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towr
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Re: various physics questions
« Reply #76 on: Dec 6th, 2008, 6:21am » |
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on Dec 5th, 2008, 8:52pm, BenVitale wrote:The article doesn't really tell us anything; so there's not much to think about.
Since the article is over a year old, I'd look up the original publication and consequent responses to it. At the very least it should tell you something about what actually happened in the experiment.
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rmsgrey
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Re: various physics questions
« Reply #77 on: Dec 6th, 2008, 10:07am » |
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A quick look on Wikipedia finds:
Quote:| However, other physicists say that this phenomenon does not allow information to be transmitted faster than light. Aephraim Steinberg, a quantum optics expert at the University of Toronto, Canada, uses the analogy of a train traveling from Chicago to New York, but dropping off train cars at each station along the way, so that the center of the train moves forward at each stop; in this way, the center of the train exceeds the speed of any of the individual cars. |
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Benny
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Re: various physics questions
« Reply #78 on: Dec 7th, 2008, 10:20am » |
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Thank you Towr and rmsgrey.
Didn't the physicits at Berkeley produce an experiment where a single photon tunnelled thru a barrier and its tunneling speed was 1.7 the speed of light?
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Gabriel
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Re: various physics questions
« Reply #79 on: Apr 10th, 2009, 2:36pm » |
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Hello guys, I'm new to this forum, so I don't really know whether this is the correct place to post a physics problem, which I have to solve. I'm kinda stuck in that,and I hope someone can give me a little help. My scientific english isn't so good, but I will try to explain the problem:
We have N nodes (we can think of them as vertice of a graph). We put resistors between them randomly, but
1. Resistors have the same resistance: R.
2. Between 2 nodes the number of maximal resistors is 1.
3. The graph doesn't have any isolated vertice/ isolated sets of vertice. (So you can walk from any of the vertice to any other of them via resistors)
In result, the number of connecting resistances to a vertex can vary from 1 to N-1.
Measure the resistance between the endpoints of all of your resistors, and sum them up. What's the result you'll get?
| hidden: | | (N-1)*R, I just can't prove it in a general situation |
I would be very grateful if someone could help.
Regards,
Gabriel
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| « Last Edit: Apr 10th, 2009, 4:33pm by Gabriel » |
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towr
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Re: various physics questions
« Reply #80 on: Apr 10th, 2009, 2:47pm » |
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on Apr 10th, 2009, 2:36pm, Gabriel wrote:| Measure the resistance between the endpoints of all of your resistors, and sum them up.l |
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Does that mean between all nodes, or between all nodes that have just one resistor connected to them, or ... ?
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Gabriel
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on Apr 10th, 2009, 2:47pm, towr wrote:
Does that mean between all nodes, or between all nodes that have just one resistor connected to them, or ... ? |
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This means to measure resistance between every two nodes, which are connected with a resistor directly.
This means on my figure: Measure resistance between AE, EC, CB, BD and DC nodes, and sum them up. You will get 4R.
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| « Last Edit: Apr 10th, 2009, 4:40pm by Gabriel » |
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Eigenray
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Re: various physics questions
« Reply #82 on: Apr 11th, 2009, 12:20am » |
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Interesting problem. To phrase it in terms of matrices:
Suppose there are m edges connecting n vertices. Let A be the sum of the matrices
E(i,i) + E(j,j) - E(i,j) - E(j,i)
over each edge (i,j), where E(i,j) is the nxn matrix with a single 1 in position (i,j). Let Ai denote the (n-1)x(n-1) matrix obtained by deleting the i-th row and i-th column of A. Then the resistance between nodes i and j is given by the (j,j)-th entry of Ai-1. So let R be the nxn matrix whose i-th row is given by the diagonal elements of Ai-1, with a 0 inserted in position i; thus R(i,j) is the resistance between nodes i and j, and the problem is to show that
 Ai,j*Ri,j = -2(n-1).
This generalizes as follows: suppose we have an edge between every pair (i, j) with conductance ci,j (resistance 1/ci,j). Let
A = ci,j ( E(i,i) + E(j,j) - E(i,j) - E(j,i) ),
and R be as above. Now we need to show
Ai,j*Ri,j = -2(n-1),
the same as before. Now this is just an identity of rational functions in n(n-1)/2 variables, so it can be proven for any given n at least.
Code:n = 5;
e[i_,j_] := Array[ If[ #1==i && #2==j, 1, 0]&, {n,n}];
A = Sum[ c[i,j](e[i,i]+e[j,j]-e[i,j]-e[j,i]), {i,1,n},{j,1,i-1}];
diag[A_] := Array[ A[[#, #]]&, Length@A ];
R = Table[Insert[diag@MatrixPower[Drop[Drop[#,{i}]&/@A,{i}],-1],0,i],{i,n}];
Simplify[ Total[-Flatten[A*R]]/2 == n-1 ]
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I'm not sure why this is true yet. In the case where all the conductances are 0 or 1, the matrix A is just the Laplacian of G, and each submatrix Ai has the same determinant, namely the number of spanning trees of G. In general, I guess it is
T  (i,j)  T ci,j,
where the sum is over all spanning trees T of Kn. So this is a common denominator for the entries of R.
Suppose we wanted the sum of the resistances between every pair of vertices, not just those which are connected. That is, the (unweighted) sum, Ri,j. Now, the sum of the i-th row of R is just the trace of Ai-1, which is the sum of the roots of det( t I - Ai-1 ), or the sum of the reciprocals of the roots of the characteristic polynomial det ( t I - Ai), i.e., 1/det(Ai) times the coefficient of t. So
Sum(R) = tr(Ai-1)
= [ 1/det(Ai) * { coefficient of t in det ( t I - Ai ) } ]
= 2/det(A1) * { coefficient of t2 in det ( t I - A ) }
= 2/det(A1) *  1/
= 2 * N * 1/,
where are the non-zero eigenvalues of A, since for all i, det(Ai) = 1/N . Of course, this counts every pair twice, so the sum of all resistances is just N * 1/.
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| « Last Edit: Apr 11th, 2009, 1:59am by Eigenray » |
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Gabriel
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Re: various physics questions
« Reply #83 on: Apr 11th, 2009, 5:12am » |
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on Apr 11th, 2009, 12:20am, Eigenray wrote:| Then the resistance between nodes i and j is given by the (j,j)-th entry of Ai-1. |
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I didn't know this one, and I didn't find it anywhere. Could you prove this, or at least give some pieces of advice about how can I prove it?
edit: ok, I'm one step closer, I found: http://mathworld.wolfram.com/ResistanceDistance.html 
But in my opinion (3) there has some missing parts, the formula is incomplete (just have a look at (2)).
But still I don't know any proof of the theorem.
Some interesting things:
If we allow multiple edges, the statement is still true (I tried it numerically with a 10x10 matrix)
If we measure resistance at every two nodes, the result will be k*R, where k is the number of resistors. Even if there are multiple edges too.
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| « Last Edit: Apr 11th, 2009, 9:32am by Gabriel » |
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Eigenray
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Re: various physics questions
« Reply #84 on: Apr 11th, 2009, 2:46pm » |
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on Apr 11th, 2009, 5:12am, Gabriel wrote:| I didn't know this one, and I didn't find it anywhere. Could you prove this, or at least give some pieces of advice about how can I prove it? |
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It's the matrix form of Kirchoff's law:
Say there are m edges and n vertices. Let B be the m x n incidence matrix of G. If the i-th edge is (a,b), then the i-th row of B has 1 in position b and -1 in position a; that is,
B = e_i=(a,b) E(i,b) - E(i,a).
The choice of orientation doesn't really matter, it just needs to be fixed. Let pa be the potential at vertex a. Then the voltage drop across edge i = (a,b) is
vi = pa - pb, or
v = -B p
If this edge has conductance ci, then assuming there are no batteries, the current through the edge is
yi = ci vi, or
y = Cv = -CBp,
where C is the m x m diagonal matrix with Ci,i = ci.
Now, there is a net flow fa into vertex a, where
fa = e_i=(*,a) yi - e_i=(a,*) yi, i.e.,
f = Bt y = - BtCB p = - A p,
where A is the same matrix from before, the Laplacian of G.
Say we want to find the resistance between vertex j and vertex n. Fix the potential of vertex n to be 0, and drop the n-th row and n-th column of B. This results in the (n-1) x (n-1) matrix An from before. Now we pull a current of 1 out of vertex j, i.e., let f = ej, and find the potential there, i.e., find pj so that An * p = -ej. The result is
R(n,j) = -pj = (An)-1ej.
Therefore, the resistances from vertex n to every other vertex are given by the diagonal elements of An-1. We get a similar result by dropping any other row and column.
Quote:
Interesting. I'm not quite sure why that gives the same result, but it looks like it might be easier to work with. Also note the formulas here, which contains the result you want.
Quote:
Some interesting things:
If we allow multiple edges, the statement is still true (I tried it numerically with a 10x10 matrix)
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If we put k resistors in parallel between vertices i and j, this results in a conductance ci,j = k. So to sum the effective resistance over each resistor we just take ci,j*R(i,j).
If this sum is identically 2(n-1), as a rational function of the ci,j, then it will be true for any values of the resistances.
Quote:
If we measure resistance at every two nodes, the result will be k*R, where k is the number of resistors. Even if there are multiple edges too.
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Are you sure about that? Even if we just took two vertices with k resistors in parallel between them the sum should be 1/k. Or if we took a triangle, with c1,2 = c1,3 = 1, c2,3 = c, then the sum of resistances should be
(2c+4)/(2c+1),
which goes from 4, when c=0, to 1, when c= .
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Gabriel
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Re: various physics questions
« Reply #85 on: Apr 12th, 2009, 3:28am » |
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Thank you very much, I think I understand it now.
I found the wikipedia link later too, but it doesn't prove it's statements. If I could prove that the sum of (LML)i,j*Ri,j equals to the trace of (LM) multiplied by (-2), then I could prove statement of my problem too, since if we use ML=(1/n*J-I), then we get our statement (I is the identity matrix, and J is the unit nxn matrix, which has 1-s at all it's entries).
on Apr 11th, 2009, 2:46pm, Eigenray wrote:
Are you sure about that? Even if we just took two vertices with k resistors in parallel between them the sum should be 1/k. Or if we took a triangle, with c1,2 = c1,3 = 1, c2,3 = c, then the sum of resistances should be
(2c+4)/(2c+1),
which goes from 4, when c=0, to 1, when c=. |
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I messed it up a little bit. I calculated the sum of resistances over the edges with a for loop in octave, taking the sum of Ri,j*Ai,j (A was the adjacency matrix for me). Then, when I wanted to calculate it over all edges, I accidentally deleted Ri,j instead of Ai,j, so I calculated the sum of the adjecency matrix's entries, which is actually the spur (trace) of the Laplacian*(-1), the number of resistors multiplied by -2. Stupid mistake 
Maybe I can prove the formula of Mathworld...I give it a try.
I will use your notations, but let me use Einstein summation convention, because I can't make the Sum symbol .
Our aim is to determine resistance between vertice i and j. Hence, we pump f=(0,0...,I,0...,0, - I,0...) current in the system: I at vertice i and -I at vertice j. According to your calculations, we have f= -A*(p1, p2,...,pn). From this, we can derive the following formulas by calculating fi and fj:
I = - Aikpk
- I = -Ajkpk
Now, let me use T=A+1/n*J notation, where T is gamma at mathworld, and J is the unit nxn matrix. The following can be easily derived:
(A+1/n*J)p=Ap+(p1+p2+...+pn)*'/n*(1,1,1,1...1)
That means, we have:
-(p1+p2+...+pn)*1/n*(1,1,1,1...1)
+(0,0...,I,0...,0, - I,0...)= -T*p
so
(T^-1)((0,0...,I,0...,0, - I,0...)+(p1+p2+...+pn)*1/n*(1,1,1,1...1)) = p
Now we prove, that Tk,1+Tk,2+...+Tk,n=1
It's quite easy to prove. Assume, that I = 0. Then one of our equations changes to:
(T^-1)(p1+p2+...+pn)*1/n*(1,1,1,1...1) = p
If there isn't any current flowing into the system, the pontentials have to be equal to each other, which means Tk,1+Tk,2+...+Tk,n=1.
We almost reached our final result, because we can write
pi=(T^-1)ii*(-I)+(T^-1)ij*I+(p1+p2+...+pn)*1/n*((T^-1)i,1+(T^-1)i,2+...+(T^-1)i ,n)
And similarly:
pj=(T^-1)ji*(-I)+(T^-1)jj*I+(p1+p2+...+pn)*1/n*((T^-1)j,1+(T^-1)j,2+...+(T^-1)j ,n)
By substrabting them, we get:
(pj-pi) / I=(T^-1)ii+(T^-1)jj-2*(T^-1)ij
Ri,j=(T^-1)ii+(T^-1)jj-2*(T^-1)ij
QED
I hope I didn't make much mistakes, and that my explanation is understandable, because I'm quite tired now.
At last I think I managed to understand and prove all parts of the problem, I just need the arrange my thoughts. I will post the solution of the problem in a few days, when I will have a little time.
Thanks for help!
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| « Last Edit: Apr 12th, 2009, 4:34pm by Gabriel » |
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Benny
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Re: various physics questions
« Reply #86 on: May 15th, 2009, 4:05pm » |
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I'm curious about Benford's law
http://en.wikipedia.org/wiki/Benford's_law
Some have made strong claims. Aren't they over the top?
See articles (at the bottom of wiki page):
http://www.physorg.com/news160994102.html
And the law of digits at
http://www.physorg.com/news98015219.html
Didn't the authors just find some pattern on a small scale that somehow worked and extrapolated with the Benford's law?
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| « Last Edit: May 15th, 2009, 4:06pm by Benny » |
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towr
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Re: various physics questions
« Reply #87 on: May 16th, 2009, 2:52am » |
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on May 15th, 2009, 4:05pm, BenVitale wrote:| Some have made strong claims. Aren't they over the top? |
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The wiki article gives a number of explanations for why it occurs.
Quote:| Didn't the authors just find some pattern on a small scale that somehow worked and extrapolated with the Benford's law? |
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It looks to me that in the prime number case they showed it followed from a more general theorem. So while they might have found the pattern on a small scale and extrapolated -- as one often does in science -- they also showed it held in general.
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Benny
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Re: various physics questions
« Reply #88 on: Jul 22nd, 2009, 2:38pm » |
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I've worked on a Fermi problem. I'm gonna post a proposed solution posted on the Web ... I've reached a different solution, though.
The Fermi problem :
When you take a single breath, how many molecules of gas you intake would have come from the dying breath of Caesar?
Please read the proposed solution:
http://www.hk-phy.org/articles/caesar/caesar_e.html
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Benny
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Re: various physics questions
« Reply #89 on: Jul 22nd, 2009, 2:42pm » |
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Why did the author estimate thickness of the atmosphere to be 50 X 103?
The thickness of the atmosphere is more difficult to pin down because its density and pressure aren't the same all the way up.
How Thick is the Earth's Atmosphere? it is not a very good question
The air at the ground level is squeezed by the weight of all the air on top of it, so it's quite dense.
The air at Mount Everest's summit doesn't have so much air above it. As a result its atmosphere never really ends, and the air just gets thinner and thinner.
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Benny
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Re: various physics questions
« Reply #90 on: Jul 22nd, 2009, 2:50pm » |
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We need a number for its thickness.
I read that if we think in terms of an "effective thickness" of the atmosphere, that is to say, how high it would be if its density all the way up were the same as it is at the surface.
I was told that there's some math that shows that the effective thickness of the atmosphere is the height at which the pressure of the real atmosphere has dropped to 37% of its surface value. That turns out to be about the height of Everest, which is close to 6 miles.
Anybody familiar with this?
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rmsgrey
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Re: various physics questions
« Reply #91 on: Jul 22nd, 2009, 3:22pm » |
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on Jul 22nd, 2009, 2:50pm, BenVitale wrote:| We need a number for its thickness. |
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We don't need a number for the depth of the atmosphere - just for the number of gas molecules it contains. The equivalent volume of the atmosphere if it contained the same number of molecules but at uniform pressure is only useful as a way of finding that number...
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Benny
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Re: various physics questions
« Reply #92 on: Jul 22nd, 2009, 4:21pm » |
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Quote:
do you agree with the author's estimation of the volume of earth's atmosphere?
I had trouble posting the rest of my work .. I'll do it later
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| « Last Edit: Jul 22nd, 2009, 4:30pm by Benny » |
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Re: various physics questions
« Reply #93 on: Jul 22nd, 2009, 10:49pm » |
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A deep breath amounts to 1 liter.
Then there's the volume of earth's atmosphere:
that's the area of our planet multiplied by the height of the atmosphere.
I multiplied the earth's area by the effective thickness of the atmosphere gives a volume of 1,200 million cubic miles.
1200 X 4.16818183 = 5,001,818,200
~ 5 X 1021, or in shorthand notation 5e21
So if Caesar's last breath had a volume of 1 liter, it forms one part in 5e21 of the air we breathe.
If your last breath was 1 liter, you just breathed in 1 X 1/5e21 = 1/5e21 liters of Caesar's last breath.
All that remains is to estimate how many molecules there are in this tiny volume.
It's done using an important number : the Avogadro number, that is, 1 liter of any gas contains around 2.7e22 molecules.
The lungful of air you breathed contains 2.7e22 X 1/5e21 = 2.7e22/5e21 of the molecules that were in Julius Caesar's last gasp, and that works out to be 5.4
It's an average.
However, the author found that the number of the molecules that comes Caesar's last exhalation is 1 molecule
What do you think? Am I wrong?
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| « Last Edit: Jul 22nd, 2009, 11:02pm by Benny » |
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towr
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Re: various physics questions
« Reply #94 on: Jul 23rd, 2009, 12:30am » |
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on Jul 22nd, 2009, 10:49pm, BenVitale wrote:| A deep breath amounts to 1 liter. |
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Actually, I think it's up to 4 times as much. Vital capacity is 4.6L for men, 3.6L for women.
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SMQ
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Re: various physics questions
« Reply #95 on: Jul 23rd, 2009, 5:03am » |
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on Jul 22nd, 2009, 10:49pm, BenVitale wrote:Then there's the volume of earth's atmosphere:
that's the area of our planet multiplied by the height of the atmosphere. |
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But the atmosphere is much denser near the planet than higher up! A better estimate would be to multiply average barometric pressure by the surface area of the planet to get the total weight of the atmosphere, then convert to mass and finally to mols.
--SMQ
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towr
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Re: various physics questions
« Reply #96 on: Jul 23rd, 2009, 5:09am » |
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on Jul 23rd, 2009, 5:03am, SMQ wrote:| A better estimate would be to multiply average barometric pressure by the surface area of the planet to get the total weight of the atmosphere |
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But objects weigh less the further they are from the center of gravity of the attracting body!
(Yeah, I know, it's a negligible difference in this case; certainly considering all the uncertainties we are dealing with already)
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Eigenray
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Re: various physics questions
« Reply #97 on: Jul 23rd, 2009, 5:27am » |
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on Jul 23rd, 2009, 12:30am, towr wrote:
Actually, I think it's up to 4 times as much. Vital capacity is 4.6L for men, 3.6L for women.
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But how deep a breath can you really take after being fatally stabbed in the chest?
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towr
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Re: various physics questions
« Reply #98 on: Jul 23rd, 2009, 6:13am » |
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on Jul 23rd, 2009, 5:27am, Eigenray wrote:
But how deep a breath can you really take after being fatally stabbed in the chest? |
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There are two breaths involved in the calculation. I wasn't considering Caesar's; but if I did, I'd hazard to guess his dying breath would have been considerably less than 1L. But shouldn't we consider his entire dying lung-volume?
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Benny
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Re: various physics questions
« Reply #99 on: Jul 23rd, 2009, 10:52am » |
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So, let's suppose it is 4 liters.
We would get: 5.4 molecules / 4 = 1.35 molecules
It doesn't make sense to a fraction of a molecule.
This a Fermi problem, and Fermi wants us to deal with estimates.
So, we end up with 1 molecule. What do you say, guys?
P.S. I did multiply the earth's area by the effective thickness of the atmosphere.
The effective thickness of the atmosphere being around 6 miles (Everest's altitude).
The thickness of the atmosphere is more difficult to figure outsince its density and pressure are not the same all the way up. The air at the ground level is squeezed by the weight of all the air on top of it, so it's quite dense.
So if we think "effective thickness" as how high it would be if its density all the way up were the same as it is at the surface. This number turns out to be 6 miles.
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| « Last Edit: Jul 23rd, 2009, 11:02am by Benny » |
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