studiot

To me, the most eloquent enunciation of this principle was "Gentlemen, shall I refuse my dinner because I do not fully understand the process of digestion ?"

Or you can turn this the other way round and observe that alloying materials harden a material eg the addition of carbon to iron to make steel. Impurities are also known to be an impediment to dislocation spread. But what was your question? Edit : Have you also thought about the transition temperature?

Since this query now appears genuine a comment or three. 1) Swansont has told you that it is not possible for heat to flow from a colder body to a hotter one (without an external agent). So your plate cannot ever get hotter (reach a higher temperature) of its own accord than the stove hotplate. The only way I could see some overshoot would be if the heated plate was insulated, but the heating plate had a much larger uncovered area so whatever was heating the heating plate (which is obviously at a higher temperature than the heating plate itself) was pumping heat into a local variation of temperature. In other words the heating plate was actually hotter in some places than others. How do you know that your sensor was measuring the temperature of the heating plate at the contact surface with the heated plate, rather than at some other colder part of the heating plate. 2) The actual temperature recorded by a sensor will depend upon the (thermal) coupling between the sensor and the object. 3) Sensors, especially thermocouples, do not have a fast response time. This needs to be allowed for in measurement recording.

Hint look up osmosis

Sensei has offered a good method of studying the magnet, but not the electric part. However, have a care here. The point of the compass needle that points to the Earth's North Pole is actually the south pole of the compass needle magnet. It is often called a 'north seeking pole' as a result. The part of the Wiki article called "how a magnetic compass works" explains this.

Hello chris, One problem is that you need to wind a large number of coils to achieve an observable effect without special equipment. Why are you using an iron bar? This is normally used to concentrate the magnetic field generated by an electromagnet. You do not gain anything by trying to energise it with a permanent magnet. In order to generate a voltage you need to take the iron bar away and move your magnet back and fore along the axis of the coil windings. A good apparatus to study this is the inside of an old house doorbell, as in the photo. The white plunger is mounted on a magnet and sent side-to-side when the bell switch is closed allowing current to flow in the coil, generating a magnetic field. The upright black metal plates are freely mounted so that they vibrate audibly when struck by the plunger, a bit like wind chimes. I have removed the batteries for clarity. You can also create a voltage at the terminals by manually moving the plunger side-to-side. this can be measured with a voltmeter or used to illuminate a small bulb (not an LED).

All magnets are polarised and have two poles, so try and explain your question more. From the looks of your picture your magnet would work better at right angles to its present position. The poles are usually at the narrow ends and the magnetic field is strongest near the poles.

Preferably a hoop painted on a brick wall.

Not really You have the basic equation [math]F = \mu R[/math] It doesn't mater whether we are talking static or kinetic here. Your attempt to account for the mu factor works on summing all the contributions from F. The plastic theory works by summing all the contributions from R. You do not need to do both, just one or the other. Plastic theory suggests that [math]{\mu _s} > {\mu _k}[/math] simply because the bonds don't have time to fully form before the surfaces move on. You should also note that sinx is never greater than 1 so the product of two sines is smaller than either.

Really? For steel, Youngs modulus is 2 x 1011 N/m2 Modulus of rigidity is 0.8 x 1011 N/m2 Thus the plastic approximation to the coefficient of friction is 0.8/2 = 0.4. Pretty close I'd say.

That's clever how does that happen as it is flying along, icing? If you look back at Bignose'e post you will see he mentions circulation. This has physical dimenstions L2T-1. Multiply this by the mass within the circulation countour and you get the ML2T-1, which has the dimenetins of moment of momentum. Yes this circulation is going with the stream over the top, down at the rear, against the stream underneath and up in the front. The rule is The thrust at right angles to the flow acts from the side where the circulation and uniform flow oppose towards the side where they reinforce. Hence the lift.

No offence meant Delbert, but their difference would not be the subject of a physics exam or homework question if they were. Look at the wording of the question, "distinguish between......" that is examspeak.

That is a very good question, except that you have not posted any actual thermocouple readings. Certainy 'accepted practice in the lab' sounds more like a practical joke than a thermocouple reading.

Is this an exam question? A lot of folks are doing physics exams at the moment. Here is a hint, to explain the difference you should think about energy, both energy sources and energy sinks (dissipators).

OK, plastic theory works like this. Take an irregular (bumpy) horizontal surface. Because it is bumpy it must have a highest point Now place a second perfectly smooth flat surface onto the bumpy one. I am only using a flat second surface for simplicity of explanation the pricciple is the same for an irregular surface. Clearly the flat surface will touch the highest point of the bumpy one first. Now the stress equals the contact force (weight of the second object) divided by the area of contact. But this area is very small. So the high point will deform plastically until contact is made with the second highest point. And so on to the third, fourth etc highest points Until the area of contact is sufficient to support the contact force without further deformation. This plastic deformation joins (bonds) the two objects across the area of contact. Breaking this bond requires a shear force equal to the frictional force calculated from the coefficient of friction. I do not have the time tonight but you can derive that [math]\mu [/math] [math] \approx [/math] [math]\frac{{{S_s}}}{{{S_y}}}[/math] That is the coefficient of friction is approximately equal to the ratio of the shear strength to the yield strength.

I've made my point that I think this belongs in speculations. Yes there are many ways to calculate an average. Again I ask why kinetic friction? Perhaps I have misunderstood, but I thought your model involved considering (adding) all the side thrusts from the local irregularities pressing against each other at random angles (hence the average angle). This would obviously also work for static friction. But my point was that the meshing of the local irrecularities would vary as the two surfaces moved across each other. This has nothing to do with conventional plastic theory. I will explain that if you wish. You can arrive at a reasonable estimate of the coefficient of static friction for metals with it and account for the drop between kinetic and static friction.

This is the first time I have seen you present a plausible theory, so +1. A few questions to help you develop it and compare it with conventional theory. Why have you not posted this not in speculations since that is what it is? Why have you chosen kinetic friction? surely your 'average angles' will change as the object slides? What predictions does your theory make? I can immediately see a testable one, but will leave you the honour of stating it, since it is your theory. How does it compare with conventional plastic flow theory of friction? The two theories are not actually incompatible. I look forward to the development of this speculation with interest.

Suppose I take a journey from A to B and plot a graph of the distance travelled en route. Can I say that this is the shape of any physical object? Of course not. It has a definite shape on paper, that's all. If pushed I could observe that it must never be decreasing, unless I went back towards A at some stage. A shape is a surface in 2 or three spatial dimensions. Like the surface of a football tells us it is round, or more accurately a sewn together polyhedron of flat (ish) sides. My graph above is a shape on paper in two spatial dimensions. I am just using one to represent time. This is the source of my query about what you are asking. Now can you see why the question "What is the shape of a photon?" makes no sense in that context?

It is still essential that if you draw a graph (which your sine curve is) you understand (and preferably label) the units on each axis. You have not done this. I repeat understanding this is the key to understanding your difficulty. Are you with me thus far?

Good evening Ophiolite, Now that it's tea time and you are therefore in a really good mood, please sign the attached bearer cheque for $1million and return it to me.

So I am wasting my time, since you do not wish to address any of my comments.

DParlevliet, Am I wasting my time or do you want an answer to your original question? Swansont et al have told you the truth about waves and slits but their discussion is not about shape. The vibrating string analogy is useful in some respects but fallacious in others. And 'shape' is one of these others. The wave variable in a vibrating string is displacement. Displacement in say the y direction if propagation is in the x direction. So both x and y are distances. This allows the equation to directly display the two dimensional shape (in the x-y plane) taken up by a vibrating string. In sound waves, however, the wave varaible, is pressure at a point. This does not take on a shape in space in the same way. Can you imagine the 'shape' of a sound wave? You can only draw pressure contours of the sound field. The Schrodinger equation is similar in this respect to the sound field equation. The wave variable is a derived quantity, not even as recognisable as pressure. The axis variable is certainly not a distance, so again the solution does not take on a 'shape' in space.

The force field depends upon the source of the force, for example gravity or electrostatic attraction/repulsion. Do you know any that depend upon temperature and if so the temperature of what?

I'm really sorry to rain on your parade, but how many axes did your picture have and how many did you label? I keep nagging at this because the answer is the crux of your question about shape.

That's a pretty wavelet picture. But you need to label your axes. What are you plotting against what, which was in effect my orignal request about your question and the proper basis (I think) for your answer.