Recent content by brotherbobby

I don't understand what symmetry really means. I am going by what I suspect.

Text excerpt : For context, I copy and paste the paragraph on "symmetry" from the text, underlining the relevant lines in red. I hope it's readable. Problem situation : This concerns the last three lines from the text excerpt. They say that : "... if a particle of mass ##m_1## travels at a...

As I have found out only recently, and I confess I should have done so a lot earlier, it is best not to think of electricity as the flow of electrons ##(e^{-})## at all. The convention : ##\small{\text{electricity flows from positive to negative}}\; (+ \rightarrow -)## is more than an just an...

Thank you and apologies. A silly error on my part. Something else if I may. There's an agreement to take the conventional flow of current ##I## from positive to negative. In this problem I raised, namely the directiom of force in a current carrying conductor, the agreement becomes crucial. If...

I don't see how. The RHR (Right Hand Rule) mind you. The magnetic field is from left to right. Charge is moving down. Hence force points into the page, not out of it

Positive.

Situation : I have drawn the image to the right. A current ##I## flows "down" along a wire in a magnetic field ##\vec B## directed to the "right". By Flemings's Left Hand Rule (LHR), the wire should experience force out of the page, as shown by the green bullet ##\color{green}\bullet##. However...

I have solved the problem I mentioned in post #1 at the top, which was about the direction of coriolis deflection of a rocket fired "up" from a point in the southern hemisphere. It would be deflected west. Addtionally, I raised two similar cases of the deflection for a body dropped from a height...

Brilliantly explained. Thank you and sorry for the delay in responding.

I should reply to you later when time and the advantage of having done some mathematics first permits. For the case is involved. The particle has acquired a speed (and deflection) due east projected up, after having reached its highest point (P in the northern hemisphere). On its way down, it...

Due to the coriolis force, the particle acquires a velocity due east (along X) on its way up. For a place in the northern hemisphere. Likewise, doesn't it acquire an equal and opposite velocity (along -X towards) west on its way down, that should result in no net horizontal displacement?

Yes good question. Sorry I was under the impression that, precisely for my thinking up there, namely that "going up in the northern" is the same as "going down in the southern" and vice versa, that I believed that the coriolis force would disappear in places on the equator. I was also "helped"...

Isn't going "up" in the south hemisphere, far as the coriolis force is concerned, the same as going "down" in the northern? (up and down being relative to the place(s) in question, relative to land)

Thank you all for your comments. I agree with the corrections to my drawing and with the expression of the coriolis acceleration, namely that the mass of the object must be absent : ##\vec a_C = -2(\vec\Omega_0\times\vec v)##. However, crucial thing is that the answer, that the coriolis force...

Yes. The Z plane goes vertically up from the place. Y is along the north direction from the place P. Rotation of the earth is in the Y-Z plane, same as the North-Up plane.