Cap'n Refsmmat

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html#c2 http://blogs.scienceforums.net/swansont/archives/6

So I guess the issue wasn't what breed they are. It's that they're not really dangerous regardless of what you call them.

Can we drop the religious discussion and stick to radioactive decay? This is making me nervous.

So can we conclude that dogs people tend to call "pit bulls" are dangerous?

I moved that thread over to the Math forum. If you want to continue the discussion there you can.

The Space Shuttle, if you want an example, has Reaction Control System rockets that fire to maneuver it around, and it uses the same system to stop the motion once it is in the attitude it wants. You fire the rockets once to start moving, and again to stop moving.

Perhaps sharks, while carnivorous, aren't always hungry. They may not always want a taste of that tempting scuba diver, no matter how much of an easy target he is.

The Wikipedia article on VeriChip has various links to PubMed articles on the subject if you're interested in perusing further.

I'd also say that because of the current male bias, there are greater barriers of entry for women on the Internet -- too many nerdy guys fawning over them. http://xkcd.com/322/

It is, actually. The key in relativity is the "relative" bit: time is relative. Time has "slowed down" for one person and is faster for the other, but it isn't just a perception: both observers are correct. Neither is seeing an illusion. All inertial reference frames are valid.

As far as I know, implantable RFID tags are passive (because they can't easily carry a power source), so they emit no more radiation than the tag reader does. The research about RFID tags causing cancer was apparently a test with rats in which 10% of the rats developed malignant tumors at the site of the injected tag. VeriChip (who makes the chips) says it's a rat-only phenomenon. (PDF warning)

What exactly is L there? The number of atoms (in moles) times the molar mass of silver is the mass of the pile of atoms. I don't see the point of L unless you're dealing in inconsistent units.

6.02 x 1023 is one mole, correct? So you need to look up the molar mass of silver to see how much one mole of silver weighs.

Is it appropriate to blame this "lack of creativity" on peer review or on the lack of funding made available to researchers to do what they want?

The requirements to have a blog should be listed on that same page. You need 50 posts to be eligible for a blog.

I personally find a flamethrower far more effective.

Yes, I think that'd be the correct way to do it, assuming your acceleration remains constant.

And so you make the leap that the Nazis had something to do with it? Correlation [math]\neq[/math] causation.

If we're going to talk physics, let's use the accepted physics definitions of words. Whoops, sorry.

Says who?

No, it means a change in speed.

Lesson 3: The formal definition of differentiation and some basic shortcuts So now we've figured out how to differentiate a basic function using the methods above. You'll notice how tedious and boring they were to work out (if you don't think it was tedious, wait until you try more difficult functions). Surely there are some shortcuts. There are. But first, we'll formalize what we know about differentiation into a simple equation: [math]\frac{d}{dx}f(x) = \lim_{h \to 0} \left( \frac{f(x+h) - f(x)}{h} \right)[/math] You're probably now wondering "what does that [imath]\frac{d}{dx}[/imath] thing mean and why is it up there?" In short, that's the notation to describe a derivative of a function - it means "the derivative with respect to x of the function f(x)." (The d is not a variable, it's an operator, so you can't cancel it out of that fraction.) You may also see things like f'(x) (pronounced "f prime of x"), which also indicates a derivative. Anyway, the fancy limit above is essentially the same as the equation we used before. You'll see why we used the limit - you can't plug in 0 for h, but you certainly can evaluate the limit as h gets infinitesimally close to 0. (Re-read lesson 2 if you don't quite understand.) Shortcuts In Lesson 2, we plugged numbers into an equation and differentiated it. Suppose we want to find the slope at several points on the same graph. We'd have to do all that tedious factoring and simplifying every single time - or not. It turns out that the formula above can work on the equation you're differentiating even without real values in it. In other words, you can leave the x variable in and differentiate and get an equation that will give you the slope at any given point on the curve. Let's try it for the function f(x) = x2. We can see that this is true: [math]\frac{(x+h)^2 - x^2}{h} = \frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h[/math] That means that the limit simplifies rather nicely: [math]\frac{d}{dx}x^2 = \lim_{h \to 0} \left( \frac{(x+h)^2 - x^2}{h} \right) = \lim_{h \to 0} (2x+h) = 2x[/math] (If you don't get that last step, remember that we're making h approach 0. When h is no longer on the bottom of a fraction, we can safely make it zero without "breaking" the equation.) So what's this mean? It means that at any point on the curve x2, the slope of the curve is 2x. You could say that [math]\frac{d}{dx} x^2 = 2x[/math] But I want to do it faster! But wait, there are yet more shortcuts! If you try differentiating a few simple equations, you might notice a pattern. Take a look: [math]\frac{d}{dx} x^2 = 2x[/math] [math]\frac{d}{dx} x^3 = 3x^2[/math] [math]\frac{d}{dx} 2x^2 = 4x[/math] Notice a pattern? Basically, if you have a function of the form axn, the derivative is [math]\frac{d}{dx} ax^n = anx^{n - 1}[/math] This rule also applies for longer equations. Suppose I have the equation [imath]f(x) = x^3 - 2x^2 + 2x - 3[/imath]. The derivative of that equation is equal to the derivatives of all the parts, added together, like so: [math]f'(x) = 3x^2 - 4x + 2[/math] You may have noticed that the term "- 3" vanished from the equation, and you're right: it has no variable in it, so we can leave it out of the derivative. The 2x became a 2 for similar reasons. Watch what happens when we take the derivative of 2x with our rule: [imath]\frac{d}{dx} 2x^1 = 1 \times 2x^0 = 2[/imath] (because x0 = 1). Remember, if you need help understanding any of this, you can just ask in our calculus forum.

Get yourself a monkey and a small fire extinguisher. Train the monkey to use the fire extinguisher... you'll see where I'm going.

What you have is a hypothesis -- an explanation not yet experimentally tested for some strange phenomenon.

Audacity might do the trick.