Pink Poogle Toy Forum

I think thoese are the ancient wonders of the world...I'm looking for the natural

The natural wonders of the world are :

-Grand Canyon in Arizona, USA
-Northern Lights
-Mt. Everest in Nepal
-Paricutin Volcano in Mexico
-The Harbor at Rio De Janeiro
-Victoria Falls in Zambia/Zimbabwe
-The Great Barrier Reef in Australia

I hope I helped in some way.

They can be found here, with a brief article on each :

http://www.cnn.com/TRAVEL/DESTINATIONS/9711/natural.wonders/

I'm doing a report on "Working in the Computer Industry" and stupid me, I forgot to write down notes about the "Hardware" part of the Industry. Can anyone tell me anything about it?

Yep

Let me know if you need any more information, since I live in Melbourne

thx so much!!!

It is the stuff inside computers like...

Computer hardware typically consists chiefly of electronic
devices (CPU, memory, display) with some
electromechanical parts (keyboard, printer, disk drives,
tape drives, loudspeakers) for input, output, and storage,
though completely non-electronic (mechanical,
electromechanical, hydraulic, biological) computers have also
been conceived of and built.

I am doing a take home assignment for Physics and I don't understand how to do the first question. Here it is:

1. A 2.0-kg object is sliding across a smooth surface at 4.0m/s when it collides with a stationary 3.0-kg object. The collision lasts for 0.80s after which the smaller object has slowed to a speed of 1.0m/s. Using this diagram below,
A) Determine the acceleration of the smaller object (After collision).
B) Determine the force that the smaller object exerts on the larger one (After collision).
C) Determine the speed of the larger object immediately following the collision.

Diagram



All the speeds are going towards the right (--->).

Part A)
Well, you know it's a frictionless surface, which is hugely important.

You know that the small object decelerated from 4.0 m/s to 1.0 m/s in a span of .80 s, so you can figure out the acceleration on the small object. (What is the formula for figuring acceleration? You know Δv and Δt...)

Part B) You know acceleration, you know mass. You can then know force.

Part C) You know the force (because the force exerted is equal in both directions) and you know mass....

Hey, smart math-inclined people, any chance I can get some help on this? I'm sure I'm doing something stupid here, but as I don't know what is, I guess it doesn't hurt to ask.

I'm supposed to find the intercepts of the equation "y=x^3 + x^2 - 3x".
Obviously, the y intercept is 0.

To find the x intercept, I started by factoring out an x, getting "y=x(x^2 + x - 3)
Therefore, one intercept has to be x=0. However, although I know that (x^2 + x - 3)=0 to find the other intercept, the quadratic can't be factored, so how do I find it?

Hope that wasn't too convoluted.

I would use the quadratic formula, which I'm sure you've learned. On that page, scroll down to where it shows you the example, as a refresher. Before that is just the proof, which you don't need.

My arithmetic is horrible, but I got: -1 +/- sq. rt. of 13 all over 2. Now if only I knew how to show that to you. Erm...
(-1 <u>+</u> square root of 13) divided by 2.

Maybe we should both ask Scott for help.

I graphed it and got intercepts approximately 1.3 and -2.3, I know you need it more specific than that, but at least it will give you starting points to work toward.

That is correct. And don't forget x = 0 as the third x-intercept.

I need a little help in French. Making some kinda news broadcast skit, and I don't exactly know how to translate a few phrases. Please don't use an online French-English dictionary unless you're sure of one that helps.



Ummm...ignore the weird storylines.

Cinquant citoyens sont mort, et plus de citoyens ont disparu/sont disparu. -- I'm pretty sure the first one would be more appropriate.

---

I would do the second one but my dicitonary is upstairs and I'm too lazy to get it.

Ah, that reminds me...I have a French-English dictionary. I've fixed a little grammar and got the terms, thank you.