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BannedUser
BannedUser
01:45
I will try to digest that. I haven't learn dynamics yet.
And kinetics.
I imagine friction changes since looking at a rough surface like road the microscopic bumps are not even pattern. But it seems that most physics people just approximate it.
But carefully measuring them is waste of time thus I think taking it as a constant is better.
 
3 hours later…
Robin Singh
Robin Singh
04:31
@JohnRennie Hello sir :)
John Rennie
John Rennie
@RobinSingh Hi :-)
Robin Singh
Robin Singh
user image
Sir, how do i get started with this one?
John Rennie
John Rennie
Suppose we had a free rod i.e. one not attached at the top end, and we struck it at the point x. Then two things happen:
1. the rod starts moving right
2. it start rotating
OK so far?
Robin Singh
Robin Singh
Yes sir
John Rennie
John Rennie
If we strike it below its midpoint then it will rotate anticlockwise so the top end will rotate left and the bottom end will rotate right. Yes?
Robin Singh
Robin Singh
04:37
Yes sir
John Rennie
John Rennie
And at the same time the whole rod is moving right.
Robin Singh
Robin Singh
Yes sir
John Rennie
John Rennie
So the motion of the top end is to the left because of the rotation and to the right because of the linear motion.
And for some value of x these two motions will exactly cancel so the top end of the rod is momentarily stationary.
Robin Singh
Robin Singh
Ahh I think I can do it now
John Rennie
John Rennie
That means we could pin the top end of the rod, as shown in the picture, and there would be no impulse delivered by the pin.
@RobinSingh OK :-)
Robin Singh
Robin Singh
04:39
Thank you sir :-)
Lllt
Lllt
04:52
@JohnRennie Sir when a bond is formed between two H atoms, say x amount of energy is released which we call as bond energy, My question is in which form does this energy is released out
Buraian
Buraian
The potential energy of the chemical systemreduces , hence heat I guess
John Rennie
John Rennie
In particle reactions the energy is almost always released as kinetic energy, or sometimes by radiating a photon.
Lllt
Lllt
Can heat be a suitable option??
John Rennie
John Rennie
It's actually very hard for two isolated H atoms to form an H₂ molecule. The problem is that the attractive force between them accelerates them together so their KE increases. When they meet they are travelling too fast to stick to each other and they just fly past each other and zoom away again.
Lllt
Lllt
Yes but what if we consider ideal case?
John Rennie
John Rennie
04:56
Typically in a gas of H atoms there are many other particles around e.g. other H atoms or H₂ molecules. What happens is as the two H atoms start to form a molecule they collide with some third body and they can transfer energy to that third body as KE.
So in this case the energy is released as the KE of that third body.
@Lllt You need to consider what we mean by "heat"
In an ideal gas "heat" is just the motion of molecules i.e. heat = KE. Yes?
Lllt
Lllt
One minute sir
Yes!! @JohnRennie
John Rennie
John Rennie
The only other common mechanism is that when two charged articles interact, e.g. an electron and proton combining to form a neutral H atom, then they can emit a photon and lose energy that way. In that case the energy is released as photons i.e. light.
Lllt
Lllt
Yes
John Rennie
John Rennie
But in general the energy is released as the KE of some third object.
This applies to pretty much everything e.g. in nuclear power stations the energy is released as the KE of the fission products.
In a chemical reaction A + B ⟶ C + D the energy is usually released as the KE of the product molecules C and D.
Lllt
Lllt
Yes
Buraian
Buraian
05:06
@RobinSingh you may find this interesting : en.wikipedia.org/wiki/Center_of_percussion
Wolgwang
Wolgwang
06:05
> There is another kind of friction, called dry friction or sliding friction, which occurs when one solid body slides on another. In this case a force is needed to maintain motion. This is called a frictional force, and its origin, also, is a very complicated matter. Both surfaces of contact are irregular, on an atomic level. There are many points of contact where the atoms seem to cling together, and then, as the sliding body is pulled along, the atoms snap apart and vibration ensues; something like that has to happen. Formerly the mechanism of this friction was thought to be very simple, …
-RF's volume 1
 
4 hours later…
Aladdin
Aladdin
10:14
@JohnRennie Hello
Can u join the room i created
John Rennie
John Rennie
@Aladdin Hi :-)
 
3 hours later…
Ashish Ahuja
Ashish Ahuja
12:52
user image
why must the torque acting on the cylinder always be zero? I understand that the buoyant force passes through the geometrical center of a body so in the case of a uniform body it exerts no torque, but here surely as the height changes, the point of application of force changes (which is not neccessarily through the center of the cylinder)?
Samyak Marathe
Samyak Marathe
Hi
I have a chemistry doubt
If anyone could help.
Ashish Ahuja
Ashish Ahuja
just post it, someone may respond.
Samyak Marathe
Samyak Marathe
Caesium ion is larger than sodium ion. Both are dissolved in water seperately. More water molecules can surround over Caesium than in sodium and hence more bonds form in caesium than in sodium case. Is this argument correct. And if yes then why caesium has less hydration enthalpy than sodium
Ashish Ahuja
Ashish Ahuja
because the charge gets distributed further on Cs than on Na. As the atom is larger, the charge over size ratio decreases.
Samyak Marathe
Samyak Marathe
13:17
@AshishAhuja So you mean that even if the cs is large the oxygen atoms dont feel that much attraction due to large size of caesium, and hence oxygen atom are farther than nucleus of cs than in case of sodium
Ashish Ahuja
Ashish Ahuja
yes
 
2 hours later…
hansika
hansika
15:41
@AshishAhuja For a rigid body to be in equilibrium(both translational and rotational), net force on it should be zero, and we have to find the height of 2p liquid such that it is in equilibrium, of course the torque changes as the height changes, that is why we have to find "that" height for which it is in equilibrium which we can find out be taking net torque on it to be zero.
Please correct me if I am wrong
I had a doubt if I want to differentiate mirror formula for finding relation between velocity of object and image, then am I supposed to substitute signs of object and image before differentiating or not?
Wolgwang
Wolgwang
15:57
@hansika Mirror formula? $\frac1f=\frac1u+\frac1v$?
@hansika IG no as you are finding a generalised formula and one can put values in it by gauss or cartesian convention...
hansika
hansika
Okay
A thief is running away in a car with velocity of 20 m//s. A police jeep is following him, which is sighted by thief in his rear view mirror, which is a convex mirror of focal length 10 m. He observes that the image of jeep is moving towards him with a velocity of 1 cm//s. if the magnification of mirror for the jeep at that time is 1/10. Find the actual speed of jeep
In this question, on on differentiating mirror formula, I am getting Velocity of I wrt M=-m^2(Velocity of O wrt mirror), so on using that I am getting answer as 19m/s,(I differentiated the equation without substituting the signs) but in the solution Velocity of I wrt M=m^2(Velocity of O wrt mirror) and on using that answer is coming to be 21m/s, kindly help!
Here I=Image, M=mirror and O=Object
and m=magnification
 
3 hours later…
hansika
hansika
18:54
Got one more doubt,
An optical system consists of a thin convex lens of focal length 30 cm and a plane mirror placed 15 cm behind the lens. An object is placed 15 cm in front of the lens. The distance of the final image from the object is
In this question, I considered the second image formed by the plane mirror to be the final image and got the answer as 75cm, but in the solution the third image, i.e. the image formed by the lens (second time) as the final image. How will I get to know what image is to be taken as the final image? Is there anything I am missing or there is ambiguity in the question only?

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