Physics: Post your doubts here!

[Thought blocker]

I feel honored

You deserved so.

 

[Thought blocker]

is Thermal Physics included in As level ??

NO

 

[leadingguy]

cool!! then too light syllabus it is .. The thing is k I forgot all and messed up. with what is in A level and As level. Thank u for the assistance

 

[Thought blocker]

cool!! then too light syllabus it is .. The thing is k I forgot all and messed up. with what is in A level and As level. Thank u for the assistance

Anytime

 

[usama321]

Can someone please help me with q4 part c. Thanks

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

 

[Thought blocker]

Can someone please help me with q4 part c. Thanks

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

stress = force / area
For max area, we need to know min area so,
Force= 1.9*10^3 , Stress= 9.5*10^8 We got stress from b part
so min area= F/S
=(1.9*10^8)/(9.5*10^8)
=2*10^-6
So for max area= Cross-sectional area-min area
=(3.2*10^-6)-(2*10^-6)
=1.2*10^-6 (FINAL ANSWER)

 

[usama321]

stress = force / area
For max area, we need to know min area so,
Force= 1.9*10^3 , Stress= 9.5*10^8 We got stress from b part
so min area= F/S
=(1.9*10^8)/(9.5*10^8)
=2*10^-6
So for max area= Cross-sectional area-min area
=(3.2*10^-6)-(2*10^-6)
=1.2*10^-6 (FINAL ANSWER)

Is that the minimum area of the wire?

 

[usama321]

Btw, i still don't understand the concept behind it. How does the sphere contribute to making the wire unbreakable

 

[Thought blocker]

Is that the minimum area of the wire?

Minimum area is 2*10^-6
Maximum area = Cross-sectional area - Minimum area
so Cross sectional area= 3.2*10^-6
so max area we got is (3.2*10^-6)-(2*10^-6)= 1.2*10^-6

 

[Thought blocker]

Btw, i still don't understand the concept behind it. How does the sphere contribute to making the wire unbreakable

IT WAS BRITTLE SO, ITS RELATED TO THAT ( I AM NOT SURE ABOUT THIS )

 

[Muskan Achhpilia]

L = 1/2 aT^2

Only the time changes now, from T to 0.5 T..

s = 1/2 a 0.25T^2

L/s = 1/0.25
0.25L

Thank you so much can you please explain the following question too-


Answer is B

Thank you once again!

 

[Muskan Achhpilia]

Please can someone explain the following question



The answer is B

 

[Muskan Achhpilia]

View attachment 38192

Thank you so much!

 

[Suchal Riaz]

Please can someone explain the following question

View attachment 38252

The answer is B

 

[usama321]

Suchal Riaz
Can you please help me with this question

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
4c. I can't understand the concept behind it.

 

[Suchal Riaz]

Suchal Riaz
Can you please help me with this question

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
4c. I can't understand the concept behind it.

Very simple.
Find the minimum area for which the rod doesn't break then subtract that from the total area of wire to find the maximum area of bubble.
This might help you:

 

[Thought blocker]

Very simple.
Find the minimum area for which the rod doesn't break then subtract that from the total area of wire to find the maximum area of bubble.
This might help you: View attachment 38254

This was what I said, usama321 or I was wrong ??? Suchal Riaz ??? I am not getting, I told was same only, wasn't it ?

 

[usama321]

Very simple.
Find the minimum area for which the rod doesn't break then subtract that from the total area of wire to find the maximum area of bubble.
This might help you: View attachment 38254

Ohk, thanks For the bubble to have the maximum area, the wire needs to have the minimum area, and the rest is occupied by the bubble. Thanks

 

[unique111]

What is the concept behind potentiometer? What are the formulae?

 

[unique111]

Help needed.