@JohnRennie Sir I have a doubt.In these problems.I am free more or less today.If you would ping me up at your free time that would be extremely kind of you.
@Rover @RobinSingh why is option A not right? As you said block B does not play any role since the tensions cancel, so it is a situation of two blocks connected by a string on an incline where the upper one has friction, thus the upper block has a tendency to go slower; the string will always remain taut, not slack, hence A, C should have the same acceleration?
@JohnRennie you mean that whole moving on CD the pattern is like maxima and minima of double slit experiment hence I'll get some points with maxima and other with minima?
If you put a charge near to a conducting plane then the charge induces charges in that plane and those induced charges then exert a force on the original charge. Calculating this would be hideously complicated but luckily there is a short cut.
The conducting plane in effect acts as a mirror, so it behaves as if it creates an image of the original charge just like a mirror creates an image of an object
So if you put a charge q a distance d above a conducting plane then the charge experiences the same force as if there was a charge -q a distance d below the mirror.
Now it says the line OC is rotating at 2ω₀ i.e. if we had a solid bar joined at one end to O and at the other end to C then that bar would be rotating about O at 2ω₀.
And it's rotating about O so it has an upwards centripetal acceleration of (2R)(2ω₀)²
So the net acceleration is 92Rω₀² downwards.
So B is false.
For part (C) we do basically the same calculation as part (B), but now the centripetal accelerations are in the same direction so they add. The upwards acceleration due to the rotation about C is 100Rω₀² ...
And the upwards acceleration due to the rotation about O is (4R)(2ω₀)² = 16Rω₀²
But the question asks for the acceleration relative to the point on te edge of the larger cylinder, so we have to subtract off the upwards centripetal acceleration due to the rotation of the large cylinder.
Ohk. Sure sir. I will come around to check in a few min then. My Q is like I have written an explanation. Just wanted to confirm if it’s right or wrong.
In absorption spectrum . So , what happens here is that when white light is passed through a gas. The photons of the white light are absorbed by the electron in the gas. This means that the electron has received the required amount of energy from the white light needed for it to jump from the g...
Suppose for month n the sales in that month are Sn and the sales in the previous month were Sn-1. Then the percentage increase in tat month is (Sn - Sn-1)/Sn-1. OK so far?
e.g. look at March. The sales in March were 9000 and the sales in Feb were 8000, so the growth in March was 1000 and the increase was 1000/8000 = 12.5%. Yes?
@SrijanM.T yes, an emission spectrum is black everywhere except for the few bright lines where the wavelength corresponds to the transition between levels.
Tis is essentially the same as Q9, except that instead of finding the percentage increase we are just finding the increase. So look at the three months that the question mentions and calculate the increase in each month. e.g. in March the increase was 1000 because it was 8000 in Feb and 9000 in March.
When we talk about the emission spectrum or absorption . We are talking about only one electron ? . So , only electron had so many different colorful lines. How does that happen ? Why is it giving out so many colours ?
Each atom produces only one photon, and that one photon has only one wavelength.
But in our gas we have many atoms, and in each atom the electron can start and finish in different orbitals.
So although any one atom can produce only a single wavelength, different atoms can produce different wavelengths, so all the atoms together produce the many different lines in the spectrum.
@JohnRennie I have a doubt regarding Bernoulli's theorem. when we derive the equation of velocity of efflux the outside pressure at the opening of the orifice is atmospheric pressure.
But in this question (Problem 1.324 on Irodov), it is explained that the centrifugal force creates a pressure till the end of the of the tube. I can understand the second case but not the first one.
I think both the cases are almost the same. Please correct me