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Topic: various physics questions (Read 16835 times) 

towr
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Re: various physics questions
« Reply #76 on: Dec 6^{th}, 2008, 6:21am » 
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on Dec 5^{th}, 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 6^{th}, 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 7^{th}, 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 10^{th}, 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 N1. Measure the resistance between the endpoints of all of your resistors, and sum them up. What's the result you'll get? hidden:  (N1)*R, I just can't prove it in a general situation  I would be very grateful if someone could help. Regards, Gabriel

« Last Edit: Apr 10^{th}, 2009, 4:33pm by Gabriel » 
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towr
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Re: various physics questions
« Reply #80 on: Apr 10^{th}, 2009, 2:47pm » 
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on Apr 10^{th}, 2009, 2:36pm, Gabriel wrote:Measure the resistance between the endpoints of all of your resistors, and sum them up.l 
 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 10^{th}, 2009, 2:47pm, towr wrote: Does that mean between all nodes, or between all nodes that have just one resistor connected to them, or ... ? 
 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.

« Last Edit: Apr 10^{th}, 2009, 4:40pm by Gabriel » 
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Eigenray
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Re: various physics questions
« Reply #82 on: Apr 11^{th}, 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 A_{i} denote the (n1)x(n1) matrix obtained by deleting the ith row and ith column of A. Then the resistance between nodes i and j is given by the (j,j)th entry of A_{i}^{1}. So let R be the nxn matrix whose ith row is given by the diagonal elements of A_{i}^{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 A_{i,j}*R_{i,j} = 2(n1). This generalizes as follows: suppose we have an edge between every pair (i, j) with conductance c_{i,j} (resistance 1/c_{i,j}). Let A = c_{i,j} ( E(i,i) + E(j,j)  E(i,j)  E(j,i) ), and R be as above. Now we need to show A_{i,j}*R_{i,j} = 2(n1), the same as before. Now this is just an identity of rational functions in n(n1)/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,i1}]; 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 == n1 ] 
 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 A_{i} has the same determinant, namely the number of spanning trees of G. In general, I guess it is _{T} _{(i,j) T} c_{i,j}, where the sum is over all spanning trees T of K_{n}. 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, R_{i,j}. Now, the sum of the ith row of R is just the trace of A_{i}^{1}, which is the sum of the roots of det( t I  A_{i}^{1} ), or the sum of the reciprocals of the roots of the characteristic polynomial det ( t I  A_{i}), i.e., 1/det(A_{i}) times the coefficient of t. So Sum(R) = tr(A_{i}^{1}) = [ 1/det(A_{i}) * { coefficient of t in det ( t I  A_{i} ) } ] = 2/det(A_{1}) * { coefficient of t^{2} in det ( t I  A ) } = 2/det(A_{1}) * 1/ = 2 * N * 1/, where are the nonzero eigenvalues of A, since for all i, det(A_{i}) = 1/N . Of course, this counts every pair twice, so the sum of all resistances is just N * 1/.

« Last Edit: Apr 11^{th}, 2009, 1:59am by Eigenray » 
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Gabriel
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Re: various physics questions
« Reply #83 on: Apr 11^{th}, 2009, 5:12am » 
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on Apr 11^{th}, 2009, 12:20am, Eigenray wrote:Then the resistance between nodes i and j is given by the (j,j)th entry of A_{i}^{1}. 
 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.

« Last Edit: Apr 11^{th}, 2009, 9:32am by Gabriel » 
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Eigenray
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Re: various physics questions
« Reply #84 on: Apr 11^{th}, 2009, 2:46pm » 
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on Apr 11^{th}, 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? 
 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 ith edge is (a,b), then the ith 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 p_{a} be the potential at vertex a. Then the voltage drop across edge i = (a,b) is v_{i} = p_{a}  p_{b}, or v = B p If this edge has conductance c_{i}, then assuming there are no batteries, the current through the edge is y_{i} = c_{i} v_{i}, or y = Cv = CBp, where C is the m x m diagonal matrix with C_{i,i} = c_{i}. Now, there is a net flow f_{a} into vertex a, where f_{a} = _{e_i=(*,a)} y_{i}  _{e_i=(a,*)} y_{i}, i.e., f = B^{t} y =  B^{t}CB 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 nth row and nth column of B. This results in the (n1) x (n1) matrix A_{n} from before. Now we pull a current of 1 out of vertex j, i.e., let f = e_{j}, and find the potential there, i.e., find p_{j} so that A_{n} * p = e_{j}. The result is R(n,j) = p_{j} = (A_{n})^{1}e_{j}. Therefore, the resistances from vertex n to every other vertex are given by the diagonal elements of A_{n}^{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) 
 If we put k resistors in parallel between vertices i and j, this results in a conductance c_{i,j} = k. So to sum the effective resistance over each resistor we just take c_{i,j}*R(i,j). If this sum is identically 2(n1), as a rational function of the c_{i,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. 
 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 c_{1,2} = c_{1,3} = 1, c_{2,3} = 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 12^{th}, 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}*R_{i,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*JI), then we get our statement (I is the identity matrix, and J is the unit nxn matrix, which has 1s at all it's entries). on Apr 11^{th}, 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 c_{1,2} = c_{1,3} = 1, c_{2,3} = c, then the sum of resistances should be (2c+4)/(2c+1), which goes from 4, when c=0, to 1, when c=. 
 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 R_{i,j}*A_{i,j} (A was the adjacency matrix for me). Then, when I wanted to calculate it over all edges, I accidentally deleted R_{i,j} instead of A_{i,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*(p_{1}, p_{2},...,p_{n}). From this, we can derive the following formulas by calculating f_{i} and f_{j}: I =  A_{ik}p_{k}  I = A_{jk}p_{k} 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+(p_{1}+p_{2}+...+p_{n})*'/n*(1,1,1,1...1) That means, we have: (p_{1}+p_{2}+...+p_{n})*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...)+(p_{1}+p_{2}+...+p_{n})*1/n*(1,1,1,1...1)) = p Now we prove, that T_{k,1}+T_{k,2}+...+T_{k,n}=1 It's quite easy to prove. Assume, that I = 0. Then one of our equations changes to: (T^1)(p_{1}+p_{2}+...+p_{n})*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 T_{k,1}+T_{k,2}+...+T_{k,n}=1. We almost reached our final result, because we can write p_{i}=(T^1)_{ii}*(I)+(T^1)_{ij}*I+(p_{1}+p_{2}+...+p_{n})*1/n*((T^1)_{i,1}+(T^1)_{i,2}+...+(T^1)_{i ,n}) And similarly: p_{j}=(T^1)_{ji}*(I)+(T^1)_{jj}*I+(p_{1}+p_{2}+...+p_{n})*1/n*((T^1)_{j,1}+(T^1)_{j,2}+...+(T^1)_{j ,n}) By substrabting them, we get: (p_{j}p_{i}) / I=(T^1)_{ii}+(T^1)_{jj}2*(T^1)_{ij} R_{i,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!

« Last Edit: Apr 12^{th}, 2009, 4:34pm by Gabriel » 
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Benny
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Re: various physics questions
« Reply #86 on: May 15^{th}, 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?

« Last Edit: May 15^{th}, 2009, 4:06pm by Benny » 
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towr
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Re: various physics questions
« Reply #87 on: May 16^{th}, 2009, 2:52am » 
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on May 15^{th}, 2009, 4:05pm, BenVitale wrote:Some have made strong claims. Aren't they over the top? 
 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? 
 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 22^{nd}, 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.hkphy.org/articles/caesar/caesar_e.html


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Benny
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Re: various physics questions
« Reply #89 on: Jul 22^{nd}, 2009, 2:42pm » 
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Why did the author estimate thickness of the atmosphere to be 50 X 10^{3}? 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 22^{nd}, 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 22^{nd}, 2009, 3:22pm » 
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on Jul 22^{nd}, 2009, 2:50pm, BenVitale wrote:We need a number for its thickness. 
 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 22^{nd}, 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

« Last Edit: Jul 22^{nd}, 2009, 4:30pm by Benny » 
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Benny
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Re: various physics questions
« Reply #93 on: Jul 22^{nd}, 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 10^{21}, 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?

« Last Edit: Jul 22^{nd}, 2009, 11:02pm by Benny » 
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towr
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Re: various physics questions
« Reply #94 on: Jul 23^{rd}, 2009, 12:30am » 
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on Jul 22^{nd}, 2009, 10:49pm, BenVitale wrote:A deep breath amounts to 1 liter. 
 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 23^{rd}, 2009, 5:03am » 
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on Jul 22^{nd}, 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. 
 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 23^{rd}, 2009, 5:09am » 
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on Jul 23^{rd}, 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 
 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 23^{rd}, 2009, 5:27am » 
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on Jul 23^{rd}, 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. 
 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 23^{rd}, 2009, 6:13am » 
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on Jul 23^{rd}, 2009, 5:27am, Eigenray wrote: But how deep a breath can you really take after being fatally stabbed in the chest? 
 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 lungvolume?


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Benny
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Re: various physics questions
« Reply #99 on: Jul 23^{rd}, 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.

« Last Edit: Jul 23^{rd}, 2009, 11:02am by Benny » 
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