Schoolwork Help Thread

[ThatGamer]

I'm kidding, of course I know what Google is, been using it for 2 years.

[blueflower999]

I'm kidding, of course I know what Google is, been using it for 2 years.

You've only been using Google for two years? 

[BlackDragonSlayer]

I'm kidding, of course I know what Google is, been using it for 2 years.

Well, you must not use it too often, then...

[ThatGamer]

 

You've only been using Google for two years? 

Is that weird?

[Bubbles]

Super quick question for all you physics buffs

If I double the diameter in an electric wire, will the resistance decrease? By what factor? This isn't even that hard but I've been stuck on it for hours

For reference the equation I'm using is:
R = p(L/(pi•r^2))

[Dudeman]

Doubling the diameter is equal to quadrupling the radius. Does that help a bit?

[Bubbles]

I'm just trying to wrap my head around the concepts. I know that doubling the diameter quadruples the radius and a bigger radius=bigger area=less resistance, but for some reason when I put it all in the equation I got a higher resistance? Unless I'm misunderstanding everything and 4R is less resistance than R (please tell me I'm wrong here)

[Nebbles]

Er, is anyone here particularly good at statistics?

[Dudeman]

Considering that r is in the denominator, the resistance is inversely proportional to the length of the radius. So if radius increases, resistance conversely decreases by a square factor relative to the length of the radius.

TL;DR, resistance should get smaller.

[Bubbles]

Er, is anyone here particularly good at statistics?

I might be. I can do pretty much any type of math except this goddamn physics

[Nebbles]

'kay, I just have a final next month and my test grades aren't too stellar and I need a lot of help.

[Bubbles]

Considering that r is in the denominator, the resistance is inversely proportional to the length of the radius. So if radius increases, resistance conversely decreases by a square factor relative to the length of the radius.

TL;DR, resistance should get smaller.

I think I might have a simplifying problem then. If R=x/2r, wouldn't 2R=x/r? Or maybe I'm starting the problem off wrong ughh

'kay, I just have a final next month and my test grades aren't too stellar and I need a lot of help.

If you're looking for an almost-complete review or going over specific chapters, I'd try out Khan Academy on YouTube or the AppStore. Mr Khan helps me keep my gpa lol

[Nebbles]

Ahhh, gotcha.

[Dudeman]

I think I might have a simplifying problem then. If R=x/2r, wouldn't 2R=x/r?

That's exactly right.

[Mashi]

Super quick question for all you physics buffs

If I double the diameter in an electric wire, will the resistance decrease? By what factor? This isn't even that hard but I've been stuck on it for hours

For reference the equation I'm using is:
R = p(L/(pi•r^2))

If you're having trouble doing questions like this using variable letters, something you can do is replace in a few numbers (as in, say that the p = 1 and l = 1.  Then say the Original Radius = 1 meter and figure out that the New Radius is 2 meters, so when you plug in, you figure out the New Resistance is 4 times less than the Original Resistance).

The basic idea for proportions like this though, is that you want to figure out what the Original Resistance (R_o) is equal to in terms of the Original radius (r_o).  Then, find the New Radius (r_new) and express r_new in terms of r_o.  When you figure that out, plug that into the equation and figure out what the proportion is by making it some number times R_o (in this case, 1/4).

Sorry if I'm being confusing, but if you need further clarification, I'll try to explain as best as I can.

Spoiler

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