Recent content by Kyle Nemeth

Understood, the substitutions I made were clearly not necessary and the fact that any equation with quotients can be written as an integer equation by multiplying the least common denominator is an approach I did not think of. I do not assert that this equation has no solutions, but rather any...

Understood. I am quite unsure exactly of the nature of a "rational solution," whether each ##a,b,c## have to all be rational or if a rational solution could be constituted by something like ##(a,b,c)## where only ##a,b \in \mathbb{Q}## and maybe ##c\in \mathbb{R}## for example. Either way, I...

My attempt at 14 (I'm a novice so bare with me!)

I should ask rather, instead of a vague "why is this the case," is it true that if the standard deviation of the Hamiltonian is non-zero for a particular state, that state is not a stationary state?

I am currently reading Griffiths Introduction to Quantum Mechanics, 2nd Edition. I am aware that, in light of considering potential functions independent of time, the Schrödinger equation has separable solutions and that these solutions are stationary states. I am also aware (If I stand correct)...

ASYMMETRIC, okay, well understood, thank you for your response.

Awesome, thank you, I've thought about this also, so if we choose a circle as our loop, or even a rectangle (so that the entire cross-section of the sheet is enclosed) it is true that both would be anti-symmetric with the geometry of the magnetic field formed by the sheet?

I am familiar with this problem, the Amperian Loop is chosen as a rectangle within the slab with a height smaller than the thickness of it (to find B at a point within the slab) and can then be chosen as a rectangle with a height larger than the thickness of the slab (to find B at a point above...

We are asked to find the magnetic field at point P, all of the quantities in the figure are known values and the current density is uniform. One way to solve this problem is by modeling the sheet as a collection of infinitely long wires, with each wire contributing an amount of magnetic field dB...

Awesome, thank you for your useful information.

Thank you for your input.

The two main principles you should have in mind for this problem are $$\sum{F}=0$$ $$\sum{F=ma}.$$ These two principles, provided by Sir Isaac Newton, can be applied to free body diagrams (individual graphical vector layouts of all of the forces and their components acting on a particular...

So upon investigating the units, it has turned out for me that, for the x-component of the speed, which would according to the solution be $$ v_x(t) = \sqrt{k}x_0 cos \sqrt{k}t, $$ the units would work out as $$ L\sqrt{\frac{C^2KgL^3}{KgT^2C^2}}= \sqrt{\frac{L^5}{T^2}}=\frac{L^2\sqrt{L}}{T}$$...

Honestly, my first approach was with polar coordinates and I found that to be much more cumbersome; my approach in regards to that aspect may have been off but that was what originally lead me to the cartesian analysis.

Hmm, okay so then, $$ -x_0ksin\sqrt{k}t = \frac {x_0ksin\sqrt{k}t}{\sqrt{(x_0^2sin^2\sqrt{k}t+y_0^2cos^2\sqrt{k}t)^3}}$$ and if we solve for the square root in the denominator, $$\sqrt{(x_0^2sin^2\sqrt{k}t+y_0^2cos^2\sqrt{k}t)^3} = -1 $$ so hence we have a solution to the system (the other DE...