physics quiz 2

[Sarahisme]

here i another 2 questions, how'd i go for these ones?

[Sarahisme]

my answers:

1) c

2) c

 

hows that? (cricket anyone )

 

-Sarah

[swansont]

Explain why.

[Sarahisme]

for 1) it is because resistance is proportional to the length of the wire

 

and for 2) it is because the power drop across a resistor goes as V^2/R

[5614]

But (I'm guessing here) because the wires have the same mass but are different lengths (and made out of the same material) the shorter one must be thicker to match the mass of the longer one... diameter of the wire now comes into question.

[swansont]

But (I'm guessing here) because the wires have the same mass but are different lengths (and made out of the same material) the shorter one must be thicker to match the mass of the longer one... diameter of the wire now comes into question.

 

 

And that's the key. R=pL/A, where p is the resistivity (material specific). Same mass => same volume. What happens to A?

[5614]

This is what, well, it doesn't annoy me, it just kinda bugs me... what does R and L and A all stand for?!?

[Sarahisme]

R is resistance, L is the length of the wire and A is the cross sectional area of the wire

[5614]

Cool, thanks, that makes sense!

 

I like this:

[Sarahisme]

hahah thats cool, where'd you get that?

[5614]

Honestly don't have a clue, found it years ago on the internet somewhere and just saved it as a image file onto my computer.

[Sarahisme]

lol, well it is now also saved on my computer too

[Sarahisme]

oh btw, were my justifications correct? or i suppose did they justify my answers??

[swansont]

You only allowed for a change in length for the resistance. If you double the length for constant mass, what happens to the area? What effect does that have on resistance?

[Sarahisme]

cross sectional area doesnt change, so resistance only depends on length..?

 

it was more the ssecond question which i was unsure about

[Anjruu]

Woah! That thing is awesome! I wish I had that for my electrodynamics tests...

 

1) Anyway, I may be wrong, but if the mass is the same, and the length is doubled, that would be R=p (2L)/(1/2 A). 2/(1/2) is equal to 4, so, the formula becomes R=p (4) L/A. Since p L/A=p L/A, they cancel, leaving 4. Would it not then be b? In order to acount for the mass remaining constant, the cross-sectional area must decrease by the same factor that the length increased...?

 

2) I agree with you on that one.

[swansont]

1) Anyway' date=' I may be wrong, but if the mass is the same, and the length is doubled, that would be R=p (2L)/(1/2 A). 2/(1/2) is equal to 4, so, the formula becomes R=p (4) L/A. Since p L/A=p L/A, they cancel, leaving 4. Would it not then be b? In order to acount for the mass remaining constant, the cross-sectional area must decrease by the same factor that the length increased...?

[/quote']

 

 

Yes. If total mass stays the same, so does volume. If L goes up by a factor of 2, A has to go down by a factor of 2 as well.

[Anjruu]

Right. In that case, the correct answer to question one would be "b". What am I missing, because Sarahisme said that the answer was "c".

[5614]

Yes but as you would see if you read the thread was that she said it was c and asked us to check, the whole thread has been discussing how to derive the correct answer, we wouldn't have needed to do that if she was right! (Basically she was wrong, everyone makes mistake, the last few posts have been explaining why)