Ok, thanks for your suggestion.
Right, I need to solve equations (1) and (3) for two omegas only. That is, two of the omegas would be zero, whilst the other non-zero.
And the sum of the omegas must always be 1.
What I've written so far in Mathematica is:
sol = NSolve[r2[z] ==...
Homework Statement
hello, I have to solve a couple of equations on Mathematica for a project, and since they are not really working out, I wondered if you could help me out a bit.
I need to solve equation (1) (you can find the equations in the .doc attached) and plot the solution for a...
Homework Statement
Ok, I'm given a formula for the cross section of the scattering and I've been told that the detector is a column of water of depth 10m. I need to find the probability of the scattering within the detector.
Homework Equations
σ = E(in MeV) x 1.5x10-44 cm2
The...
Homework Statement
Ok, the problem is simple enough, I think. I just think I'm missing something obvious.
I have an equation involving the scale factor R(t) and need to integrate it.
I am at the first equation and need to get to the second by integrating (with respect of R, I suppose)...
because if you use <p> = m d<x>/dt, you would determine the expectation value of the momentum via that of x.
But <p> is given only by ∫u p u* dx - it's a definition.
From U=Q+W
U + -> Internal energy gained by system
U - -> Internal energy lost by system
Q + -> Heat entering the system
Q - -> Heat leaving the system
W + -> Work done on the system
W - -> Work done by the system
R.
parity is something very simple.
If you have an object (or curve, equation..) in x, y, z, let x-> -x, y-> -y, z-> -z.
If the body (or equation or curve) looks exactly the same as before, then parity is conserved.
Otherwise, it is violated.
R.
Ah, of course.
You do need to use energy conservation, but I told you the wrong thing.
total energy of system = potential energy + kinetic energy
1/2 k \Deltax2 = mgx + 1/2mv2
However, the velocity is maximum when the potential energy is 0.
(\Deltax is the displacement, x is the position...
It's you again :P
I recognised your "delta" !
For part a), how about using energies?
Set the initial potential energy, with h=\Deltaxextra and let mgh=1/2 mv2.
R.
You are absolutely, right. I beg your pardon.
However, my answer for m is completely different from yours.
I get m = 738kg.
Try solving again the second equation.
I would suggest you to square both sides so to get rid of the square root, if you hadn't done so already.
R.
c) doesn't require calculations, as it specifically asks to write the answer in symbols only.
Part d) is entirely down to tensions, weights and coefficient of friction.
For a start, think about what would actually happen when the bodies start moving. it should be easy enough to extrapolate the...
use k\Deltax = mg to find k. To do this, use the information given when the car is at rest. So m=120kg, \Deltax = 7.36cm .
Once you found k, plug the value in the other equation ( T = 2\pi ...) to find m.
R.