Orthogonality has something to do with it for sure but I'm not positive about the relationships it has with the solutions.
The final wave functions for the different bonding/antibonding molecular orbitals must all be symmetric in some fashion.
There is a reason why c1 or c2 or c3= 0. I know...
Yeah I see what you're saying. There actually are more laws for MO theory. They revolve around the requirement that the systems be symmetric for homoatomic constructs, of which this is one. The solutions you provided are mathematically valid but they don't abide by that.
This was my main...
Right. Also by the Frost model you can see that there should be two degenerate orbitals. I understand that part.
Are you saying that it's essentially okay to solve it intuitively instead of mathematically? Because I have a system of two equations each with three variables, and obviously as you...
Homework Statement
Solve for the wavefuntions using the Secular Determinant, of cyclopropene anion/cation/whocares.
Homework Equations
Secular determinant:
| x 1 1|
| 1 x 1|
|1 1 x |
The Attempt at a Solution
I'm literally understanding every aspect of this concept, but...
This is going to sound like weird advice, but you're not thinking about the problem right. Don't think "I know that in the end I have to..." Just work through it.
Your equations are sound. Let's start with this: you know the maximum height right? What can you find with that if you know the...
qvB = mv^2/r
qB = mv/r
r = mv/qB
period = 2pir/v
period = 2Pm/qB
so, as mass increases, period increases, as charge increases period decreases.
each have the same mass and charge, so each have the same period in the same field. Period is independent of radii.
All your equations are good. But I'm not sure what you're asking. Also as gneill said you didn't state the orientation of the field, but I'm assuming that's your variable.
Fb = qv CROSS B, v and B are vectors. Meaning that the least force your particle will experience due to B is when B is...
Uh okay, but as gneill said, the displacement is opposite to the work being done on it, therefore the work is negative. I don't know what your teachers want (maybe the magnitude, but that's kinda silly), but regardless, gravity does negative work on the astronaut in this problem.
I was pointing out that you over complicated things. Sorry if I over complicated my response LOL :)
Anyway, here's what I'm trying to say.
The astronaut originally has 0 energy (0 U and 0 K) when in the water right? The helicopter does work on the astronaut, which is countered by the work...
You're over complicating things.
W = SFdx, or for a constant force W = Fx
Wgravity = mgh, where m is the mass of the astronaut, g is the acceleration of gravity, and h is the distance traveled. Work will be negative, though, since the force of gravity is acting opposite to the displacement of...
Yeah I'm an idiot, completely confused myself with voltage equations and stuff.
Its just -kQq/2r (orbital energy), and the difference is what I'm looking for.
Sorry for annoying you guys with this... thanks
Homework Statement
Hi everyone and thank you in advance for your time. I just had this problem on a physics exam (that everyone in the class bombed, and I mean everyone, including the best students). I honestly couldn't care less about the grade, but I really want to understand where I went...
You can solve this one of two ways:
Energy approach: Energy at highest point is equal to the kinetic energy at the lowest point, since all potential energy is converted to kinetic as the ball is about to hit the ground.
Kinematics approach: Use the time independent kinematics equation.
Post...
I assume "by the same point as the landing point" means the same y coordinate. If the point was the same the gun would be firing straight up.
Anyway, since the gun is at a 45 degree angle, you can find the y component of the initial velocity, and you know the acceleration of gravity. Just use...