studiot

What I am trying to get you to consider is the distribution of mass gain as a function of radial distance along with its implications, with both my earlier and more recent comments. Perhaps I haven't expressed myself very well, if so I'm sorry.

The first thing to realise is that pipes have curved bends, not abrupt changes of direction as you have drawn. This is because smooth profiles make for smooth flow and smooth flow changes. It is not clear whether your pipebend is vertical or horizontal. Pipes do not normally twist in two directions at once. So the analysis can usually be two dimesional as I have shown below. So fig1 shows the flow entering horizontally along the x axis. If the bend is in a horizontal plane the y axis is also horizontal and the weight of the fluid in the bend (W) is omitted from the following calculations. If the bend is in a vertical plane then the y axis is vertical and W appears in the vertical summation. The fluid passes through a bend in the pipe and is deflected through angle d in the positive y direction. I have also shown the bend pipe diameter reducing as the most general case. If the pipe has constant diameter then the velocity and area do not change. We consider a segment of fluid between sections S1-S1 and S2-S2, which are edit :where the pipe is tangent to the entry and exit directions respectively. For a constant volumetric flow rate, Q we have [math]Q = {V_1}{A_1} = {V_2}{A_2} = const[/math] and a mass flow rate of [math]\frac{{dm}}{{dt}} = wQ = w{V_1}{A_1} = w{V_2}{A_2} = m[/math] Which is also constant or fluid would accumulate in the pipe. Where w is the fluid density. If we consider the fluid inside the bend between sections S1-S1 and S2-S2 and draw a free body diagram we obtain fig2. It is very important to realise that the following is not an equilibrium analysis, since the fluid is in motion. Further the fluid in the bend segment is not a differential (small) element it is the whole amount. So we have the basic equation that Net force acting on the fluid = Sum of all forces acting = rate of change of momentum = mass flow rate x velocity change Now both the forces and the momentum are vectors so may be resolved along the x and y axes as is done in the following calculations. So R is the resultant pipe force on the fluid such that [math]R = \sqrt {F_y^2 + F_x^2} [/math] and [math]\tan r = \frac{{{F_y}}}{{{F_z}}}[/math] r is the angle the resultant makes with the x axis. We calculate the forces on the fluid becasue this is easier than calculating directly the forces on the pipe. Of course the force exerted by the fluid on the pipe is equal and opposite to R. The other forces acting on the fluid are the weight if vertical (otherwise W=0 as previously noted) and the pressure forces S1 and S2 at each section. [math]{S_1} = {P_1}{A_1}[/math] and [math]{S_2} = {P_2}{A_2}[/math] Where P is the fluid pressure and A the section area at that section. So in the x direction [math]{S_1} - {S_2}cosd - {F_x} = wQ\left( {{V_1} - {V_2}\cos d} \right)[/math] and in the y direction [math]{F_y} - {S_2}\sin d - W = wQ\left( {{V_2}\sin d - 0} \right)[/math] Where V the fluid velocity at the section. These may easily be solved for Fx and Fy This analysis is useful if we wish to provide a thrust block to receive -R as a direct force. If, however, the pipe is fitted to brackets then the brackets resist the flow force in bending. This can be seen as R applies an unbalanced counterclockwise moment to the fluid to turn it, so the fluid applies a corresponding clockwise moment to the bracket. Does this help?

Your problem is that your object (a disc) has components that possess differential tangential velocities with relastivistic implications, particularly for the mass summations needed to evaluate a 'centre of mass'. You did not answer my question about relative velocity, which should help you answer this,as should an appeal to symmetry.

What is the centre of mass in a relativistic system?

You have stated the acceleration version of Newton's law. another version is that Force = rate of change of momentum. This is the version used in the momentum balance or so called momentum equation in fluids. It's too late to do this tonight, but tomorrow I will draw a diagram to show how this is applied to forces acting at bends in pipes.

Fluid in a pipeline does indeed exert (side) thrust on any bend or change of direction. This is why thrust blocks are placed at changes of direction along a pipeline, and why pilelines should be built straight except at defined changes of direction. You are also correct in thinking that this force is calculated by using a momentum balance for the fluid. Do you understand the principles of momentum balance in fluid dynamics?

Thank you for your link. Did you follow up the references I gave? Those who believe that the irridium came from the interior of the earth have to explain how it appears in non igneous rocks, in particularly high concentrations, around the perimeter of impact craters. The concentration decreases rapidly with distance from the epicentre of the impact. This is one of the distinguishing tests to try to determine whether a 'crater' (filled in or otherwise) was meteoric in orign or otherwise.

Don't think it is a different isotope, just a variation in the relative abundance, since irridium is a vary rare element on the Earth's surface. It is called the irridium spike or irridium anomaly. https://www.google.co.uk/#q=alvarez+iridium+anomaly

Did I say fluid? Anybody baking bread is making a disperse system as is anybody lighting a fire (to generate smoke). A disperse system is simply one material in one phase (usually finely divided) distributed randomly but more or less evenly throughout material of another. The phases can be any so for normal temperatures you could disperse your powder in a solid, liquid or gas. Of course you would need a container if you chose either of the last two. I have a spirit level, made of a clear plastic block, which contains flourescein in the vial. I would imagine that you could disperse your powder, ground to a suitable fineness, in a (fast setting) clear resin to form such a block. There are many trinkets available in souvenir shops based on this principle. Some of them incorporate flourescent grains. I did ask if you were prepared to experiment a bit.

Surely not. Perhaps you do not understand the meaning of disperse systems, which are one of the commonest systems in Nature.

I hesitate to offer advice here. John Cuthber et al have said most of the science and it has not been well received. I am not sure if you have any of the powder you refer to, but if you are prepared to experiment there may be a way. You do not need a solution for this you need a disperse system (sometimes wrongly called a colloid) with your powder as the disperse phase and something transparent to the input and output wavelengths as the disperse medium. Note that most substances that retain energy for later release as light take in energy at one wavelength and release it at another.

There are almost no 'standard' notations. Many conventions have grown up from convenience. Some stem from typographical constraints, eg superscript, greek letters and special symbols. Some from the desire to fit in with a larger scheme eg the use of x, y, z or x0, x1, x2 or x1, x2, x3 for axes, Some from the simple fact that there are substantially more constants and possible variables in physics and maths than there are in several alphabets put together. Some from convenience in use eg the D operator v Leibnitz dy/dx v Newton's prime notation in calculus. And some from the fact that historically different symbols for the same quantity evolved in two different geographical regions or disciplines so your notation depends where and in what subject you are writing. Finally in the teaching on mathematical subjects some consistency is desirable so as students progress through they are not constantly presented with new and different notation for something familiar. Failure to implement this impedes learning and understanding.

Dell batteries have a built in battery tester on the bottom http://www.youtube.com/watch?v=m65yBJylgZ8

Are you studying impulsive forces now?

You really are a glutton for the extreme. I haven't ever had need to think about it but the usual thing in those circumstances would be to generalise the binomial expansion with the multinomial expansion. http://en.wikipedia.org/wiki/Multinomial_theorem

Indeed there are. You have uncovered Leibnitz rules. Well done. This develops the nth derivative of a product of two functions u and v via the binomial coefficients of lower derivatives, For example [math]\frac{{{d^2}}}{{d{x^2}}}\left( {uv} \right) = u\frac{{{d^2}v}}{{d{x^2}}} + 2\frac{{du}}{{dx}}\frac{{dv}}{{dx}} + v\frac{{{d^2}u}}{{d{x^2}}}[/math] [math]\frac{{{d^3}}}{{d{x^3}}}\left( {uv} \right) = u\frac{{{d^3}v}}{{d{x^3}}} + 3\frac{{du}}{{dx}}\frac{{{d^2}v}}{{d{x^2}}} + 3\frac{{{d^2}u}}{{d{x^2}}}\frac{{dv}}{{dx}} + v\frac{{{d^3}u}}{{d{x^3}}}[/math] http://en.wikipedia.org/wiki/General_Leibniz_rule

Personally I would start with the Readers Digest book Rocks and Fossils A visual guide by Robert R Coenraads For this application. go well

You have missed out one piece of information - the relationship between x and y on impact. This appears as my third equation. You appear to like messing around with trig, so have a go yourself and post agian if you need further help.

Pardon? The units of viscosity are fairy easy to look up on the internet eg this pdf http://www.hydramotion.com/pdf/Website_Viscosity_Units_V2.pdf You should be using the SI units. You will note that they have quoted two types of viscosity. Dynamic viscosity, which is the one I have given you as f Formulae in fluid dynamics are so often divided by the density that a second form of viscosity is often given called the Kinematic viscosity, which is simply the dynamic viscosity divided by the density. [math]h = \frac{f}{p}[/math] This has SI units of metres squared per second or m2 s-1. To see how the units of dynamic viscosity are derived we use Newton's law of viscosity which says that the shear stress in a fluid is proportional to the velocity gradient. [math]Shearstress = \frac{{Force}}{{Area}} \propto velocitygradient = f\frac{{dv}}{{dz}}[/math] Where the dynamic viscosity, f, is the constant of proportionality [math]f = \frac{F}{{A\frac{{dv}}{{dz}}}}[/math] If you are familiar with the following it is called dimensional analysis. we can represent a formula by inserting M for mass, L for length and T for time as the basis of mechanical units [math]\frac{{ML{T^{ - 2}}}}{{{L^2}L\frac{{{T^{ - 1}}}}{L}}} = M{L^{ - 1}}{T^{ - 1}}[/math] So the units work out to mass divided by (length x time) or kg m-1 s-1 However this is normally expressed in as Newton seconds per metre squared, (N-s m-1) which is equivalent (you should check this for yourself).

It is a pity that this thread has gone off track since I think Schneibster started off with the germ of a good point about book references. Some threads are about the cutting edge of science and one would not expect material to have made it to the textbooks. Also the old practice of 'pamphlet writing' (some were as big as a large book) has dwindled. So the only sources left are current papers, online or offline. Sf has a dedicated forum for this types of material. However we also have forums dedicated to lower level or older subjects in science and I see references to textbooks and even selected popsciworks are wholly appropriate. Indeed there are periodic requests for lists of such material. So surely it is a matter of (good) judgement as to what is appropriate in any given situation, as is so much in this life. So perhaps a less dogmatic title, coupled with a subtle change of emphasis would permit something good to come out of this thread?

It is also important to distinguish between the rate constant and the the equilibrium constant when meeting with this stuff for the first time. I now deserve a cup of tea so here comes my tea break constant.

Hello, function. The equilibrium constant is just that - constant. It does not change when you change the concentrations of the constituents - these are not independent from each other. You look K up in standard tables of chemical reaction constants. K does vary with temperature so you should always not this and take it into account. Now look at the equation you have. If K is constant and you increase [H] on the bottom line (first power) then you must increase [HI]2 to compensate. The equation tells you by how much and will be larger since it s effect if proportion to a second power. Does this make sense?

Yes and No it is not normally called a sub harmonic, but it could be. Like the louspeaker, the Helmholtz resonator is complicated (although not as much as the loudspeaker since there is no electric to mechanical energy conversion stage). The key word is resonator. You can excite an oscillator with any of the harmonics of its fundamental frequency. So with respect to the resonant (fundamental) frequency of oscillation of the air inside the bottle the effect works best (is at its most efficient) when the rush of passing air contains harmonics of the resonant frequency of the bottle. This is easy to achieve since the passing air approximates to white noise which contains all frequencies. So the bottle 'picks out' or selects energy at is resonant frequency. Incidentally, I don't know how much you understand about the mechanical mechanism of the action but the whole of the air in the bottle does not resonate. A plug of air in the neck is bounced up and down between the pressure of the passing air stream and the restoring force due to the springiness of the air in the large volume of the body of the bottle. This plug of air thus acts as a piston like a loudspeaker cone. It is this that generates the tone you can here, transferred to the general air in the room.

Remember also that we are used to seeing 2D curves on flat paper when we talk about curvature. In that case there is only one direction available to curve in and only one plane for the radius of curvature. When we go to 3D, there are two possible radii of curvature at any point on a curve and if we go to 4D spacetime then there are three, with time as one of the plane's axes. This is really the province of differential geometry in higher dimensions, which I think is ajb's speciality.