NeonBlack

Is this part correct? Does the rate in have anything to do with the amount of salt already in the tank?

Haven't you started enough arguments like this already?

I swear I saw this exact same question about 2 weeks ago... http://en.wikipedia.org/wiki/C_standard_library#ANSI_Standard

Basically, how many ways can you write the number 64 in the form c^d. Using 16 as an example: [math]2^4=4^2=16^1=16[/math]

Then you may be interested in this previous discussion on how beams and moving charges create electric and magnetic fields. http://www.scienceforums.net/forum/showthread.php?t=34610

I think I understand. In principle, you can use any arbitrary A. Though for realistic models, you want to make sure that [math]\nabla \bullet A=0[/math] If you want to study a particular system, the magnetic vector potential is defined as [math]B=\nabla \times A[/math], or maybe a more useful form, for line currents is [math]A=\frac{\mu_0 I}{4 \pi} \int \frac{dl}{r}[/math] This probably wasn't as simple as you had hoped. If you want to, you can start off by considering DC currents (I know), where [math]\frac{\partial A}{\partial t}=0[/math]

I'm not an expert on statistics and probability, but I don't think this question is well-formed. For example, what does it mean for the statement "on day one, 45% of people are in favour of x" to be "false"?

I also do not understand your confusion. As bignose said, t is (pretty much) no different than x y or z. Maybe it would help if you noticed that the "del" operator acting on a scalar is typically written as: [math]\nabla V = [ \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} ][/math]

Did a real doctor tell you that? Has he heard of quackery?

I have noticed sometimes that the "Battery remaing: xx%" tool on my laptop sometimes gives unstable results. For example it might be 100% one minute, drop to 80% and then go back up to 100. When this happens, I think it gets a little warmer than usual too. I know that laptop batteries have a circuit to tell the power supply when to stop charging. I am worried that if there is something wrong with this, my battery might be damaged by overcharging. Is there a way I could diagnose/fix this?

Which version?

How much heat needs to be applied to salt before it decomposes?

The rules of the homework forum say that before you receive help, you need to first show that you have attempted the problem. Okay, here's my attempt at 4. std::cout<<(n*n*n*n+n*n*n+n*n*n+n*n)/0x04;

[math]\epsilon_0[/math] is a rather silly constant isn't it? Seriously though, the only silly thing here is arguing about unit systems. It is absolutely untrue that a scientist only uses SI units. I took a course last year where the only SI unit we used was the Kelvin. A scientist will use whatever units are convenient: electron-volt, parsec, angstrom, etc...

Yeah, one time I had a couple of pennies that were heavily oxidized. You can probably clean them with some vinegar. Edit: Sorry, I didn't actually read your post. In response: 1. This was thought of years ago. 2. Maybe we can build a giant Hoover with a HEPA in it.

[math]\cos^2x[/math] is an easier way to write [math](\cos{x})^2[/math] edit: the advantage is that you will not confuse it with [math]\cos{(x^2)}[/math]

Not very many options, has some scrolling issues, but my biggest complaint is how it looks. One of the things I really hate is when an app has complete disregard for window styles.

For simplicity, we're going to do this calculation without 4-quantities These are the lorentz transformations for electromagnetic fields if the motion is in the direction of the x-axis. (I lifted these from wiki) [math]E'_x = E_x[/math] [math]E'_y = \gamma \left ( E_y - v B_z \right )[/math] [math]E'_z = \gamma \left ( E_z + v B_y \right )[/math] [math]B'_x = B_x[/math] [math]B'_y = \gamma \left ( B_y + \frac{v}{c^2} E_z \right )[/math] [math]B'_z = \gamma \left ( B_z - \frac{v}{c^2} E_y \right )[/math] Now take a look at the picture- I warned you it wasn't going to be pretty. We have two positive charges some distance d apart in the y-axis. We'll let frame S be the frame in which they are at rest. In frame S', they are moving together with some speed, v. Let's calculate the electric and magnetic fields at the point of the top charge in frame S. (The bottom charge will see the same thing in the opposite direction) [math]\vec{B}=0[/math] [math]\vec{E}=\frac{1}{4 \pi \epsilon_0} \frac{q}{d^2} \hat{y}[/math] We can also write [math]E=E_y[/math] Now, let's use the field transformations to calculate E and B in the S' frame as they pass the origin. [math]E_y'=\gamma E_y[/math] [math]B_z'=\gamma \frac{v}{c^2}E_y[/math] We can calculate the force on the top charge, [math]\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})[/math] With the quantities that we know, [math]F_y=q(\gamma E_y - \gamma \frac{v^2}{c^2} E_y)[/math] To find the speed at which the magnetic attraction overcomes the electrostatic repulsion, we can let F=0, and we find v=c. Every step of this is not explicit, but I think it should be enough for you to be able to follow. Also, please let me know if you see any mistakes.

Your first questions sounds more like psychology to me, but I don't know much about that. I think "fitness" in biology is used to mean reproductive success. p.s. You should consider forgetting about the lolcats if you want people to take you seriously.

I will post something tomorrow. I will make some pretty pictures (okay, maybe not so pretty) pictures and doing latex takes a long time for me. In the mean time, take a look at the field transformations, which were mentioned earlier by Pete. http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Lorentz_transformation_rules_for_fields We can treat electric and magnetic fields as different aspects of the same thing, in much the same way as space and time.

Swanson, I did not read the wiki pages in entirety, but they just look like an alternate treatment of electromagnetism using 4-vectors and treating magnetism as a relativistic effect. I still haven't seen anything suggesting a relativistic form of Maxwell's equations. Still the only thing I can think of is the field transformations http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Lorentz_transformation_rules_for_fields Did you mean this: http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Covariant_formulation ?

I am not sure what you mean. I have never heard of a "relativistic form of Maxwell's equations." The usual Maxwell's equations are already consistent with relativity. I used the Lorentz transformations for fields in the calculation, if that's what you mean. That the charges separate at different rates in different frames is no paradox. Simply put, say the rate at which the charges move with respect to each other is dy/dt. Observers in different frames will see different dt's so this should not be surprising.

You are almost correct. However, if you tried to calculate the speed at which the pinch force overcomes the electrostatic repulsion, you would find that "a value" is the speed of light.

I think there may be some confusion regarding the electric/magnetic fields involved and I'm sorry. There needs to be an electric field (change in potential) inside the electron gun. This will be the 10000 volts. Without this, the electrons cannot move at high speeds. Outside of the electron gun, there will be a magnetic field (created by your electromagnet). Although the electro magnet is powered by electric current, there will not be any electric field unless you have regions of non-zero charge density (nominally, charged plates or something similar).

Sorry, yes I was being lazy with negative signs but in the end, the sign does not matter. I hope you have seen [math]E=\frac{1}{2}mv^2[/math] before. If you have trouble with this, you will not be able to do the math required for the rest of this project. There needs to be some type of plates inside the electron gun. A battery and coils will not make an electric field. Is it magnetic field you want to calculate? Magnetic field will not affect the speed of the electron, only the direction.