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Posts posted by nicobudini


Ok, let's see if this helps.
(Question 1)Show that a number which has 0 and 9 as fenders has at least four more fenders.
If a number has 0 and 9 as fenders then it must be divisible by 10 and 9, because you can't divide by 0 and 9 is the first number that ends with 9. Now:
 divisible by 10[math]\Longrightarrow[/math]divisible by 2 and by 5
 divisible by 9[math]\Longrightarrow[/math]divisible by 3
Until now you have 3 more fenders {2,3,5}, and the other is 1.
So the fenders are {1,2,3,5}, four at least.
Question 2, coming soon
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Corrected version (using LaTeX) of my previous post .
[math]
L=T+m_1gx_1+m_2gx_2+\frac{k}{2}(2x_1^2+x_2^22x_1x_2)
[/math]
where [math]T[/math] is the same kinetic term you wrote, and the other terms are minus the potential energy [math](U)[/math].
I calculate this supposing that the origin of coordinates is located at the roof and is positive downwards. I called [math]x_1[/math], and [math]x_2[/math] the positions of the masses from the origin.
For the first mass you have: [math]U_1=m_1gx_1\frac{k}{2}(x_1l)^2[/math].
And for the second mass: [math]U_2=m_2gx_2\frac{k}{2}(x_2x_1l)^2[/math].
Expanding these terms and just keeping terms of the form [math]x_ix_j[/math] you arrive to the expression I wrote above.
Observe that there is a term that couples both masses.
Greetings.
Nicolas
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[math]x^2_1[/math]
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Ok, I tried to calculate de Lagrangian and I arrived to this expression:
L = T + m1·g·x1 + m2·g·x2 +1/2·k·[2·x1^2 + x2^2  2·x1·x2]
where T is the same kinetic term you wrote, and the other terms are minus the potential energy (U).
I calculate this supposing that the origin of coordinates is located at the roof and is positive downwards. I called x1, and x2 the positions of the masses from the origin.
For the first mass you have: U1 =  m1·g·x1  1/2·k(x1  l)^2.
And for the second mass: U2 =  m2·g·x2  1/2·k·(x2  x1  l)^2.
Expanding these terms and just keeping terms of the form xi·xj you arrive to the expression I wrote above.
Observe that there is a term that couples both masses.
I am sorry, i don't know how to insert a formula as an image.
Hope this helps.
Greetings.
Nicolas
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Yes! I got it!
I briefly discussed your explanation, Severian, with a friend (he is studying fields theory) and I understood.
So it's indeed a relativistic effect. I first oriented my question to the quantic nature of the photon. Anyways, all is connected
Thanks again!
Nicolas.
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Thanks Severian! I understood so far... I've got some notions on relativity, fourvectors and all that. Although I didn't understood the whole equations I do see the physics behind.
But your demonstration doesn't finish right there, does it? If so I will have to read it several more times to get the point. If not, I will be waiting anxiously the last part of it.
Thanks again!
Greetings.
Nicolas.
0 
The Gibbs function is a Legendre Transformed of the internal energy function, G=UTS+PV where the the extensive variables, entropy and volume, are replaced by their corresponding intensive variables, temprerature and pressure, respectively.
Its differential: dG=S dT + V dP + mu dN
Hope this helps.
Nicolas.
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Once someone said: "If you don't get confused with quantum mechanics... then you didn't understand it"
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Ok, ok... do not applause anymore
That was just a question I kept in mind before a modern physics class.
Thanks a lot Tom, but could you explain why exactly does that imply the supression of the J=0 state?
Thanks in advance...
Nicolas.
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Hello...
I have a question about the spin of the photon.
I have heard that the photon is considered a spin=1 particle, and that is the electromagnetic forcecarrier.
My question is, why does the energy of a photon have a twofold degeneracy instead of a threefold given by the projections of the spin=1? Does this have to do with the fact that the photon is a massless particle?
And another question: Do all forcecarriers particles have to be bosons?
Greetings...
Nicolas.
0 
(1) how do I take an equation like 3cos(5 theta) and figure out how big an interval is for one loop. I know that there are 5 loops' date=' i just don't know how to get the interval.
[/quote']
First, the equation is r(theta)=3*cos(5*theta), isn't it?
The polar plot of this equation gives a 5 loops flowerlike graph. So you are right, there are 5 loops or petals. Each of the petals converges in r=0.
To get the interval of one loop you have to figure out the first value of theta for which r(theta)=0 and then multiply it by 2.
For example, the maximum value for r is r(0)=3
Now, r(theta)=0 for theta=Pi/10, cos(theta) is an even function so r(Pi/10)=0 too ==> the theta interval for one loop or petal is 2*Pi/10=Pi/5.
Hope this helps...
Greetings
Nicolas
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the reason of what?
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suppose we wait until random combinations of elements catch the right combination to create life. well, the time this process would take exceeds the age of the universe. some basics calculations of probabilities throw a total time of about 100,000 million years while the age of the universe is supposed to be of 15,000 millions.
by the way, some basics principles of physics tell us that the entropy tends to increase or the "order" tends to decrease in every process. what about that?
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diff. eq. how to solve this?
in Mathematics
Posted
(Dy)^2 stands for [math](\frac{dy}{dx})^2[/math] or [math]\frac{d^2y}{dx^2}[/math]?