Section I: Part B - Short Answer Responses (1 Viewer)

[ewasemma]

Okay well here goes. Pls tell me if theyre wrong (they probably are) coz i want to (roughly) estimate what i got.
Im not going to do the wordy qs and some of them i havent written down.
16-
17-
18- a) downwards
b)
19) a)
b) 514.39 m /s
c) 1/64 : 1
20 -
21-
22-
23- a
b) - 3.204 times 10 ^-16
c) 20V
24-
25-
26- arrangement A
27a)
b) r = 3.75
c) wire x
d)

Sorry! I only wrote down some of my answers. Please criticise etc

edit: james is right for the orbital ratio. i forgot to sqare root it. so the correct answer for 19c) is 1:8

 

[JamesTockuss]

What did people get for the orbital ratio? I thinkkk I got 1:8? or 8:1 whichever way it asked for it

 

[karnage]

What did people get for the orbital ratio? I thinkkk I got 1:8? or 8:1 whichever way it asked for it

aw fuck, i made a mess of that question so i didnt evaluate it and just moved on :/

 

[Alvik]

What did people get for the orbital ratio? I thinkkk I got 1:8? or 8:1 whichever way it asked for it

I also got 8:1

But I really don't even know if I did it right.

 

[kerlon44]

i said wire x coz the inverse gradient givs u the resistance, so yea i said some shit about how pure metals hav less resistance than the alloys which is seen in the lower resistance

was wire x the one we drew or was that y ?

 

[dolbinau]

I got 8:1. Honestly, with just perfect numbers how could it NOT be right lol.

 

[JamesTockuss]

I got 8:1. Honestly, with just perfect numbers how could it NOT be right lol.

haha yeh as soon as I divided one by the other and got 8, I just couldn't help but smile. Such a nice, whole number haha

 

[Kearnzo]

screwed up the voltage, used cm for some reason. LOL.

What do you guys think the physical reason behind the outlier was? I put that the ammeter didnt go past 2.5. Couldn't think of anything else.

 

[mynamewastaken]

What did people get for the orbital ratio? I thinkkk I got 1:8? or 8:1 whichever way it asked for it

same. some people in my class got something with 64 in it, others said 4:1 (or 1:4) because the radius was 4 times the other one. but yeah i got 1:8/8:1

 

[dolbinau]

Inaccurate measurement (for outlyer)

 

[JamesTockuss]

Inaccurate measurement (for outlyer)

I think I said mechanical failure of the voltmeter or ammeter would give erronous results

 

[adosh]

same. some people in my class got something with 64 in it, others said 4:1 (or 1:4) because the radius was 4 times the other one. but yeah i got 1:8/8:1

i got 1:8..... think the direction matters,, 8:1 would be wrong maybe because it asked the new period to the original period which is Tnew/Toriginal which is 1:8.....but xcorrect me if im wrong

 

[Kearnzo]

i got 1:8..... think the direction matters,, 8:1 would be wrong maybe because it asked the new period to the original period which is Tnew/Toriginal which is 1:8.....but xcorrect me if im wrong

It asked for the comparison between the orbital periods of the original orbit to the new orbit.

At the top it said that it moved from orbital 1 to orbital 2.
Therefore it was 1:8.

So yeh you are right.

 

[independantz]

It asked for the comparison between the orbital periods of the original orbit to the new orbit.

At the top it said that it moved from orbital 1 to orbital 2.
Therefore it was 1:8.

So yeh you are right.

''The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period? 2 marks''
the new orbit was orbit 2, the original orbit was orbital 1

thus it's 8:1

 

[Kearnzo]

''The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period? 2 marks''
the new orbit was orbit 2, the original orbit was orbital 1

thus it's 8:1

Ah, dam. You have the paper infront of you huh.
Well 1 mark.

Edit: Actually above it i wrote T1:T2 then below 1:8 . So I should get the marks, hopefully.

 

[pdoueihi]

''The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period? 2 marks''
the new orbit was orbit 2, the original orbit was orbital 1

thus it's 8:1

lol whered u get the paper from? or how the fuck do u remember that much detail lol

 

[independantz]

I have the best memory in the world .... nah here:

http://community.boredofstudies.org/showthread.php?t=190708

 

[hon1hon2hon3]

How did you guys get 1 : 8 or 8 : 1 ?

I got orbital velocity of orbit 2 is half of orbital velociy of orbit 1

Since G (m1m2) / r sqaure = mv^2 /r
Then from that we get v = square root ( GM / r )

Let orbit 1's radius be r1 and orbit 2's radius be r2

We then get

v=square root ( Gm / r1 ) and v=sqaure root ( Gm / r2 )

But r2 = 4r1

So

v = square root ( Gm / 4r1 )

and when you get that 4 outside of that sqaure root , it becomes half

Wouldnt that mean, the secound orbit velocity is half the orbit 1's velocity ?

ANy suggestion what i did wrong here?

 

[dolbinau]

Nothing like that, use Keplar's law of period (you are finding ratio of period IIRC). Just use different values for R.

 

[nty]

I used Kepler's Law as well. made T the subject. substituted values for R and got numerical answers.. but then it wasn't exaclty a nice ratio so i rounded it a little bit and got 8:1. anyone do that? i hope i m not so confused that i have no idea what i'm talking about.