studiot

Would you perhaps like to explain these two statements first? Is the track curved?

You should not be working with those substances unless you know what you are doing, including proper strorage handling and other safety requirements. It is a legal requirement in most, if not all, countries. Since you are working with gold you can obviously afford to pay a proper rate for a competent technologist to advise.

But why stop at triples? The only other thing I could suggest is to look into the theory of quaternions and dyadics.

Yeah, product is good since the output is also a vector you could go further outer product or vector product might be better. Even if I can't, draw yourself a few diagrams. They help.

You need to include a statement of when the function is to return 1, 0 or -1 ie what is the specific condition on x for f(x) = 1 etc

I would not use the term resultant, as that is reserved for the addition of two vectors. But you have added two of these stretched, rotated vectors, each representing the square of a complex number. This does have a geometric resultant in the proper sense of closing the triangle in the complex plane and I think ( though I am not sure) that the closing resultant is the complex number squared on the other side of your equation.

Well multiplication of two complex numbers can be viewed as a stretching and rotation of the vectors for each on an Argand diagram. So each square will produce a new vector representing a new complex number which when added to the other on in the normal way in an argand diagram produces a third in a vector triangle. Sorry i can't so any drawing whilst I am still in Scotland.

Do you mean squares as you have indicated or multiplication by the complex conjugate, it makes a difference. For example your first bracket (-13 + 6i) (-13 + 6i) = (132 -156i - 62) but (-13 + 6i) (-13 - 6i) = (132 + 62)

Not a single resistor you have a controlled current source on the LHS. Thevenin would replace it with a voltage source in series with a resistor. Norton would 'replace' it with a current source in parallel. Except you already have one. Look up source transformation.

I assume you are looking for a Thevenin equivalent? Does this help? http://www.ittc.ku.edu/~jstiles/312/handouts/Example%20An%20Analysis%20of%20a%20pnp%20BJT%20Circuit.pdf

That is what perfectly elastic means. Of course, that is what happens. What difference do you think that makes? No the physics of the water impact is nothing at all like that of little balls, and much more complicated. The ball, mass M, has velocity +V momentum +MV as it approches the wall. It hits the wall and bounces back along the same trajectory, but in the opposite direction with velocity -V So it now has momentum M(-V) = -MV. So (Momentum before - momentum after) = (MV) - (-MV) = 2MV. KE before = 1/2 MV2 KE after = 1/2 M(-V)2 So KE before = KE after. If you are happy with this then we can proceed to the more complicated situation of the water.

That is what I have been trying to tell you. Yes if there is change of momentum a force is involved/generated. Yes in reality some of the KE is lost to the wall, especially if the water damages it, but once again you are trying to mix up KE and momentum. Don't, it doesn't work like that. And there is a difference between the momentum change for the water which is 1mwater and the momentum change for a ball bouncing off, which is 2mball. There is no reason for mwater to be the same as mball.

However all credit for making the experiment and posting the results. +1

Yes, you are right thank you MacSwell.

You get a medal if you can tell me what the centripetal forces are when these cogs rotate.

No one fully understands friction, but certain observations can be drawn that facilitate calculations with it. 1) Friction always acts to oppose motion or the tendency to move. 2) Friction has two separate zones of operation. A variable zone called static friction and a fixed zone called dynamic friction. 3) Friction can change (reverse) direction at the point of beginning to move, you mass on a slope has this quality. 4) The force of static friction on an object is only ever enough to oppose an applied force to prevent movement. In the case of zero applied force it is zero. 5) The force of static friction can do no work. 6) The force of dynamic friction does work. 7) The force of dynamic friction is constant regardless of the applied force and less than the applied force. Note on (3) above Suppose the applied force is pulling or pushing the mass up the slope, but the mass is not moving. While the mass on the slope is not moving but being pushed up the slope by the tendency is to slip down the slope, pulled down by gravity. This is static friction and it acts up the slope As soon as the appled force becomes so strong that the mass starts to move upthe slope, the force of friction reverses (opposing the motion) and acts down the slope The friction is now dynamic friction.

So ask, that's how we find things out. If a (perfectly elastic) ball travelling from left to right hits a wall and bounces horizontally back, It starts with a momentum m in the left to right direction. After bouncing it now has a momentum of (-m) ie in the right to left direction. This is a change of 2m. But its speed has not altered so its KE has not altered. Thinking about the same wall I have already directed a fire hose at it. All the momentum from left to right is destroyed, the water has zero left to right momentum, upon impact. But since it must move away (it does not stick on the side of the wall), it still has some or all of its KE. Please don't get into objects carrying a force. That is the realm of sub atomic physics and so called exchange particles that does not belong in ordinary common or garden classical physics.

I like this answer, short and sweet. +1 (And no maths involved. )

I don't have the benefit of knowing the history of your question, but a point is a mechanical term whereby the dimensions of the point are negligable compared to the dimensions of the system under consideration. So for instance you would be considered as occupying a point in the surface of the Earth, but the Earth itself would be considered a point in comparison with the Solar System or the Galaxy.

Think something travelling round a circle. In the time of one revolution it travels 360o or 2[math]\pi[/math] radians or one circumference ( = 2[math]\pi[/math]r) in distance. The time of one revolution is called the period. Time is the linking quantity between the quantities. So let us say it travels X linear units or Y angular units. (This may be more than one circle or part of one) in a time t. So the objects travels n = Y/360 (degrees) or Y/2[math]\pi[/math] (radians) or X/2[math]\pi[/math]r (linear measure) revolutions in time t. So the object travels n/t revolutions or cycles per second which is the definition of frequency. Equally the object takes t/n seconds to complete one revolution, which is the definition of period.

Father to son "What do you want to be when you grow up?" Son "A full time layabout"

Yes that is a frequency, although your equation is an unusual rearrangement. Can you also relate this to angular velocity in radians per second ? Going back to dimensional analysis. Work is Force times distance = (MLT-2) x (L) = ML2T-2 Potential energy is mass times height times g or M x L x (LT-2) = ML2T-2 This shows that work and energy have the same units, a useful result.

I'm glad it helped and I see no reason for the discouraging negative so +1. No after checking carefully (since I make too many typos) this is correct. All I have done is said mass times velocity and put in M (for mass) times (the dimensions I worked out earlier for velocity). Is that any clearer?

The three basic quantities in mechanics are Mass - symbol M Length - symbol L Time - symbol T All other mechanical quantities can be expressed by combinations of these three. These three are called 'dimensions' and combinations are called 'units'. This makes it easy to compare them and check for consistency. so for instance velocity = distance/time which has the dimensions LT-1 and units of miles per hour or metres per second. volume is L3 and area L2 or cubic metres and square metres. As a little extra: Force is rate of change of linear momentum Torque is rate of change of angular momentum Linear momentum is mass times velocity or mass times (LT-1) or MLT-1 So force the dimensions of force is momentum per unit time or MLT-2 How are we doing?

Once again see my post2