Physics: Post your doubts here!

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
Q18

 

[..sacrifice4Revenge..]

Chemistry ?

*facepalm*
I still got chem on my mind -_-

 

[ZaqZainab]

*facepalm*
Im still got chem on my mind -_-

And now that i am done i am here like wha is chem

 

[..sacrifice4Revenge..]

Chemistry ?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Q18

K.E. gained = P.E. lost - Work done against friction
K.E. gained = 50 - 10
K.E. gained = 40 J

Final K.E. = Initial K.E. + K.E. gained
Final K.E. = 5 + 40
Final K.E. = 45 J

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

5)
time for 8.5 cycles is 6 x time base setting i.e. 6x(0.01) = 0.06
time period = 0.06/8.5
frequency will be 1/Ans
its close to 140

15)
Efficency=Useful Output energy/Input Energy multiplied by 100
Now Input energy=Force*Distance=Fs
For output energy if u read the question carefully the car is moving with constant speed v so it is not gaining kinetic energy.It is only gaining Potential energy.Now u have to calculate height.Whenever u need to calculate h and distance of slope and angle of slope with horizontal is given,use the formula
sin alpha=h/s so h=s sin alpha
Gain in P.E=m*g*s sinalpha
m*g*s sinalpha divided by Fs giving u D as answer

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q5

 

[Thought blocker]

Q5

You are just

 

[ZaqZainab]

You are just

You are
i am

 

[crazytaylorfanXD]

22, it is C. Congrts
23, youngs modulus for one material is always the same no matter what length u take etc, so its steel in both cases and it remains E

haha got it thanks

 

[Thought blocker]

You are
i am

We both are
and no one is

 

[crazytaylorfanXD]

Q5 B as d has a square and when we write the percentage uncertinty
we will write it as 2(uncertinity of d)/d *100

Q6 the time for it to move from X to Y will be less than time for Y to Z
so B and C are wrong
the acceleration from X to Y is NOT equat to Y to Z and so D is wrong

Q12 why not D cause when the ice-hockey puck moves back it havs a negative velocity and forward was positive and so momentum=m*v will be negative

Q16 it is inelastic collision because they travel together
Ek=0.5mv^2
after collision the mass will be doubled
0.5*2*mv^2 =mv^2
so mv^2=0.5mv^2/2=Ek/2

thank you soo much

 

[..sacrifice4Revenge..]

We both are
and no one is

you can be , i'm already
Jk

Thanks both of you

 

[Thought blocker]

you can be , i'm already
Jk

Thanks both of you

DOUBTS PUCHONA KOI!!!

 

[Thought blocker]

I forgot how to do this ???????????
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Q37
Mene kisiko sikhaya tha... but m not getting where did I explained........
Anyways..........
H E L P !

 

[..sacrifice4Revenge..]

DOUBTS PUCHONA KOI!!!

K, see you in 4 hrs with 'doubts' if there are some

 

[Thought blocker]

K, see you in 4 hrs with 'doubts' if there are some

4 hours ???????
I have to go for walk ._.

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Q17,40

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf

Q35

 

[..sacrifice4Revenge..]

I forgot how to do this ???????????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
Q37
Mene kisiko sikhaya tha... but m not getting where did I explained........
Anyways..........
H E L P !

In the new connection,. The new circuit also links the green lamp, so it will also light, 12 V was enough to light the red lamp normally previously, but now since the green is also in the circuit, potential divides and ao both lamps light dimly,

 

[Thought blocker]

In the new connection,. The new circuit also links the green lamp, so it will also light, 12 V was enough to light the red lamp normally previously, but now since the green is also in the circuit, potential divides and ao both lamps light dimly,

I got it....... thanks though!