The reason that light travels more slowly in a dielectric is because it interacts with the electrons in that dielectric.
Light has an oscillating electric field, and if any charged particle is in the path of the light that particle will feel an oscillating force due to the oscillating electric f...
Is the string pulling the block along a tangent to the track? From the diagram there's an angle $\theta$ and I'm not sure what the significance of the angle is.
@Abcd If the string was along the tangent to the track then it would be easy because the tension in the string, the velocity of the block and the frictional force are all in line.
I think the diagram differs from that only in that the string is being pulled at an angle
The hand does work on the string, and the string does the same work on the block, and the block does the same work on the track. And that's the negative of the work the track does on the block (i.e. the frictional force).
Hi everyone. I already asked this question before, but as I do not have it clear yet... why there is no torque in the following perfect inellastic collision ball-rod? Thanks
@Abcd on the basis of energy balance I agree. At the end of the day all the work goes into friction - there isn't anywhere else for the work to go. So I don't see how the work done by the hand/string can depend on the angle.
@MadhuchhandaMandal if the sphere is conducting then the potential is constant everywhere in/on the sphere.
That's because if there were any differences in potential the charges would move to the region with a lower potential.
You're quite correct that there will be induced charges, and those charges will arrange themselves so they exactly cancel out the potential due to the dipole.
@JD_PM I'm not sure I understand what the diagram is supposed to show? Does the ball fall and hit the rod then stick to it?
I had a quick look at it but I can't remember anything about inductors. However it should be obvious what happens. The inductor has energy due to the magnetic field it has stored. When you close the switch the inductor uses that energy to charge the capacitor.
@MadhuchhandaMandal you've just agreed that the potential is constant everywhere in the sphere, and that includes the surface of the sphere. Since the point A is on the surface if the sphere it is connected to the rest of the sphere and must be at the same potential as the rest of the sphere.
@JohnRennie ...My question : A body moves up an inclined plane with constant velocity and reaches the highest peak . The inclined plane consists of friction . I was told to calculate the work done by the body to move up the inclined plane . First , the body gained potential energy . Then , my book calculates work done against friction $Fs$ , as $s$ is given . Finally , they add $mgh + Fs$ ... I am confused of the addition part ... Why do they add ? Shouldn't we subtract ?
I've figured out that in order to maintain constant velocity , force equal to that of friction is to be applied and so we add it right ?
@JohnRennie I just found out that it's a non charged sphere .. So , (V induced)= Sum(k dQinduced / R) = Sum(dQinduced) *(k/R) and by Conservation of Charge we get Sum(dQinduced)=0 so potential due to induced charges at Centre = 0
@NehalSamee When you say the work done by the body presumably there is a motor or something similar inside the body to propel it up the incline? So is it the work done by this motor or whatever that we are trying to calculate?
@NehalSamee so if the incline isn't flat and the friction isn't zero the motor has to do work both against the friction, $W_f$, and to go up the incline, $W_h$, so the total work done by the motor is $W_f + W_h$. That is, you add the two works together.
@NehalSamee If $W_t$ is the total work then: $$ W_t = W_f + W_h + (KE_f - KE_i) $$
where $KE_f$ is the final KE and $KE_i$ is the initial KE
@MadhuchhandaMandal Hmm. The induced charge will be symmetric i.e. the charge in one direction will have an equal and opposite charge in the other direction.
@JohnRennie ... However , I watched in lecture of Walter Lewin that during a body falls from top to bottom in frictioned incline plane , he wrote: $mgh-W_{friction}=\Delta K$
@NehalSamee When you're going up an incline the frictional force and the gravitational force are acting in the same direction. When you're going down an incline they are acting in different directions.
Remember that a frictional force always acts opposite to the direction of motion. It doesn't have a fixed direction like gravity does.
@JohnRennie ... Last but not the least , isn't work done to overcome friction , irrespective of upward or downward movement , so why do we subtract in downward and add in upward ?
Work is force times distance. If you are interested in the work done by the body (or on the body) then calculate the net force on the body.
If the body is moving upwards the friction and gravtational forces both act in the same direction i.e. down the slope. So they havethe same sign and simply add together.
If the body is moving downwards the the gravitational force acts down the slope and the frictional force acts in the other direction up the slope. So they have different signs.
The question is: The displacement of the particle at $x=0$ of a stretched string carrying a wave in the positive $x$ direction is given by $f(t)= A\sin (x/a)$, where $A,a$ are constants. The wave speed is $v$. Write the wave equation....
@JohnRennie Could you please explain what this question is trying to convey?
Why do different wavelength get impeded more or less when in different materials? Moving with the same speed, but a longer physical distance would imply that the fields oscillate less times in the material, but I don't know why a difference in the number of oscillations would impede the wave- I d...
@NehalSamee wavelength, colour and frequency are all related, so the refractive index depends on them all. I would probably say it's frequency dependent as it's really down to the photon energy and that depends on the frequency.
If you write $g(t,x) = A \sin(kx - \omega t)$ then your function depends both on time and position. You can ask, suppose I take a moment in time - e.g. take a picture - what does the wave look like, and you get this simply by setting $t$ equal to a constant.
If we make that constant zero, i.e. choose the moment in time when $t=0$, we get $g(x) = \sin(kx)$. And that's what your question has done.
The question says: The shape of the string at t=0 is given by g(x) = etc
@MadhuchhandaMandal imgur.com/a/g9493 see if this helps (I swapped the radii, but the rest of the procedure is correct) @Rick sorry I wasnt myself sure of that question either. I'll take a closer look tomorrow and let you know. Ping me if I forget.
@Abcd I think we are mis-communicating somewhere. If you graph $g(x) = \sin(x/a)$ it's just a sine wave. Put $x=300$ and you get some point on that sine wave.
@Abcd ah, OK, waves can be really confusing at first because generally speaking when students first encounter them they aren't used to functions of several variables.
OK, suppose I take a long bit of wire and bend it into the shape of a sine wave, then I place my wire on the $x$ axis.
The distance of the wire from the $x$ axis will then be given by $g(x) = A \sin(kx)$ for some constants $A$ and $k$ that depend on how I bent the wire. OK so far?
OK, so if I sit on the $x$ axis at some point, e.g. $x=300$, then I'll just see a stationary bit of wire at a distance from the $x$ axis of $A\sin(300k)$
Now suppose I give my piece of wire a velocity along the $x$ axis. I'm not changing anything about the wire - I'm not bending or reshaping it - I'm just moving the whole wire along the $x$ axis.
@Abcd I kind of get the feeling that you've suffered from brain overload and decided to call it a day :-) If you want to come back to this I'm happy to attempt more (probably not very good) explanations.
So $T = \lambda/v$, or rearranging gives $v = \lambda/T = \omega/k$, which is how we get the equation linking $v$, $\omega$ and $k$ that I mentioned earlier.
@Abcd if we go back to the example of the wire that I have been using, then at any time $t_0$ the wire is just a sine wave and all that changes is how far the wire has moved along the $x$ axis at the time $t_0$.
The distance moved since time zero is just $v t_0$