We usually consider the object to be small enough that we don't need to worry about how the object curves underneath the object. That is, we take the normal force to be in the same direction everywhere that it acts on the object. Certainly this is true for the JEE.
In your example where the object is large compared to the radius of curvature of the surface it is true that the normal force acts in different directions at the two points where the block touches the surface.
At each point where the block touches the surface the velocity is tangential to the surface. It has to be otherwise that point on the block would either go below the surface or rise above it.
Yes and since point of application of force is moving normal to the direction along which force is acting , therefore work done by normal force is zero
We could calculate the time the 12kg block takes to move 10m but actually we don't need to because for a fixed time the distance is proportional to the acceleration. Yes?
And the acceleration of the 4kg block is 2.6 m/s² which is half the acceleration of the 12kg block, so that means in the time it takes for the 12kg block to move 10m the 4kg block will move 5m. Yes?
This question tells us that the 12kg block accelerates at 5.2 m/s². It doesn't tell us what the external force is, and we don't care what the external force is. The acceleration is 5.2 m/s². That's what the question says. Yes?
The answer is : perpendicular to the diameter. I don't understand how to arrive at that answer. I'm taking a high school physics course and have covered Electric field and Gauss's law for simple symmetries such as the sphere. Can anyone help me on this?
Edit: The solution says that the field com...
A bowl, or at least that's how I interpret the question. It is just half a sphere i.e. take a sphere and cut it in half and you get a hemisphere. So there is no base to it.
also can we say the "direction of the electric field at a point on a diameter away from the center of a uniformly (positively) charged hemisphere" is perpendicular
for both the bowl and the hemisphere
for the bowl the answer is given and for the hemisphere, the electric field is always perpendicular to the surface, here there is a base
yes right? electric fieldlines is always perpendicular to the surface , so here base is part of the surface and hence the electric fieldlines is always perpendicular to the base as well?
Oh thank you sir. i got a couple more questions from the ncert exemplar but ill ask later seeing that there are a couple of others in the queue and seeing that I am an awfully slow typer😂
wait in a stable equilibrium if i displace an object will it (a) come to rest at the original position or will it (b) be in a SHM with the center of the SHM at the original position?
Energy is always conserved, and if you displace an object you have increased its PE so its energy relative to the equilibrium position is now greater than zero.
So when you release the object it will oscillate with a constant total energy. Whether the oscillation is SHM or not depends on the potential. You only get SHM when the potential is quadratic.
Although in some systems there may be damping, in which case we get a damped oscillation and the object will eventually settle back to the equilibrium position.
Wait..To prevent relative motion and bottom is smooth, so if they will move together a=10/12 =5/6( since net force on the system of three blocks is 10N ) and on each block if we check friction in each case provides the required force. I thought this way ..
And if any friction is not sufficient to provide the the required force then there will be sleeping and acceleration will be different of that block
If we consider 3kg and 7kg blocks as a system, net Fext =10-f2 and since acc of 2kg is different I.e there's slipping friction will be maximum I.e 2(0.2)10 =4
@Ali And there is friction between 2kg and (3+7)kg system , the friction will try that acceleration of 2kg is also 0.6? And $f_1$( friction acting on 2 kg)max =4 > 2(0.6) =1.2
So, here friction is efficient to avoid slipping at 2kg and 3kg junction, so that contradicts our assumption
Oh ! So you are saying that friction is sufficient to avoid slipping between the blocks. So the answer should be 5/6 m/s² and not 3/5 because that induces slipping ?
Ok thanks I get it now
Thank you very much @Rover
So that answer is superior where there is no slipping ?
@JohnRennie Can you check my question and confirm the reason given by Rover
That's what I am saying. The values do satisfy everything
But according to Rover ,as the tendency of static friction is to stop relative motion. The values of acceleration in which they are equal is the better solution.
@Wolgwang as the photoelectrons "ejects", the surface of the metal becomes +vely charged and attracts the electron, so you need some work to overcome that. That is reflected in the work function ($\phi$) for that particular metal, (which I would guess will depend strongly on the ionisation potential). So anyways, when the electron finally detaches from the metal, it has an energy of $h \nu - \phi$, which we associate to its K.E i.e $1/2 mv^2$
Photoemission in the usual experiment is a two step process. First the incident photon creates a photoelectron in the bulk of the metal:
The quantum efficiency for this is almost $100$% i.e. almost every photon that hits the metal creates a primary photoelectron. However this photoelectron is ...
I have to figure out what happens to the brightness of each bulb after shorting with the wire.How do I start? I have a basic knowledge of induced emf and electric fields
Photoemission in the usual experiment is a two step process. First the incident photon creates a photoelectron in the bulk of the metal:
The quantum efficiency for this is almost $100$% i.e. almost every photon that hits the metal creates a primary photoelectron. However this photoelectron is ...
