darkenlighten

 

It for example says that practically never we will get perfect (classical) circulation from the topic ... and that our knowledge about particle's position along magnetic field decreases with time ...

It is not quantum mechanics that causes a "non-circle", but the properties of Electrodynamics.

Where is the probability or uncertainty in this specific subject? It's only a matter of [math] \mathbf{F} = q\mathbf{v}\times\mathbf{B} [/math] and the decaying circle in non-ideal situation is due to the radiation decay, as stated before.

OK, I agree.

 

So,

 

1) Where do the photons come from for fusion.

2) Why do they perform more work at the infrared level.

1) I'm not sure if fusion uses specifically photons, but if it does then we either create them using a laser or the are produced from the collision of particles due to conservation of energy/momentum.

2) What example do you have for this, do you have a link to an article or what not. Its possible that what you're referencing about infrared is because its the lowest energy photon they use for this process, being the most efficient, but I'm not sure.

You have nice answers.

 

But, if you recall, Einstein for the photo electric effect posited a work function as primitive.

 

So, all logic must obey this method.

 

( I am amazed by the factory of photosynthesis, may I suggest you look at it. You may have, but again, it is something to me)

 

Anyway, to analyze the photon, we must posit Einstein's work function to correctly operate on the problem.

 

Do you agree or disagree?

I'm not too familiar with the details of the process of photosynthesis, of course the general idea, but not in detail.

But I'm not sure exactly what you are meaning by your question. If you are asking that we have to assume that the work function is correct, then yes I do agree. It has been used and calculated for a lot of materials.

Well, I know there aren't really "photo-electric effect photons". I used this phrase to refer to photons in the light spectrum. And yes, I know only certain materials are sensitive to the photo electric effect.

As a side note, did you know if a plant did not operate in the pico second range for the photo electric effect for photosynthesis, there would be no life on this planet? We just hit this range in the last decade.

http://photoscience.la.asu.edu/photosyn/education/photointro.html

 

Do you find that amazing? The plant knows to use the photo electric effect to break water into H and O. It then reasembles them into a carbohydrate for energy storage. Naturally, higher level plants use N etc for protein construction using this same method.

Yea this is interesting how they can do that and I know studies are being done in the area to figure out how the plants recombine the cells to prepare for more photosynthesis.

Anyway, this issue under consideration is the work function.

 

Clearly, photons operating in the infrared range perform more work than do photons that operate in the higher frequency ranges. Do you agree or disagree?

The work function has to do with energy, not work, regardless of the name. So when you ask if the infrared does more work than higher frequencies it doesn't really make sense for this application. So we are only worried about their energies and momentums.

vuquta, you are thinking about this all wrong. First there aren't specific "photo-electric effect photons", just certain materials that only release electrons due to the energy required to kick on out of the material. This has to do more with the separation and atomic bonding of the material. Any photon can release electrons, as long as it has enough energy.

So has stated before it does have less energy, but you are not moving a particle that is in free space when it comes to the electron, it has a certain energy that it needs to overcome to be released from the material.

While a molecule might be in "free space" needing less energy to be moved in the desired direction.

Thank you so much!! I feel extremely enlightened; when I read it all of the sudden every road block in my mind was cleared. But, then when I looked at my trig homework, I saw a problem where the angle theta and the opposite and adjacent are all unknown, and the only thing given is the hypotenuse and the alpha angle. I'm guessing that weird sign that my teacher squeezed between the opposite and hypotenuse is the alpha sign, at least... it looks a lot like a Jesus fish... .... okay, sudden religious/mathematical connection just smacked me upside the head but I don't think this is the right place to talk about theology x)

 

What do I do? D:

I'm glad that it helped.

For your problem, remember that those equations don't have to be independent from each other and can all be used for the same problem. Look at the knowns and see which equation can apply in order to move on to the next part of the problem.

Okay so first things first: A function is anything that takes input and does something with it to give an output. For example, [math] f(x) = x^2 [/math] or stated as [math] y = x^2 [/math]. Say [math] x = 2 [/math] then your output is [math] f(x) = 4[/math] or [math] y =4 [/math], either is equivalent nomenclature.

So if you understand a function, we can now move to [math] Sin(\theta), Cos(\theta)[/math] and [math] Tan(\theta)[/math].

The best thing to remember SOH, CAH TOA, which stand for [math] Sin(\theta) = \frac{Opposite}{Hypotenuse} [/math], [math] Cos(\theta) = \frac{Adjacent}{Hypotenuse} [/math] and [math] Tan(\theta) = \frac{Opposite}{Adjacent} [/math]. You should have learned what the Hypotenuse, Opposite and Adjacent sides of a triangle are. With this you should be able to use it properly for trying to find lengths and angles.

Each equation is going to have 3 variables, the angle: [math] \theta [/math] and the two lengths corresponding to that angle. So you would use each one when you are given 2 out of the 3 variables and the angle is the angle residing in between the two sides.

We'll start there, go.

Though a good idea, this would never be practical in anything else than an empty room. When you are dealing with electronic components, signals, etc., large magnetic fields are not going to be friendly and cause a lot of problems and interference and damage.

The best you could do would be to do this on the bottom of the feet of a suit and the magnetic fields you would want to be really weak comparatively, but I still don't think it would be practical.

This one is annoying, I've seen this misunderstanding multiple times. Its not because that we personally are aware of the observation but the method in which we observe, such as Mr. Skeptic is stating. It's not our mind or consciousness that has anything to do with collapsing the wave function or affecting an event, but due to the fact that we have to change the system in order to find out what is going on.

Paper thin! lol I found it to be around .1 - .0254 mm

I was going to try and break this down (I'm kind of tired), but I'll start here. If you have learned anything about quantum mechanics it is very useful when studying semiconductors and how they work. Essentially its due the materials potential. Imagine your three materials lined up, NPN, they each create a potential well corresponding to the material and its length. So these wells placed next to each with also create "walls" if one is higher or lower than the other, this causes the blocking.

Another thing to remember is that these are semiconductors, not metals. In a semiconductor you don't have a "sea" of electrons, that is why it is a semiconductor, sometimes it acts as an insulator and sometimes as a conductor, but criteria has be met for this to be done, such as applying a voltage, or current flow.

Hopefully this is a start to your understanding.

Voting on the post doesn't help anyone unless it's explained why there is a disagreement...

Is there any sort of higher amount of photons emitted by a particular set of frequencies?

The amount of photons emitted is based on its intensity and the frequency corresponds to its energy, if I'm not mistaken.

Well, for instance, a piece of paper does not interact with gamma radiation because the radiation passes through the paper with no interaction. So what i mean by this, is what types of substances interact with alpha, beta, and gamma radiation to excite the material and emitt photons?

Interaction is based on wavelength. If the wavelength is smaller than the separation of the atoms, it will barely interact, but the amount it interacts just decreases as the wavelength gets smaller in comparison to the material.

But which substances emitt large amounts of photons?

In other words, which types of substances when excited by various means, specifically gamma radiation, emit more photons than other substances?

Once again, though gamma radiation has much higher energies, since it is so much smaller than other EM radiation the chances of interaction is very small, usually causing insignificant results.

Then why don't we concentrate a given area with photons, then release them?

Photons are energy carriers, so you can concentrate them, but it's not like storing them for later use. There are devices that use light (lasers) to somewhat confine materials.

Here were some links I found: Laser Plasma Interactions Inertial_confinement_fusion

It would still be sinusoidal.

If you can do 750 physics problem a week, I doubt you would need to be doing them. I mean for an upper level physics course, we were doing at max 8 a week. In my opinion, that does not seem anyway realistic or practical.

If you can do a physics problem in less than 5 minutes, it most likely wasn't worth doing, with that 750 at 5 minutes per problem on average is about 9 hours per day. Heck at 2 minutes per problem is still 3.6 hours a day every day, which is fine, but why waste your time with that easy of problems.

No, the longer strap will not make the bag heavier. The downward force of it just hanging there will be identical. Moment of force, i.e. torque, comes into play with "twisting." In the bag example, because it isn't rigid, it always hangs straight down, in the direction of force. You would only get a torque if there were an orthogonal ("sideways") component.

 

So, for example, if he bag was on the end of a rigid stick held out horizontally, it would be exerting more torque if the stick were longer, and it will be more difficult to keep it horizontal. The force here is applied downwards (by gravity), perpendicular to the stick.

My point was that it can act as a moment, of course when it is hanging straight down, the length will not make it heavier, but as soon as that starts swinging, forces are gonna change. I see it more like a pendulum though.

You use conservation of momentum to get the horizontal velocity after collision.

And yea since you know that the arrow is horizontal when hitting the bird, I believe it is safe to assume that is when the arrows vertical velocity is 0.

If you can find the height of the tree (where the bird is at) that means you can find how long it takes for them to fall. And if you know that you can find how far away from the tree they went.

Just to be clear, it's not the change in voltage that gives you 0's and 1's, but either a high or low voltage, for example 0V could be a 0 and 5V could be a 1. So imagine a sinusoidal wave at whatever frequency oscillating from 5V to -5V, passing through zero (sometimes 5V and -5V can both be read as 1). So if you are changing the frequency you are only changing the rate at which you are receiving those 0's and 1's. Hence why computer processors are given in frequency, the higher the frequency the faster the rate of information sent: A 3Ghz processor is slower than a 4Ghz.

So doing it by changing the frequency, to me, doesn't seem like it would be advantageous and most likely would not provide fast results.

A moment is a Force multiplied by the perpendicular distance from the object. So if you consider the shoulders the pivot point and have the weight (force) farther away then yea it might appear to be "heavier" and should act like a moment.

Edit for clarity: It won't only appear to be heavier it will cause more of a force.

Actually just because there is a force does not mean that force is doing work. The best example is the magnetic field, which NEVER does work, but yet there is a force involved.

I still don't understand why you guys are trying to solve this using impulse. It almost seems like a battle of egos at this point lol.

And to add, your answer is most likely a conversion error, because it will be exactly 10-lbs (if impulse is not considered). There will be no loss, given the simplistic scenario.

This is has gone on for way longer than it needed to imo, so I'm gonna settle this.

There can only be two scenarios:

1) You consider the impulse of the impact, in which the question does not have enough information for it to be solved.

2) You don't consider the impulse and the fact the the acceleration will be the same from either height when the bowling-ball hits the scale (ignoring change from the distance from the center of the earth, edit: which then still doesn't matter since the scale is at the same height ).

There isn't much more to this problem than this.

I would assume more of an impact loading since most loads on the bike would be short stronger loads. If you have access to any FEA programs that would be ideal. What I would do is maybe some research on what the typical loads on a bike of your size would be at the front and rear and then give a Factor of Safety of at least 2, if you are worried about it.