Physics: Post your doubts here!

[Suchal Riaz]

I dont get XY part ._.

when the resistance of variable resistor increase it's PD increase because V=IR
so the p.d across XY decrease as V1=R1/(R1+R2) * E
so the p.d per unit length will decrease
so it must more towards right to make the same amount of p.d as it did before.

 

[Thought blocker]

WE HAVE TO FIND IT FOR T=3S AND ACCELERATION IS NOT CONSTANT!!!!!!!!!!!!!!!!!!!!!

Han so make a tangent and it should touch the point at 3s on graph as I did... and find the gradient of the tangent you'll get acc at 3 sec...

 

[ahmed faraz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_13.pdf Q 15 Plz Full Explaination

 

[Thought blocker]

9702/12/M/J/10 Q-9 PLs solve dis 1

I solved na nai ayi kya shamajh ?

 

[Thought blocker]

when the resistance of variable resistor increase it's PD increase because V=IR
so the p.d across XY decrease as V1=R1/(R1+R2) * E
so the p.d per unit length will decrease
so it must more towards right to make the same amount of p.d as it did before.

ty

 

[Suchal Riaz]

9702/12/M/J/10 Q-9

at time t=3 draw a tangent and find it's gradient. thought blocker was using correct method.

 

[Asad Moosvi]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

Question 8 please!

 

[ahmed faraz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf Q 11 and 14 or any of them Pleaseeeee

 

[Rockstar RK]

at time t=3 draw a tangent and find it's gradient. thought blocker was using correct method.

 

[ahmed faraz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf Q14 this is Last Doubt!! Thanks In ADVANCE PEEPs!

 

[Suchal Riaz]

rockstar PK

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_13.pdf Q 15 Plz Full Explaination

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf Q14 this is Last Doubt!! Thanks In ADVANCE PEEPs!

resultant force = 10-4=6N
acceleration = force / mass
mass = 20/10=2Kg
acceleration = 6/2 = 3m/s²

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_11.pdf

Question 8 please!

LOT OF TIMES
8)
Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.

 

[ahmed faraz]

Thanxxxxxxxxxxx

 

[ahmed faraz]

resultant force = 10-4=6N
acceleration = force / mass
mass = 20/10=2Kg
acceleration = 6/2 = 3m/s²

Wah You Made it Look simple !

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_11.pdf

Question 8 please!

let the distance of one-side of the flight be s
for going: speed = s/t1= 600 so t1=s/600
for coming back: speed = s/t2 = 400 so t2 =s/400
now if we find overall speed as total distance/total time = 2s/(s/600 + s/400) = 2s/(s/240)=480km/h

 

[Suchal Riaz]

Wah You Made it Look simple !

experience of 14 years of papers all variants

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf Q 11 and 14 or any of them Pleaseeeee

11)
theta as 0
The tension in the rope needs to balance out the weight of the student acting downwards, so since the total tension needs to be opposite and equal to the weight, Let tension in rope on left side be T1 and the tension on rope on right side T2 and :. Total tension = T1sino + T2sino
add them up... W= 2Tsino, rearrange to get T= W/ 2sino. Answer: B

14)
find the hypotenuse force 1st.
200/sin30. that will give u 400N.
Then 400*1.5 gives 600J.
Add frictional force and u get 750J.

 

[ahmed faraz]

experience of 14 years of papers all variants

Wow..! i feel Worried now...! i Ony did Variant One 2002 onwards !