Recent content by Kharrid

The work is in the photo. Essentially, I think the emf of the two horizontal sides is 0 and the emf of the two vertical sides is vBL. Since the vertical sides have the same direction and magnitude, they cancel and the total emf is zero around the loop. Is this correct?

I am having trouble figuring out if the circular loop has an induced current. One explanation is ∫ E ds = -d Φ / dt. Since flux = B ⋅ A, a change in the magnetic field would require a change in the magnetic field, a change in the area, or change in direction of either vector. Since none of...

I already have the solution in front of me, I am wondering why there is a difference in the formula for path difference. I've attached the problem as well to show the Figure. What I am struggling to grasp is why the path difference for the angles closer to A is dsin(Θ) = (m+1/4)λ while the path...

Haha yea, I focused on the problem and forgot the basics. Also, I "forced" it to work, which should have been a big sign that it wasn't right. The error was that I multiplied the bottom by 2 instead of squaring. ##v = 2.89 * 10^8 \frac{m}{s}## Substituting the new value in: ##E_{p1} =...

At this point, I really do. That's on me. I should be more accurate and expand the equations fully. I understand how each equation is derived, so now moving to solve. If I am looking for ##E_{p1}##, then the top equation becomes ##E_{p1} = E_{tot} - m_pc^2##. From the equation below...

So from the discussion, I assume that I should stay away from velocities? I'm not sure what next step I should use if I go down this path. If I do need the velocity of the particle, I need the speed of the zero momentum frame relative to the lab frame. Since the velocity of the proton 2 in the...

In the CoM frame, one can separate the total energies before and after the collision. Before the collision, there are only two energies of the protons. These energies are the exact same because the momentum of the protons are equal and opposite and velocity is squared in the energy equation. So...

##E_{p1i} = \frac{1}{\sqrt{1-\frac{(v)^2}{(3*10^8)^2}}}*(2.294*10^{-10}+(2.27*10^8)(1.67*10^{-27})(2.27*10^8)(\frac{1}{\sqrt{1-\frac{(2.27*10^8)^2}{(3*10^8)^2}}}))## The mystery is what ##v## is. If the lab frame is stationary, then the CoM frame is moving at some velocity ##u##. I'm not sure...

Quick aside, the equations that PeroK listed are not written in my textbook. I assume these are standard equations that students should know? A similar looking equation in my textbook is ##E^2 = (mc^2)^2 + (pc)^2##. ##E=γ(E′+vp′)## ##E_{p1i} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}*(E'+vp')##...

##v_{p1i} = \sqrt{(3*10^8)^2(1-(\frac{1.67*10^{-27} * (3*10^8)^2}{2.294*10^{-10}})^2)}## ##v_{p1i} = 2.27 * 10^8 \frac{m}{s}## Good catch. Note to self: stop skipping steps.

Hmm ... this is what I plugged in: ##v_{p1i} = \sqrt{(3*10^8)^2(1-\frac{1.67*10^{-27} * (3*10^8)^2}{2.294*10^{-10}})}## ##v_{p1i} =1.76 * 10^8 \frac{m}{s}## Does that look right?

Given equations are ##E=γmc^2##, ##p=γmv##, and ##K=(γ-1)mc^2##. To calculate the minimum kinetic energy, the only piece I am missing is the velocity of the proton. I can calculate the velocity of the proton in the zero momentum frame, but I'd have to use a transformation to shift it into the...

This means that the speed of proton 2 is also the same magnitude but opposite direction of proton 1. Based on that, ##E_{p2i} = E_{p1i}## in the zero magnitude frame. Therefore the kinetic energy of the two protons is ##E_{p1i} + E_{p2i}= 938.3 + 938.3 + 2*(493.7)## ##E_{p1i} + E_{p2i} = 2864##...

Rest energy of Kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. Are these the outcomes of the two equations? ##p_{p1i} + p_{p2i} = p_{p1f} + p_{p2f} + 2*p_{kaon}## ##p_{p1i} + p_{p2i} = 0## ##p_{p1i} = -p_{p2i}## ##E_{p1i} + E_{p2i} = E_{p1f} + E_{p2f} + 2*E_{kaon}##...

Yes that makes sense. We are trying to find enough energy to create the kaons, but not enough to actually pass momentum to them. Since we only want enough energy to create the kaons, all of the kinetic energy would be transferred to the kaons and their creation. Therefore, the protons after the...