studiot

OK so why do we need to calculate all these kinetic quantities? Walk me through the sequence of energies (KE and PE) at the important stages in the path of the ball. See if you can puzzle it out first, before I tell you.

Most things in the real world do not have an absolute zero point. Almost everything we deal with is relative. The example in my post 37 has an absolute zero.

Let me rephrase this So how far does the centre of the ball fall, ie what is its loss of PE?

No it is not to do with activation energy. There are two main drivers in this universe. 1) The tendency of all things to move towards their lowest energy state. This is why objects fall - they lower their potential energy This is the condition for spontaneous chemical reactions and nuclear to occur and so on This is known as the minimum energy criterion. 2) The more difficult tendency of all things to move to spread out the available energy over the entire system This is why sugar dissoves in water This is why heat spreads out towards a uniform state This is why osmosis, diffusion, electrophoresis and similar processes happen This is why gases mix This is known as the maximum entropy condition. Entropy is a measure of the 'spread outness' of the energy. Sometimes these two drivers work in the same direction, sometimes they work in opposition and a balance must be struck and sometimes only one is at work. In many processes in order to commence the process energy must first be input to finally achieve lower energy. This energy is called activation energy and is always lower than the final output or the process will not happen, even with the activation energy. A simple example is a ball inside a bowl on a table. The ball would move to a lower energy if it rolled off the table and fell to the floor, but it needs to have (activation) energy input to lift it over the sides of the bowl first. The higher these sides, the greater the required activation energy. Another example is the energy of the match we use to light a fire. The energy we get out when the fire is burning is the result of the fuel moving to lower energy states, but this does not happen without the activation energy of the match. So the stick or piece of paper can carry on for a very long time, perhaps indefinitely withour burning until that activation energy is supplied.

Gosh, guys this high falutin stuff is interesting but we do not need to fly high to discuss the OP. @noonespecial would you be prepared to listen if any electrical engineer gave you a practical example of negative that we use every day? The electrical power wave supplied by your socket on the wall has positive sections and negative sections that exactly balance out. This is because the average power is exactly zero This is why we (have to) use what are known as root mean square values for current, voltage power etc.

Er Hem How big is the ball?

Calculations refer to the position of the centre (of gravity) of the ball. They give you the ball diameter because this is not negligable. For instance the ball falls from where the top is at +80 cm to where the bottom is at 0 cm. (reckoning up as positive). So how far does the ball fall, ie what is its loss of PE?

Infra is Latin for below and ultra is Latin for above, when referring to the frequency spectrum. Is the wavelength of red longer (greater) than or shorter (less than) blue-violet? So is the frequency of red less than or greater than blue-violet?

Yes, but brightness is still a vagueish term. The intensity is a measure of the energy of the wave (per uint area or volume) and, as you say it is proportional to the square of the amplitude. So [math]I \propto {A^2}[/math] [math]I = {k_1}{A^2}[/math] and [math]I \propto \frac{1}{{{r^2}}}[/math] or [math]I = \frac{{{k_2}}}{{{r^2}}}[/math] Can you now combine these two equations to find the answer?

Yes, D. The energy graph has twice the frequency of the displacement graph

What do the terms 'intensity' and 'amplitude' mean and what is the relationship between them ?

Check this. Since you got that one wrong, can you say why B and C are wrong?

Each (air) particle participating in the wave executes simple harmonic motion. That is it is an SHM oscillator. Note the particle motion is not the wave motion. A particle executing SHM has minumum (zero) velocity at maximum displacement. So at these points in the SHM cycle, the KE is zero. As you observe, the displacement - time graph starts at maximum negative displacement. So the KE - time graph must start from zero. Question for you. At what point in the SHM cycle is the velocity (and therefore the KE) a maximum ?

Good to see you checked my arithmetic, and found the missing 0.5. Students should always do this. Incidentally what are you studying Physics or Engineering Science? I chose this method because I don't know if you have any calculus, which would be the standard method.

Yes and NO. Yes you have calculated the total power dissipated in both cases by the 6 volt EMF. But in the first case half this power is dissipated in the internal resistor R and half in the external resistor R. So you lost a factor of 2 in your answer, since you are asked to compare the power in the external resistor only in both cases. Remember always read the question twice.

The energy doesn't disappear. (But that was a reasonable question) Current is rate of change or movement of charge. So if the current is zero then there is no movement of charge. If there is no movement of charge ie no current there is no change in energy. Voltage (by itself) is not energy. A battery that is not connected does not loose energy (except for leakage and old age) But once you connect a circuit, charge flows as a current and the battery looses energy to the circuit. Two equal voltages in opposition yield a net zero voltage in the same way that two equal forces in opposition in mechanics yield a net zero force. If the net force is zero, nothing moves and no work is done. The statement no work is done means there is no change of (mechanical) energy.

So what is the 'old' power and what is the 'new' power ? You know the voltage and the resistance so what formula do you think is appropriate here?

Look at the following correspondences. When all the rates of change are uniform. velocity is rate of change of distance : current is rate of change of charge So distance ( symbol s) corresponds to charge (symbol q) acceleration is rate of change of velocity : I will call rate of change of current p The kinematics equation s = ut + 0.5 (f t2) Where s is distance, u is initial velocity, t is time and f i acceleration. This equation also applies to your uniform change of current. i = initial current = 100mA the current changes from 100mA to 20 mA in 8 seconds ie (20 - 100) /8 = -10 mA per second. So p = 10 mA/s2 Changing our formula from s = ut + 0.5 (f t2) to q = it + 0.5 (pt2) = 8 q = 100 * 8 + (-10 *64) millicoulombs which I think you will find is answer A Edit q = 100 * 8 + 0.5(-10 *64) millicoulombs which I think you will find is answer C

Oh dear, you haven't got electrics yet. You need to practise lots of examples until it clicks in your mind. ~The Maximumpower theorem states The maximum power is delivered to a load when the load resistance is equal to the internal resistance. So the curve has a maximum at this point. There is only one graph that fits this description.

Since I'm sure you haven't covered differential equations here is another way. Have you done motion under constant acceleration yet?

In the old circuit there are two resistors. Are they in series or parallel and what is their combined resistance?

The Mac has always also had for video editing a full blown movie maker that works as part of the operating system.

You will probably find better support here. http://www.scienceforums.net/forum/28-suggestions-comments-and-support/

First an apology. These, of course, should have be x and y are the extensions, not the strain. However the math was correct and this oversight obviously didn't bother you. well done. Now for your second question. I'm sorry I have no experience with cellphone cameras so can't help with the rotation/posting. Try asking one of the younger members. I would say, however, that if you can, greyscale saves memory and transmission bytes, when colour is not needed. Here are your rotated and greyed images plus my working of the problem. Note I have used the radii not the diameters, since this avoids some extra fractions. I suspect you have got some (one) of the fractions the wrong way up, so the fewer the better. Once again the problem solution depends upon the force being the same in both.

You should have something like this Let x = strain in grp section Let y = strain in nylon section Then (x + y)* 10-3 = 3 * 10-3, where x and y are in mm Stress in grp = stress in nylon Egrp * x * 10-3 = Enylon * y* 10-3 Egrp * x = Enylon * (3-x) I'm sure you can finish the arithmetic.