Country Boy

The problem I have with this is that there are no error calculations given. The values you give are all very, very close together. How did you measure them and what have you done to determine the accuracy of your measurements.

Your assumption "neutrons are such particles which can not be destroyed or created" is, as Strange said, incorrect. Neutrons can be both created and destroyed.

I have no idea what point you are trying to make. Yes, there is, currently a need for wood (and wood pulp to make paper). Yes, that causes "deforestation" which is a major problem for many reasons. Yes, as long as there is a "need" for wood, there will be deforestation. With all of those "yesses", what is your point? That "artificial wood" can help reduce deforestation? Or that deforestation has to continue in order to guarantee loggers jobs? There is not, and never has been, a requirement that jobs be "guaranteed" to a particular vocation. Would you advocate that people be required to smoke in order to guarantee tobacco farmer's jobs? In the last generation tobacco smoking has been reduced sharply and tobacco farmers have switched to other crops. Would you require people to ride in horse drawn buggies in order to guarantee jobs to people who make buggy whips? Oh, and the advent of "indoor plumbing" has completely destroyed the job market for "night soil collectors". Do you even know what that is? Changes are constant in all economies and people's needs are constantly changing. There is always a need to retrain people whose previous jobs have disappeared.

Because such a small percentage of our genes ​are from Neanderthals.

If man, A, has 4% Neanderthal genes and woman, B, has 4% Neanderthal genes, it does NOT follow that their children will have 8% Neanderthal genes. It is likely (indeed, almost certain) that those are the same genes so that the child would get the gene either from the father or the mother but not both.

And how do you STOP it?

What you suggest would work if the operator group, G, were commutative. But G here is specifically said to be not comutative.

What was your purpose in simply copying what I said?

An intelligible question would be appreciated! What do you mean by a "bundle of photons"? A single photon or a "bundle" would go past the electron "at the speed of light" so while there will be a slight gravitational attraction the photons would go past too fast for it to be noticeable.

"Gallilean Addition Law"? It would help if you had said that you mean the "addition of velocities". The "Gallilean Addition Law" says that if A is moving, relative to you, at speed u and B is moving relative to A with speed v, then B is moving, relative to you, with speed u+ v. Yes, at high speeds, the "Gallilean Addition Law" is incorrect. It need to be replaced by the law from relativity. With relativity the law for "adding velocities" says that if A is moving, relative to you, at speed u and B is moving, relative to B at speed v, then B is moving, relative to you, at [math]\frac{u+ v}{1+ \frac{uv}{c^2}}[/math]. I thought that was pretty well known. I don't know what you mean by "Let say a target is at 1 second when the switch is turn ''on'' by the man in the train ; if the train's speed is X the train will arrived at the target". A target would be at some distance from a train, so I don't know what you mean by "a target is at 1 second". Do you mean that, at the current speed of the train it would take one second to get there? And what switch is being turned on? A light switch? if so the speed of light is "c" so using the formula above, with u as the speed of the train, relative to you, the speed of light relative you is [math]\frac{u+ c}{1+ \frac{uc}{c^2}}= \frac{u+ c}{1+ \frac{u}{c}}[/math]. Multiplying both numerator and denominator by c, that becomes [math]\frac{c(u+ c)}{c+ u}= c[/math]. Yes, the velocity of light is "c" relative to any inertial frame.

Why would ​anything ​"keep getting more energy" moving at constant speed?

Neither of you is talking about mathematics, you are talking about applications of mathematics.

Given that [latex]\lim_{x\to f} f(x)= L[/latex], let u= x- b. As x goes to b, u goes to 0. Since u= x- b then x= u+ b. So [latex]\lim_{u\to 0} f(u+ b)= L[/latex].

Sounds like a mild muscle strain. And, of course, after running very fast you might have a "stich in your side". That's often a sign that you are not sufficiently hydrated.

You will need, probably after a semester or year of preparatory classes, to ask a faculty member to supervise your research. Part of that selection will, of course, be based on what topics interest you- ask a faculty member who works in the field you are interested in. He or she will probably select a number of topics from that field and you will select one of those.

I think you need to say exactly what you mean by "developed".

No, mathematics is not a science. The key to being a science is the "scientific method": 1) Observe a situation. 2) Create (many) hypotheses to explain the situation. 3) Extend the hypotheses to develop tests of each hypothesis. 4) Perform the tests to eliminate hypotheses. Mathematics does not do that. It is true that mathematics can be used in (3).

You appear to conflate "religion" with "belief in Gods" which has not been historically true. There have been, and are today, religions which focus on how people should treat each other irrespective of whether on believes in the existence of one or more Gods. I think "How people should treat one another" will alway be a concern as long as there are people.

If I remember correctly, Poincare postulated that a moving magnetic field would be stronger in the direction of motion so would cause atoms and molecule's to "contract" in that direction giving the Lorentz contraction. However, if you had a series of rods, with spaces in between, that would not cause the space in between to equally contract. Einstein's theory was that "space itself" contracted. Additional experiments, such as the "Kennedy- Thorndike experiment", http://www.conspiracyoflight.com/Kennedy/Kennedy.html, showed that Einstein, not Poincare, was correct.

Your "links" don't work so it is impossible to know what you are asking. "Would it be possible to magnetize" what?

Stephen Dawkins has written a number of books on evolution. I would recommend any of them. My favorite is "The Ancestor's Tale".

From [math]dL/dt= K(L_\infty- L)[/math] I would rather write it as [math]\frac{dL}{L_\infty- L)= Kdt[/math]. To integrate the left side, let [math]u= L_\infty- L[/math] so that [math]du= -dL[/math] and you have [math]-\frac{du}{u}= Kdt[/math]. That should be easy to integrate.

This is not true. The Abel-Ruffini theorem simply asserts that a polynomial equation, of degree greater than 4, may have solutions that cannot be written in terms of radicals. Every polynomial equation, of degree n, has exactly n solutions in the complex numbers, counting multiple roots. Every such differential equation has such a "complete general solution". It may, of course, be very difficult to write down.

The direction of the inequality is 'flipped' if and only if each side is multiplied or divided by a negative number. The difficulty, of course, is that if you multiplied or divided by an expression in "x", whether the expression is positive or negative depends upon x. So basically, you need to determine when each such expression is positive or negative. For example, to solve [math]x^2- 5x+ 6= (x- 5)(x- 6)> 0[/math], I can see that, if x< 5, x- 5< 0 so dividing both sides by it will "flip" the direction: x- 6< 0 which gives x< 6. Of course, if x< 5, then we automatically have x< 6 so one solution is x< 5. If x> 5, then x-5 is positive and dividing by that does not flip the direction: x- 6> 0 which gives x> 6. Of course, if x>6 then we must have x> 5 so x>> 6 is a solution. The solution set is {x| x< 5}U{x|x> 6}. Another way: we can argue that the product of two numbers is positive if and only if the two numbers have the same sign- both positive or both negative. That is, either x-5> 0 and x- 6> 0 or x-5< 0 or x-6< 0. The first pair give x>5 and x> 6 which are both true for x> 6. The second pair give x< 5 and x< 6 which are both true for x< 5. Again, the solution set is {x| x< 5}U{x> 6}. Finally, we could use Sriman Duttas method. First solve (x- 5)(x- 6)= 0. The solution set is {5, 6}. Since the function is continuous, those are the only points at which the function can change from negative to positive or from positive to negative. That is, there must be a single sign on each of the intervals [math](-\infty, 5)[/math], [math](5, 6)[/math], and [math](6, \infty)[/math]. It is sufficient to choose one point in each interval to determine what that sign is. x= 0 is in [math](-\infty, 5)[/math] and (0- 6)(0- 5)= 30> 0 so (x- 5)(x- 6) is positive for all x< 5. x= 5.5 is in [math](5, 6)[/math] and (5.5- 6)(5.5- 5)= (-0.5)(0.5)= -0.25< 0 so (x- 5)(x- 6) is negative for all x between 5 and 6. Finally, 7 is in [math](6, \infty)[/math] and (7- 5)(7- 6)= 2> 0 so (x- 5)(x- 6) is positive for all x> 6. Once again, the solution set is {x| x< 5}U{x> 6}.

I know very little about chemistry but, as a mathematician, my first thought would be to fit a power equation, in x, the concentration of P4, and y, the concentration of H2, to this: rate= C x^ny^m. The three data points give three equations to solve for m, n, and C. 0.000320= C(0.0110^n)(0.0075^m) 0.000640= C(0.0110^n)(0.0150^m) 0.000639= C(0.0220^n)(0.0150^m) If I am reading this correctly, m is the "order" with respect to P4 and n is the "order" with respect to H2.