angular velocity while angular velocity of cylinder changes from $0$ to $w$. During this process, magnetic field inside changes from $0$ to $\mu_0 I/l$ and there is induced electric field inside. For the second point - yes, the proof is mostly as you said, apart from that the field outside is not radially ...


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