studiot

I started this reply before ajb and imatfaal posted further, however the comments are still valid. I suggest you reread ajb's post#2. This was proved in the early part of the 20th century. But we need to go back a lot further. A system of mathematics (there are many, there is no single one) is a logically self consistent construct that builds on axioms to create theorems and other results. However it is not built on axioms alone. Axioms in isolation cannot provide sufficient information. The history of geometry is a good example. The original 5 axioms (he called them propositions) of Euclid were supported by 23 definitions and 5 what he called 'common notions', without which we could not have Euclidian Geometry today. Without definition 4 (a straight line) the rest is nonsense. If the analysis is not restricted to straight lines many of the results can be negated by a curved line as a counterexample. Exactly as ajb has indicated. In the 18 century (I think) one of the axioms was changed and projective geometry was born. (Some of) The theorems and results of projective geometry are at variance with standard Euclidian geometry, but the new system is consistent within itself and its altered axiom. In the 19th century, the fifth axiom was removed altogether to found Riemanian geometries. In the 20th century Geometry moved from discussion of figures and shapes as being the fundamental to discussion of sets, symmetries and groups.

Yes. Incidentally a 'series' is already a sum. A 'sequence' of points is just a list.

None of the laws of Thermodynamics refer directly to efficiency. Mechanical systems (machines) Efficiency is output work divided by input work. This also equals the mechanical advantage divided by the velocity ratio. http://www.slideshare.net/jbishopgcms/mechanical-advantage-and-efficiency and http://www.learneasy.info/MDME/MEMmods/MEM23041A/dynamics/simple_machines/simple_machines.html

Agreed. Don't forget that General Relativity (which deals with gravity) operates in 4D Minkowski space. In Newtonian physics (that you are describing) the gravitational effect and contant G is due to mass density in 3D space. In General Relativity these effects are due to momentum density in 4D Minkowski space.

If you weren't so commited to attack me for reasons that I don't understand, I'm sure we could cooperate a great deal better. This should not be contest as to who can pick the most holes.

Enthalpy post#89 At very high pressure swing, acoustics get nonlinear because P*V is a product. It stays nonlinear whatever the gamma. Anyway, sound is linear at the sound pressure levels produced by a bell. Using the product P*V is incorrect, although even Newton made this error. That is a pity since there is inverse proportionality ie if you double the pressure you halve the volume and so on. In this sense the P*V relationship is linear. The ratio is independent of the starting values and is constant. Using the correct adiabatic expression does not enjoy this cosy relationship of ratio, it depends upon the actual values. Enthalpy post#89 It is perfectly known that flexural waves become shear waves at higher frequencies, at about the frequency where both propagation speeds get equal. You almost got away from compression waves, go on, that's the right direction. Sound is slower in usual austenitic stainless steel, and by less than 20%. Wrong figures somehow. 6000m/s would be very much. Anyway, compression waves are irrelevant in a bell. 5000m/s would give a resonance outside ear's range, so it's about time to change your opinion. No compression wave. Extract from Kaye & Laby, for speeds of sound in solids all in m/s Mild steel Longitudinal bulk waves 5960 Waves in thin sections 5196 Shear waves 3235 Rayleigh Waves 2996 Stainless Steel Longitudinal bulk waves 5980 Waves in thin sections 5282 Shear waves 3297 Rayleigh Waves 3045 Penn University rated its sample as mild steel with a velocity of 5000m/s I keep repeating this, the action of the bell is a (complicated) vibration, not a wave. But you do not hear this action. You only hear the sound in the driven air. As to the frequency in the air, I was suprised that so many higher frequencies were present, but measuement shows that they are, and not all harmonically related. This implies that the mode of vibration of the bell is acting in 'panels', with small panels of bell wall producing the higher frequencies. This is consistent with the images in the articles I linked to. Enthalpy post#89 If a tuning fork is alone in air, it's a bipolar souce. Though, its bottom is commonly put on a wood part that radiates more strongly for being bigger - and sometimes at a resonator. The bottom (middle of the U shape) vibrates with a small displaceent that fits the wood part better. This movement can result in a monopole source if a resonator is used, often a quarterwave box. The ouptut from a tuning fork can be biploar, quadripolar or more up to full spherical. Clearly if there is a close plane or other boundary things will be different. Here is a good discussion. http://www.acs.psu.edu/drussell/publications/tuningfork.pdf

The analogies presented in post #1 are logically inconsistent. For example if I stand on a thick steel plate on the elevator floor, both the plate and I are pressed to the floor, the addition of the plate does not provide any shielding or interference effect. Suppose further that gravity were a 'universal field'. Then in order to "interfere with this field" as stated, a massive object would perforce have to produce a counterfield of its own, contrary to the stated premise. Actually I prefer the term interact, rather than interfere, which has some specifically different connotations. Finally in suggesting that a massive object can interfere with the field to the side of the body and at a distance from it is contrary to the suggested notion that massive objects just block it by being in the way. Note that we know light is deflected in passing massive bodies to the side.

Bearings are measured clockwise from the vertical axis on paper. This is the north axis on the ground and on the paper. I have started your sketch off by drawing a north axis from P due north and positioning Q on it 20km north. Then I have drawn a line from Q at 140 clockwise from this north axis towards R. Then I marked R on this line (which is the bearing of R from Q) at a distance of 8km. At this point I will leave you to carry on with the diagram to draw in the three extra lines needed to complete both parts of your problem, which is then just an exercise in trigonometry. Post your results when you have done this.

We have to be careful not to be too all embracing. What would happen if all the schoolkids were exactly the same height and you told them to line up in order? Or, since I can take any members what if I ask four hydrogen atoms to line up in order? How about a mathematical example, the set of 4 rotations by degrees {0, 360, 720, 1080} ?

You are talking about the phenomenon called 'real and apparent depth'. Look here http://www.physicstutorials.org/home/optics/refraction-of-light/apparent-depth-real-depth

I'm trying to avoid the thorny isse that strictly speaking the answer to the question "What work is done by the oil ?" is zero. So strictly speaking the question is faulty since it does not offer zero in the list of options. The next part is really a language question. Although the oil does not do any work, there is work done. This work is done by the ball. Both the ball and the oil cannot do work on each other or the two amounts of work would cancel out and there would be not net energy exchange. If you wish to consider this in terms of forces then the ball rubs against the oil as it falls and in so doing applies a frictional force to the oil. Also in falling the ball moves its point of application of this frictional force. Remember there are two conditions for a force to do work. Both these must be met. Not only must a force be applied but it must also move its point of application.. Now the oil also applies a frictional force on the ball But the oil does not fall. So the oil does not move the point of application of the frictional force on the ball. So the work done by the oil = force times zero movement = zero work. So the only work done is by the ball and energy passes from the ball to the oil. What the question wants is for you to calculate this energy and call it negative since it is from the ball to the oil.

According to your original sketch the 5A current is flowing in the opposite direction from the 20A current. Therefore the magnetic field circles the wire in the opposite direction. Note that you should not say "the 5A conductor"; rather say "the conductor carrying 5A", because both conductors could be identical, just one is carrying more current than the other.

The force and distance act along the same line in this case so the cosine is not invoked. That is the angle between the line of movement and the line of action of the force is zero and the cosine is therefore unity. The positive or negative sign is relative to our statement of which way the energy flows or what force does work on what body. This depends upon our statement. In this case the question was what work does the oil do on the ball? Well it doesn't. Where would it obtain the energy? What actually happens is that the ball looses some potential energy in falling. This energy is transferred to the oil as work done by the ball on the oil. Since the oil does no work, we can call the energy it receives, by way of work done on it, negative work. This idea is important because it is the cause of many errors in the application of the first law of Thermodynamics.

Here is what I think you your professor might be saying. Two long straight parallel wires pass currents in opposite directions of 20amps and 5 amps. Given the distances apart find the null points in the combined field. Now the fields are additive and the field around a single wire is circular and inversely proportional to the diatance from it and given by the equation at the bottom of my sketch. The sense of the fields is given by the right hand screw rule and is as shown. This allows a section or plan to be drawn through any plane at right angles to the wires where the resultant field may be calculated. Can you see how to do this? Note the cross and point should really only be used for the current in the wires. (imagine looking at an arrow and seeing either the feathers or the tip)

Yes in the work equation delta x is a vector. The work = force . distance ........... where . is the dot product Thus the cosine is unecessary, since it is inherent in the dot product of two vectors. If you keep the cosine, the force and the the delta x have to be considered as scalars, but then the cosine can be positive or negative depending upon its quadrant.

Sorry you will need to explain your diagram a little further. What are the continuous lines with 20A and 5A by them? What sort of conductors do the cross and point represent? Are they straight wires only or are they part of a loop? And are they connected to each other?

There are two ways to look at this. "Calculate the frictional work done by the oil" Firstly does the oil receive energy or give energy? If the oil gives energy the work done by the oil is positive. If the oil receives energy the work done by the oil is negative. Alternatively. Work equals force times distance moved against that force. The direction of the force applied by the oil is positive ie upwards, but the distance moved is downwards ie negative. positive times negative make negative.

It's a measurement, Perhaps you would like to explain further what you mean.

In space, the air or on land we deploy base stations that transmit EM signals for positioning purposes. Many base stations these days are satellites. Underwater, these EM signal do not propagate well. Luckily acoustic signals do. So we do the equivalent. Setting up acoustic transmitter base stations and using these signals. A classic text on this subject is Underwater Acoustic Positioning Systems by Milne.

I am going to watch with interest, certain individuals arguing with themselves. I wonder, if they told each other something would they have to kill themselves?

Not quite since the free end of a tuning fork is not edge stiffened, however function's chalice may not be and so the analogy would be apt. Whatever, the following animations are interesting. Scroll down to the end of the apge and click on the frequency for the tuning fork animations. http://www.ibp.fraunhofer.de/en/Expertise/Acoustics/Fundamentals-and-Software.html But I keep saying this mantra. What we hear is the sound in the air not the vibrations of the oscillator. The air is being driven by the oscillator. The oscillator provided a forcing function. When the air is unconfined as in a large space such as a room, or better a field, then there are no resonance effects the wave just spreads out. The air has no preferential frequencies so the sound follows the oscillator. When the oscillator is driving a body of air as a whole in a confined space (such as inside the bell, loudspeaker cabinet, helmhotz resonator or whatever), the physics is different since the oscillator is now exciting vibration modes of the air as an object. One difference is the measurable difference in frequency spectrum of the sound. The inside air, acting a a body, now transfers energy to its preferential parts of the spectrum, as with any forcing function to a resonant system. The transfer to higher frequencies is entirely consistent with the analysis performed earlier. This is a resonance/forcing phenomenon, not an interference one. Energy calculations are difficult via vibration analysis and best done via adiabatic air theory. This sidesteps the inherent non linearity due to gamma being about 1.4 for air. Our (my) ears are not particularly sensitive to this, particularly as I am hearing the outside sound(which I know is louder) in one ear and the inside sound in the other. So my brain is tricked into thinking they are the same. A typical masking phenomenon.

In the first two, the resistor and capacitor are in series So the total voltage is the sum of the resistor voltage plus the capacitor voltage and the same current I(t) passes through both. VT = VC + VR The equation for this circuit relates the circuit variables (current and voltage) to each other through the circuit constants (resistance and capacitance) [math]{V_T} = \frac{1}{C}\left( {\int {Idt} } \right) + RI[/math] This is an integral equation. We obtain the differential equation by differentiating it. This can be done term by term to obtain [math]\frac{{d({V_T})}}{{dt}} = \frac{I}{C} + R\frac{{dI}}{{dt}}[/math] The total voltage in Vin; the output voltage is across the capacitor in the first and across the resistor in the second. Can you substitute for Vin and Vout to obtain the differential equations you need?

Good on you for experimenting. My bowl has a turned over lip that considerably stiffens the rim and changes the modes of vibration compared with a plain edge. Many pan lids are the same, how is yours? Further, ask yourself why does a tuning fork have two prongs yet is considered a single point sound source? 1) I'm always happy to discuss real physics, so how does this fit with the spectrum analyser results I posted? 2) The edge stiffening makes longitudinal flexion more likely. However did you follow up the Penn State University research results I posted? They shows that flexural waves are replaced by shear waves at higher frequencies, amongst other things. They used ordinary steel which has a compression speed of 5000 m/s. All the figures I could find for stainless average some 20% higher. Some are obviously above this average and some below. That is how I came by the figure of 6000 m/s.

Are you working towards a form of string theory?

I think at this point in the discussion it is important to distinguish between different sensors used to measure the received colour.. The human eye will yield a different result from a photoshop colour calibrator, or the calibrator on your scanner. I think swansont is talking about impartial scientific instruments for measurement. Of course the human eye is connected to the human brain which, as Lizzie says, complicated matters enormously. I bow to her vastly superior knowledge in that human part, which is her particualr area of expertise.