Cap'n Refsmmat

Find the acceleration the car is experiencing. You'll need an equation that relates velocity and acceleration: [math]v = at + v_0[/math] where v = final velocity, a = acceleration, t = time, and v0 = initial velocity. (You should have an equation like this somewhere.) Just plug and chug. You should hopefully know that force = mass x acceleration, so you just stick in the numbers and get the answer.

First, you'd want to convert the vehicle's velocity to meters per second. Next, you want to see how far the car traveled before the driver hit the brakes. The equation for that is something like this: [math]x = vt[/math] The distance = velocity * time. The distance traveled after the driver hits the brakes is weirder. The equation I've used is this: [math]V^2 = V_0^2 - 2a(x - x_0)[/math] Where V2 = the final velocity (squared) (which is 0, because the car is stopping), V0 = the initial velocity of the car, a = the acceleration of the car (the breaking rate), and x - x0 is the distance it travels.

I'd be more comfortable passing judgment if someone could find out what was actually going on at their compound and if they were actually being abused. It's hard to tell right now.

I've worked with gedit for a while (Linux's default text editor). If you know what you're doing, you can get away with using any text-editor with syntax highlighting. If you want something drag-and-drop, I'd suggest Nvu or what Pangloss linked to.

What's the standard acceleration due to gravity? (Hint: It's always 9.8 m/s2)

Are you familiar with the equation [math]x = \frac{1}{2}at^2[/math] where x = displacement, a = acceleration, and t = time elapsed? The version you've seen might look different, but it should at least resemble this. You should know a (it's gravitational acceleration, which is the same everywhere), so you can just plug in the numbers and get an answer.

There are a thousand meters in one kilometer, so you can divide your number in km/h by 1,000 to get m/h. There are 3,600 seconds in an hour (60 x 60), so you then multiply by 3,600 to get m/s. Simple way: [math]\text{m/s} = \text{km/h} \times \frac{3600}{1000}[/math]

Here you go. We now have a Book Talk forum for all your book-related needs.

I'd just try it on a plate outside on concrete where a cooking oil fire will be able to burn itself out. Make sure the candle floats well above the oil-line, and if anyone asks what you're doing, you're making a sacrifice to the great Jebus under the concrete.

Pure awesome.

Good in that it encourages people to take myths and put them to the test. Bad in that it doesn't always set up proper controls and run experiments in the most scientific fashion.

Water and cooking oil are different densities, so they won't mix together. You'll be able to see the water that condenses. edit: or, as D H said fractionally quicker...

"As the candle burns, it uses up the oxygen in the air. The candle wax melts to form the elements carbon and hydrogen, which combine with the remaining oxygen to form water and carbon dioxide. As this oxygen is used up, the water level in the jar rises." That's the explanation I found in a book. I think it assumes that the water will condense somewhere. Heat wouldn't make the gas expand if we assume that we're putting the glass over some already warmed (and therefore slightly less dense) air.

We're all going round and round and round and round and round and round and round in circles. I think this thread needs a sanity injection.

It has no tangible/physical form, but it does exist.

Or, in other words, relativity defies common sense.

I quote: Where's the oscillation? You are basing your claims on the Bohr model, which is obsolete. I refer you to any good book on quantum mechanics or chemistry. For example, Zumdahl Chemistry, 5th edition, page 306: "Electrons do not move around the nucleus in circular orbits." That video explains the Bohr model. Page 306 of Zumdahl Chemistry: "Although some attempts were made to adapt the [bohr] model using elliptical orbits, it was concluded that Bohr's model is fundamentally incorrect."

Cosmic rays are high-energy particles -- some of them ludicrously high-energy.

What'll she be primarily using it for? Writing papers, email and the Internet? If there are games, video editing or major Photoshop use you'll want to get a significantly more powerful machine.

It isn't.

25cm3 or 25cm on each side? A 25cm3 box would only be about 3cm on each side -- a little small.

We issue warnings to people discussing potentially hazardous procedures. There's a sticky in the Chemistry section explaining the rules.

You'd have to account for the expansion of the water as it freezes. I'd imagine that would expand your cube a little. Your block of ice will weigh about 30 pounds, by the way.

I think we're making this a lot more complicated than it really is.

Easy. Simulate a system that can tell the difference between simulated self-awareness and real self-awareness. A simulated system's indistinguishable from the real thing, after all.