I was able to solve b) but I am confused for a). I understand that in the proton-antiproton collision, only two quarks (one from proton and other from anti-proton) can be combined to get a virtual photon that in turn creates muon and anti-muon. I don't understand what would happen to the other...
I think ##X## appears to be ##\pi^{+}## because it is light and energetically more favourable. Pion should be positive to ensure charge conservation. I am stuck at drawing a Feynman diagram for $$p+\bar{p} \to W^- + \pi^+$$.
Is this correct? Is this the leading order diagram or is there a...
I actually am not sure what equations are relevant here but I thought these are the relevant ones.
My Approach:
By Stefan-Boltzman Law, the intensity absorbed by the Earth is given as ## I = e \sigma T^4## where e is the emissivity of Earth, ##\sigma## is Stefan-Boltzman constant and T is the...
This is what I did, the cross-sectional area of the cylinder is ##\pi 10^2 = 100\pi## cm. Further, as the cylinder is moving a distance of 50cm. Thus, the total volume of vacuum, object covers throughout its journey is ##100\pi \times 50 =5000\pi## cm. Assuming that the molecules are uniformly...
@kuruman The actual solution has the same approach as that of @PeroK.
My logic was that, since momentum is additive, after inelastic collision as both the masses will be moving with the same velocity, I can take the velocity ## u## common thanks to vector algebra. And now if a second person...
This is my approach:
These quantities namely mass, length, and time, are all additive in nature. ##2 m + 3m = 5 m ##. If the argument of the functions mentioned in the problem statement is not dimensionless, then mass, length or time do not remain additive in the image space of the functions...
I already have referred to the solution to this problem. But the way I originally solved the problem is completely different from how the available solution proceeds. I wish to know if my solution is right or wrong.
My Solution:
Consider three particles undergoing one-dimensional motion all...
I think I understood your point. The change in the potential energy as the rocket moves from ##h = 0 ## km to ##h = 200 ## km is $$U = G.m_s.m_e. \left( \frac{1}{R_e+200} - \frac{1}{R_e}\right) $$ Substituting the respective values, I got ##U= -2.845 \times 10^{6}## J
So the total energy is...