i had to convert it from python to a word doc so i hope the indents are still intact!! I'm also relatively new to programming so forgive my messy set-up, the code wasn't marked just the report we've written on it so I didn't try making it 'pretty'!
I hope you can make sense of it!
I can't get it into my head why it wouldn't affect the rocket's orbital velocity as the path of the orbit can't be circular around the earth-moon system's centre of mass with the COM being about 3/4 way to the edge of the earth's radius!! Surely, the rocket must spiral slightly around the path...
Hi all! I've just finished writing a program in python that determines the position of a rocket in the earth moon system. Initially it is in circular orbit around the earth and then it is given a velocity boost at some point in it's orbit to make it travel to the moon.
I started all 3 bodies...
I strongly advise finding another editor to write your code in such as emacs, eclipse, sciTe... most are free downloads and they all feel different to use so it's a bit of trial and error to find one you like.
I prefer using SciTe on my home computer - it keeps all the colours and has options...
Thanks so far!! I have since been told that the values for peak 4 and 5 are wrong so to ignore them.
Therefore I got N values for the first three peaks of 3,4,8 which I know are right for a fcc cube so now I just have the second part which says I need to confirm my findings with structure...
Any help at all would be appreciated!!! I've spent ages searching for answers and the people that i've spolen to at uni have no clue either so i'd love to be able to pass on the knowledge!!!
Any help at all would be appreciated!!! I've spent ages searching for answers and the people that i've spolen to at uni have no clue either so i'd love to be able to pass on the knowledge!!!
1. Homework Statement
Using the powder XRD data below, show that the substance has a face centred cubic structure. (xray lamda = 0.154056 nm)
Peak No.------2(theta)
1 -------------38.06
2 -------------44.24
3 -------------64.34
4 -------------68.77
5 -------------73.07
2...
Homework Statement
Using the powder XRD data below, show that the substance has a face centred cubic structure. (xray lamda = 0.154056 nm)
Peak No.------2(theta)
1 -------------38.06
2 -------------44.24
3 -------------64.34
4 -------------68.77
5 -------------73.07
Homework...
Can anyone find a way to explain this to me or point me in the right direction? I've been searching online and my books for hours and the only examples that I find are the opposite way round!!!
I still don't get it!!
if i name the corners of the square a,b,c and d for (0,0,1), (0,1,1), (1,1,1) and (1,0,1) respectively, then will dr be:
a->b: (0,dy,0)
b->c: (dx,0,0)
c->d: (0,-dy,0)
d->a: (-dx,0,0)?
But then when I dot product with f=(2z^2,6x,0) and integrate each seperately...
Homework Statement
Let: \vec{F}(x,y,z) = (2z^{2},6x,0), and S be the square: 0\leq x\leq1, 0\leq y\leq1, z=1.
a) Evaluate the surface integral (directly):
\int\int_{S}(curl \vec{F})\cdot\vec{n} dA
b) Apply Stokes' Theorem and determine the integral by evaluating the corresponding...
Homework Statement
A 4m telescope has a 16bit controllr with a gain of 1 and must detect faint stars(M>15) without saturating the bright stars(M=8-10).
(i) What is the max no of photo-electrons that can be registered before saturating the CCD?
(ii) The zero point of the system is...
Ok, thanks for that (it was just a case of multiplying instead of dividing by a fraction!!)
Now I have: <p> = 0
and <p^{2}> = -\frac{\pi^{2}\hbar^{2}}{L^{2}}
Please let this be right before I go insane!!!
I promise this is the end!!!!
I now have a value of:
\frac{-\hbar \sqrt{1/L}\pi^{2}}{L^{\frac{5}{2}}}
for the expectation value of the momentum operator squared. Could this be right? It seems very long winded to me!!
The last part of my question says that if the wave function is...
silly me!!! So the answer is: \frac{-iL\hbar}{2}??
Now I have to find the expectation value of the momemtum operator squared so I may be back!!!!
Thanks for all your help so far! :smile:
That all makes sense to me and I think I could do it if I knew what the 'ground state' wave function was!!
Do I just use psi = root(2/L)*sin((pi*x)/L)
or do i need to use e^i..... as the expectation value includes a complex conjugate?
Homework Statement
Consider a particle of mass, m inside the potential:
V(x) = 0 for 0<x<L and infinite Otherwise
(i) Write down the normalised ground state and first excited state energy eigenfunctions?
(ii) Use the ground state wave function to calculate the expectation value of the...
I don't know how where i was going with that intgral!! I started again from scratch following your advice and every thing seemed to fall into place itself so thanks to both of you for your help!:smile:
Ok! I redid the integration for the wave function and got N=\sqrt{\frac{a}{2}} now, which looks alot better and my expectation value of x is still zero so that's good too!! My problem now is that when i try to calculate the expectation value of x^2, (using integration by parts) I just keep...
normalize the wave function and more!! Please help!!!
Homework Statement
i) Normalize the wave function
ii) Calculate <x>
iii) Calculate <x^{2}>
iv) What would happen if a < 0?
Homework Equations
\psi\left(x\right) = N\left(1+i\right)exp\left(-a|x|\right), for -inf...
Thanks for all your help everyone! I made the stupidest mistake possible and worked out that
\frac{y^3+1}{2} was 0 when y was 0!!!
Why is it always something so simple that gets you!!??
Well, it's all sorted now but I'm sure I'll be back sometime soon!
I've been going over this again and still can't come up with the answer that maple (and an online definite integral calculator!!) gives of 0.22485. I've tried having my calculator in radians (=0.306465) in degrees (=-1.10537 * 10^-4)
Can anyone see what I'm doing wrong - I've never had a...
\int_{0}^{1} \int_{0}^{y^2} \sin{\frac{y^3+1}{2} dxdy
=\int_{0}^{1} [x\sin{\frac{y^3+1}{2}]^y^2_0 dy
=\int_{0}^{1} [y^2\sin{\frac{y^3+1}{2}] dy
then by inspection:
=\frac{2}{3}\int_{0}^{1}\frac{d}{dy}(-cos{\frac{y^3+1}{2}}) dy
=[\frac{2}{3}(-cos{\frac{y^3+1}{2}})]^1_0
=...
thanks for the fast response!! I was just confused by my answer which was 0.666565 (6dp) as normally our problems nearly always end up as obvious fractions.
I obviously rounded up to 2/3 though!!
I'll try and input my working out by reading that post if you or anyone else can check it for me!