# Tag Info

-1

$y=\frac 12gt^2$ This is the best hint I can give since it is not a meta exchange site.

0

Start by taking the curl of Maxwell's third equation (for vacuum), and substituting $\vec{B}=\mu_0\vec{H}$ one can obtain, $$\nabla^2.\vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E}$$ Similarly, by taking curl of Maxwell's fourth equation, substituting $$\nabla\times\vec{E} = -\frac{\partial}{\partial t}\mu_0\vec{H}$$ one can ...

0

First of all the distance traveled in $n$-th second is the distance traveled in one second between $t = n$ and $t = n-1$ seconds lets first calculate the distance traveled in 3 seconds. $S=ut+0.5 g t^2$ (here initial velocity is zero) $S=0.5 \times 10 \times 3^2 = 45 \, \mathrm m$ now lets calculate the distance traveled in 2 seconds $S=ut+0.5 g ... 0 In order to solve this question, you have to calculate distance travelled in 4 seconds as well as in 3 seconds. And then subtract these two to get the distance travelled from 3rd to 4th second (which is 3rd second). What you did was that you calculated the total distance travelled from$t=0$to$t=3$seconds, instead you have to find the distance travelled ... 0 As the particle is moving in counterclockwise direction, the velocity of the source at point A will be along the line joining the source and detector and hence the apparent frequency will be maximum at that point. Therefore apply the equation for Doppler and maximum apparent frequency detected by detector. Moreover calculate the time taken by source to reach ... 3 Think about this from the perspective of a person in the elevator. No windows, they can't look outside. As far as they are concerned, they live on a small box-like planet where the acceleration due to gravity is 9.8 + 1.2 = 11 m/s$^2$. In a system where the acceleration due to gravity appears to be 11 m/s$^2$, a bolt drops 2.7 m. How long does it take to ... 0 Make initial velocity of the bolt equal to$2.4m/s$not$0$0 As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the ... 1 Whatever you are doing, I think you are making this too complicated. The normal way to solve this problem (approximately) is to ignore all the terms in the Fourier expansion, except the one at the resonant frequency of the swing. Then the solution is just the normal time-dependent solution for a damped oscillator forced on resonance. 0 Given two operators$A$and$B$, the pairing of the two operators forms another operator, either$Q=AB$or$S=BA$. Given the commutator $$[A,B]=j$$ we can write $$Q-S=j$$. Now you can use the standard definition of expectation values to determine which form to use because $$<Q>-<S>=<j>.$$ 2 123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ... -1 The real answer is: it doesn't really. Or rather: we can still extract some physics out of it! Lets derive $$S[\phi] = \int d^d x \mathcal{L}[\phi,\partial\phi]\\ \delta S[\phi] = \int d^dx \delta \mathcal{L} = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta ... 0 When you got a rotation movement, the change of the angle per time is characterized by the angular frequency \omega=\frac{d\phi}{dt}. You can assign every rotation movement to a oscillation. This is often done in books to introduce the mathematical concepts of oscillation. Rotations are easy to understand at that moment. Then you can assign pointers to ... 0 One cycle involves \sin(x) where x goes from zero to 360 degrees, or zero to 2\pi radians. The frequency is cycles per second, and the angular frequency which is to say, degrees/second or radians/second. Radians are a more "natural" unit, degrees are arbitrary (based on Babylonian base sixty arithmetic) 1 Okay, so I figured it out myself. Here's what I think: Take P to be the point on top of the ring, where the string is attached. Now, two things contribute to P's acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, a_P = a_{ring} + \alpha \times R, where \alpha is the angular ... 3 At the instantaneous moment shown in the diagram, we can write:$$2R\alpha_{ring}=a_{disc}$$as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration =2R\alpha_{ring}=2a_{ring} so we find that:$$a_{disc}=2a_{ring}$$Note that when the string moves to another position this will not be true, ... 1 I want to charge a 12v 100ah battery which need 20 amp of current. You can charge the battery at any current proving it is not too high and it might be that the 20 amp is the maximum charging current? If the adapter gives a constant (regulated) 12 volts then you will not be able to charge a battery of the lead-acid type it will require more than 12 V ... 0 The best way to differnciate is to look at the count-rate vs time graph. Random nature of radioactivity is indicated by the fluctuation of the count-rate. Whereas, spontaneous means that the process of radioactivity is not dependent upon any external environmental conditions like temperature, pressure, magnetic/electric field. For example: Plot a count rate ... 1 Forget anything about a capacitor and just consider the resistance of the conducting liquid. Think of the liquid as made up of thin dr concentric shells of radius r and find the resistance of a shell in terms of the resistivity, radius and thickness. Then do the integration to find the resistance of all the liquid. 2 You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime :$$ds^2 = -dt^2 + d\theta$$Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ... 0 If it's a positive charge, the particle moves in the direction of the electric field and vice versa for a negative charge. However, there also exists another force called the Lorentz force. It essentially states that, if a particle is moving with a velocity perpendicular to a magnetic field, a force is produced in a direction mutually perpendicular to the ... 1 Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is:$$dE=dK+dU=0,$$that is$$mvdv=-\frac{K}{x^2}dx,$$where K\equiv Gm_1m_2. Integrating from (x_i,v_i) to (x,v) we get$$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$This is the correct ... 1 The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field V with ... 0 There are few choice here. One length scale is the Bohr radius a_0. By using this scale, the corresponding frequency scale and energy scale can be defined as a_0 = \sqrt{\hbar/(2m \omega_0)} and E_0 = \hbar \omega_0 respectively. Substitute these back, you will get the dimensionless equation. Suppose you are trying to find the ground state, you can ... 0 Under the U(1) transform, A_n \to A_n - \nabla_R \xi_n(R), and using Stokes theorem it immediately follows that \gamma is invariant under U(1). 2 Hint for (a) : Use the Figure above to prove that : $$\tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right) \tag{a-01}$$ This explains why I find the last equality instead of the first as I post in one of my comments : we refer to ... 0 You can substitute:$$u(r) = f(\lambda r)$$in the differential equation and choose \lambda so that the left hand side becomes dimensionless up to some overall factor (you can then absorb what's left into E and f). You can also reason as follows. You know that \lambda= \frac{\hbar}{m c} has the dimensions of a length, it is the Compton wavelength of ... -1 Normal force N= mgsinα- 2mv^2/3H -1 Why do you think that It doesn't make sense to me that the normal force would be smaller than the perpendicular component of mg Yes, it is, because the ground is bending away below the car. If it would bend quicker the car would be flying and there would be zero normal force. No, it's okay. The normal force would be mg\cos\alpha without the ... 0 Alright, one issue I noticed is the incorrect Gauss' Law expression. It is actually:$$ \oint {\vec E}\cdot \mathrm{d}{\vec A} = \frac{q}{\epsilon_0} $$So only the perpendicular component of the vectors through each face surface matters. Since the electric field vectors for all the voltages you've given across each surface are acting upwards, only the ... 1 I can do it by considering m_1 and m_2 to be a system, which would give me a=F/(m_1+m_2). How can I use a free-body diagram instead to calculate the acceleration? But this is by the use of a free-body diagram. Otherwise, how would you know that it is the force F you should include? Because, your acceleration expression comes from Newton's 2nd ... 0 Determine the surface element of the cube, 2x2y2z or 8xyz, then use the integral version of Gauss's Law. 2 The name "Gaussian noise" actually has to do with the higher order correlations in the noise, such as:$$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \rangle, \langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \eta(t+\tau_3) \rangle, $$and so on. If the noise is Gaussian then all of these higher order correlations can be rewritten in terms of the two-term ... 0 Yes, the strategy is right. If one has a one-dimensional current I in the thin wire, the volume density of the current is$$ \vec J(\vec r) = I\cdot \delta^{(2)}(\vec r - \vec R_{\rm nearest}) \cdot \vec n $$where \vec R_{\rm nearest} is the point on the wire that is closest to the point \vec r. There are other ways to write the current but this is ... 1 There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient 7.181\times 10^{-16}, when converted from MeV to eV, should give 7.181\times 10^{-46}. Mega means a million, and it to the fifth power gives 10^{30}, not just 10^{15}. In this way, the OP has to add a 10^{-15} factor to his result. That makes his result 10^{-3} ... 1 Hints: The starting point is the 2-point relation$$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem$$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ... 1 You should take moments about the points r1 and r2, which then you will get two seperate equations, which is what you have done, and that's fine. Take another look at both of your equations, r1 and r2 are forces and since you are taking moments about both those points, you need to multiply r1 and r2 by their distance. What you have essentially done is equate ... 1 The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ... 1 You will need to use two free body diagrams, one for each mass, one FBD will include the force of A on B and the other one the force of B on A. Both mass have the same acceleration (do they), so you end up with a little system of three equations and three unknowns. 1 On page 68 the paragraph headed Calculating displacement given a time and acceleration includes the text: Assume that you’re on your traditional weekend physics data-gathering expedition. Walking around with your clipboard and white lab coat, you happen upon a football game. Very interesting, you think. In a certain situation, you observe that the ... 0 The continuity equation is described by: $$\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho{\vec u}) = 0$$ For incompressible and steady flow, it reduces to: $$\nabla\cdot{\vec u} = 0$$ The incompressible continuity equation in spherical coordinates is: $$\nabla\cdot{\vec u} = \frac{1}{r^2}\frac{\partial}{\partial ... 2 a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take t=1/c seconds to reach. Now, using trigonometry (the sine rule in the first equation), we get:$$ \frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct} ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$Dividing the first equation with the ... 0 Have a look at the answers to Pressure and altitude as they explain how the pressure:altitude equation is derived. There is nothing wrong with our working, but you have assumed that the temperature is constant and in reality the temperature falls with altitude (in the troposphere at least). That means the pressure falls more rapidly with height than your ... 0 The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that. Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that ... 0 S., yes, you are correct, once you measure the energy, the state collapses to one of the two, with the associated probability that you have correctly identified. Now, energy eigenkets are sometimes called ''stationary states", because they do not evolve in time, being eigenstates of the Hamiltonian (time evolution generator). Secondly yes, in the case ... 0 In all the cases, there will be same answer, the charges will equally spread among the outermost surface of the objects. It is proved by Faraday's butterfly-net experiment. And even if it be conducting or non-conducting, the charge will be equally spread over the surface and here we get charge density. And in most cases, the charge density is more important ... -1 According to law of conservation of angular momentum by increasing moment of inertia angular velocity should decrease. As given by Iw=constant 0 As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic=(1/2)mv^{2}, remains constant. Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the ... 2 When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to T (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ... 0 If m is mass of steel, T_s is temperature and M is mass of water and T_w is temperature, then in the absence of any heat losses the following heat energy balance applies:$$mc_{p,s}T_f+Mc_{p,w}T_f=mc_{p,s}T_s+Mc_{p,w}T_w$$Where$c_{p,s}$and$c_{p,s}$are the resp. specific heat capacities of steel and water and$T_f\$ is the final temperature, after ...

Top 50 recent answers are included