Physics: Post your doubts here!

[Mustehssun Iqbal]

This question or a question like this has been also answered somewhere before in this thread. You can check it too if you want to.
And I suggest you redo this question too.

 

[Mustehssun Iqbal]

View attachment 6427

Assalamu alaikum,
Young modulus = stress/ strain
therefore Young modulus = F/A/delta l/l
therefore Young modulus = Fp l/A delta l
therefore E = Fpl/delta l => Fp = A delta le/l
Young modulus is represented by E and Fp represents the Force on wire P;
therefore E = Fq (2l )/(A/2) delta l => Fq = EA delta l/4l
therefore Fp/Fq = A deltaE/l/AE delta l/4l
therefore Fp/Fq = 4 Ans D
p.s. check your marking scheme : )

 

[Mustehssun Iqbal]

View attachment 6429

Young modulus = Stress/Strain
E = F/A/delta l/l
therefore E = 60/pi r^2/delta l/l
therefore E = 60/pi (d/4)^2/delta l/l
Equating both the Young modulus, since the wire is of the same material;
therefore 60/pi r^2/delta l/l = 60/pi r^2/32/l/4
therefore l = 8mm Ans C
Note: In this question, the value of extensions are put in mm instead of metres. The answer gained is also in mm.
l = 8 mm Ans C

 

[rideronthestorm]

Thanks.in the 1st question u answered , How do we know which ones clockwise and which ones anticlockwise :/ ? and why are we doing 10x2 the pivot distance is 4 from the 10n force ...

 

[rideronthestorm]

can u answer the other questions too plz ?

 

[ABDSyed]

Any Genious Here Giving A2
PLZ Help Me

 

[rideronthestorm]

the remaining ones from before and these too

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]

D,A,D,B,C,B,B,C,B in the order uploaded .

 

[guigao1994]

Hey everyone!

this question is really anoying me. The question asks us to calculate the potential difference at B when the temperature is 15 C The marking scheme says that the potential difference at B is 2.4 V but I have calculated it to be 6.6 V, using 2200*9/(2200+800).
Consequently it asks for the output power which would be +9.0 V in case I am correct. And following which LED would be bright, if I am correct the first one (green).
All the answers in the marking scheme are the opposite of mine because of the first difference in the answer. 2.4 vs 6.6.

is the marking scheme wrong?

Thank you very much!

 

[omg]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q4 C (ii) plss helpp

 

[darknessinme]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q4 C (ii) plss helpp

Efficiency=(Useful Energy Output/Total Energy Input)x100%
It says the machine is only 25% efficient. So only a quarter of the kinetic energy of the water passing through the wheel can be used to produce a useful 110kJ output (this is every second).
So you get : 0.25 x (Kinetic Energy of water flowing every second through wheel)=110kJ
0.25 x 0.5 x (mass of water) x (15.3440...)^2=110
mass of water= 3740kg
Final answer= 3740kg/s

 

[unique840]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf
how to solve q4 part b