General Thoughts: Physics (2 Viewers)

Mr_Kap · Oct 27, 2015

RIP for that copper question.

I drew a straight line with negative gradient and then from C to D it was FAIRLY straight (but not exactly).

lol. at least my explanation was alright, maybe 2/5 lol.

Drsoccerball · Oct 27, 2015

Mr_Kap said:
RIP for that copper question.

I drew a straight line with negative gradient and then from C to D it was FAIRLY straight (but not exactly).

lol. at least my explanation was alright, maybe 2/5 lol.

Exactly this is why I used Extension 2 since I saw Integrand do it once and it actually worked

PhysicsMaths · Oct 27, 2015

Drsoccerball said:
Exactly this is why I used Extension 2 since I saw Integrand do it once and it actually worked

They can't award you marks for doing this.

Drsoccerball · Oct 27, 2015

PhysicsMaths said:
They can't award you marks for doing this.

I didn't think they would... It was for me not them.

Mr_Kap · Oct 27, 2015

Biggest regret: Dropping 4u. Should've kept it just for physics.

was this the mechanics topic where you learnt this..because that is the ONLY 4u topic i didn;t learn.

or was it the 4u integration topic ---> because i already know 4u integration so if i could have used it i'm such a dickhead for not using it.

Drsoccerball · Oct 27, 2015

Mr_Kap said:
Biggest regret: Dropping 4u. Should've kept it just for physics.

was this the mechanics topic where you learnt this..because that is the ONLY 4u topic i didn;t learn.

or was it the 4u integration topic ---> because i already know 4u integration so if i could have used it i'm such a dickhead for not using it.

Mechanics helped me alot for physics... Mainly resistance questions...

SlimiJay · Oct 27, 2015

no one has the paper??

Drsoccerball · Oct 27, 2015

SlimiJay said:
no one has the paper??

Ask Fizzy.

IamKirby · Oct 27, 2015

Drsoccerball said:
I didn't think they would... It was for me not them.

Hey man, I did 4u but I don't know how it could be used here. Can you explain? Thanks

Sent from my iPhone using Tapatalk

apurba · Oct 27, 2015

IamKirby said:
Okay haha thought so, thanks

Sent from my iPhone using Tapatalk

Okay this was my reasoning for my graph:

I said as eddy currents are induced on the copper plate, causing it to slow down.
but then, the trolley passes through the magnet, and the direction of the eddy currents are then switched, causing it to experience a force in the opposite direction (i.e. accelerating it to the right slightly).

Where did I go wrong?

Drsoccerball · Oct 27, 2015

IamKirby said:
Hey man, I did 4u but I don't know how it could be used here. Can you explain? Thanks

Sent from my iPhone using Tapatalk

acceleration = -k

$v \frac{dv}{dx} = -k$

$\frac{dx}{dv}= -\frac{v}{k}$

$[x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed

even though k is changing it gave me a rough estimate of what the graph looked like.

Thjv · Oct 27, 2015

Drsoccerball said:
acceleration = -k

$v \frac{dv}{dx} = -k$

$\frac{dx}{dv}= -\frac{v}{k}$

$[x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed even though k is changing it gave me a rough estimate of what the graph looked like.

4u motion..

Drsoccerball · Oct 27, 2015

apurba said:
Okay this was my reasoning for my graph:

I said as eddy currents are induced on the copper plate, causing it to slow down.
but then, the trolley passes through the magnet, and the direction of the eddy currents are then switched, causing it to experience a force in the opposite direction (i.e. accelerating it to the right slightly).

Where did I go wrong?

As it leaves it is also slowed down remember how a similar experiment where a magnet dropped through a metal induces an m field to always slow down the magnet whether it enters or leaves.

Drsoccerball · Oct 27, 2015

Thjv said:
4u motion..

It was for me just to see if the graph was curved.

Mr_Kap · Oct 27, 2015

Drsoccerball said:
As it leaves it is also slowed down remember how a similar experiment where a magnet dropped through a metal induces an m field to always slow down the magnet whether it enters or leaves.

if i explained all the shit correctly with CORRECT REASONING (i did) but drew the graph that i did (negative straight line from A to C, and then a fairly straight line from C to D), can i get 2/3 out of 5???

InteGrand · Oct 27, 2015

Drsoccerball said:
acceleration = -k

$v \frac{dv}{dx} = -k$

$\frac{dx}{dv}= -\frac{v}{k}$

$[x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed even though k is changing it gave me a rough estimate of what the graph looked like.

$This is basically a derivation of the well-known formula $v^2 = u^2 + 2a\Delta s$ for \textsl{constant} $a$.$

Drsoccerball · Oct 27, 2015

InteGrand said:
$This is basically a derivation of the well-known formula $v^2 = u^2 + 2a\Delta s$ for \textsl{constant} $a$.$

Agahahahaahah how many things did I derive in the exam...

Drsoccerball · Oct 27, 2015

Mr_Kap said:
if i explained all the shit correctly with CORRECT REASONING (i did) but drew the graph that i did (negative straight line from A to C, and then a fairly straight line from C to D), can i get 2/3 out of 5???

I think so. Also the question would've been more amazing if there was friction from the floor aswell... But since V didnt change there was no friction...

Thjv · Oct 27, 2015

Drsoccerball said:
I think so. Also the question would've been more amazing if there was friction from the floor aswell... But since V didnt change there was no friction...

Pls no... this is hsc physics

SlimiJay · Oct 27, 2015

Drsoccerball said:
I think so. Also the question would've been more amazing if there was friction from the floor aswell... But since V didnt change there was no friction...

wait wait no friction? why wouldnt there be friction? i did my graph incorporating friction as well

General Thoughts: Physics (2 Viewers)

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)