Physics: Post your doubts here!

[2012]

sory length

 

[confused123]

their frequencies n wavelenght plz

answering you in 2012:-
wavelength range / m
radio waves : 10^6 to 10^-1 m
microwaves : 10^-1 to 10^-3m
infra-red: 10^-3 to 7 into 10^-7m
visible light: 7 into 10^-7m (red) to 4 into 10^-7m (violet)
ultraviolet : 4 into 10^-7 to 10^-8m
X-rays: 10^-8 to 10^-13m
Y-rays: 10^-10 to 10^-16m

for frequencies see this:-
http://en.wikipedia.org/wiki/File:EM_Spectrum_Properties_edit.svg

 

[2012]

thanks

 

[omg]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf
first variat qs paper!
Q.5 part (b)!!
help needed

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
first variat qs paper!
Q.5 part (b)!!
help needed

path difference = 28 cm
frequency is increased from 1.0 kHz to 4.0 kHz
that means (33o/1000)m to (330/4000)m
= 0.33 m to 0.0825 m
in cm,
= 33 cm to 8.25 cm
Now, lets check wavelengths for minima (n + ½)λ
0.5λ = 28 --> λ = 28/0.5 = 56 cm
1.5λ = 28 --> λ = 28/1.5 = 18.7 cm
2.5λ = 28 --> λ = 28/2.5 = 11.2 cm
3.5λ = 28 --> λ = 28/3.5 = 8 cm

In the range, 33 cm to 8.25 cm
There are only two wavelenghts, which are:
18.7 cm and 11.2 cm

Therefore two minima

 

[KurayamiKimmi]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_ms_2.pdf
can anyone pretty please explain parts b and c of question 3 ??^_^

 

[Silent Hunter]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

help required in Q8 and 15 JazakAllah

 

[A.ELWY 7]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q 11, Q28, Q 32 need some explanation pls...thanx

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_ms_2.pdf
can anyone pretty please explain parts b and c of question 3 ??^_^

3 b) you see the mass strikes the horizontal ground when it reaches 4.2 m/s, it remain contact with the ground till -3.6 m/s. so the time between is the time taken into consideration while calculating the momentum.

45/1000 ( v-u) ( -3.6 - 4.2) = -0.35Ns , but ms says +.35 i dnt knw why.

c) Force = change in momentum / time when the mass was in contact with the ground which is 0.57-0.43

so 0.35Ns/ 0.14 = 2.5 N

i can't understand why is 4.2 m/s taken as the final velocity.

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

help required in Q8 and 15 JazakAllah

8) the displacement time graph shows us the velocity. its gradient telling us velocity so we can see after indicating their is a change in the velocity. hence D

15 ) i can't figure how its B? i think it shud be A ''16kj'' as the K.E will be 4 times higher then the previous. as force plus distance increases by twice. only one situation come to my mind if it has to be B option, that is maybe we are adding the possible friction present in the trolley at the second case. bcz it is NOT specified when the force is 2F and distance 2s dat it is still frictionless trolley

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q 11, Q28, Q 32 need some explanation pls...thanx

28) its d because decreasing the frequency means increasing the wavelength. v=f into lambda.
therefore there will be less diffraction according to lambda = ax /d formula. and dre will be less distance between maximas and minimas

11 and 32 i am unable to do.

 

[Mustehssun Iqbal]

Assalamu alaikum,
Can you please explain how amplitude is calculated and also part c(2) in this question??
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
There's its marking scheme:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_ms_4.pdf

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q 11, Q28, Q 32 need some explanation pls...thanx

- Question 11
Driving force = m x a = 2 x 9.81 = 19.62
Driving force - frictional force = m x a
19.62 - 6 = (8+2) x a
a = 1.362. Approximately = 1.4 m/s^2
Now the main doubt here is why are we doing 8+2. It would help if you've studied P4 mathematics, but if you haven't, then this is why:
T = tension, F = frictional force
Forces on box: T-F = 8a
Forces on 2kg mass = 2g - T = 2a (Note: we're taking -T here because in the previous equation we took T as positive when it was away from the box, now T is towards the box and away from the 2kg mass)
Substitute the first equation in the second one. 2x9.8 - 8a - F = 2a
10a = 13.6
a = 13.6/10 = 1.36 m/s^2. Approximately = 1.4 m/s^2

- Question 28 was already explained above.

- Question 32
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

help required in Q8 and 15 JazakAllah

Question 8 was already answered above.
Question 15:
F = ma1
1/2 m u^2 = 4
1/2 m v^2 = 8
That leaves us with a1 = F/m
u^2 = 8/m
v^2 = 16/m
We put these in the formula: v^2 = u^2 + 2a(1)s
That gives us: 16/m = 8/m + 2(F/m)s
8/m = (2Fs)/m
Fs = 4
----------
Now, in the next part 2F = ma2 (a2 is not 2 multiplied by a but rather a2 is a different acceleration [2 is in subscript])
a2= 2F/m
u^2 = 8/m (since initial KE remains same, so does u^2)
1/2mv^2 = E (E is the final KE)
v^2 = (E x 2)/m
v^2 = u^2 + 2a(2)s
2E/m = 8/m + 2(2F/m)2s
(2E-8)/m = (8Fs)/m
2E - 8 = 8Fs. But we know Fs = 4 from the previous working.
So 2E = 8x4 + 8 = 40
E = 40/2 = 20 J
[I know, I know not worth the 1 mark]

 

[omg]

path difference = 28 cm
frequency is increased from 1.0 kHz to 4.0 kHz
that means (33o/1000)m to (330/4000)m
= 0.33 m to 0.0825 m
in cm,
= 33 cm to 8.25 cm
Now, lets check wavelengths for minima (n + ½)λ
0.5λ = 28 --> λ = 28/0.5 = 56 cm
1.5λ = 28 --> λ = 28/1.5 = 18.7 cm
2.5λ = 28 --> λ = 28/2.5 = 11.2 cm
3.5λ = 28 --> λ = 28/3.5 = 8 cm

In the range, 33 cm to 8.25 cm
There are only two wavelenghts, which are:
18.7 cm and 11.2 cm

Therefore two minima

thanks a bunch =D

 

[XPFMember]

Any Genious Here Giving A2
PLZ Help Me

Assalamoalaikum wr wb!
i'm giving A2.....

 

[A.ELWY 7]

- Question 11
Driving force = m x a = 2 x 9.81 = 19.62
Driving force - frictional force = m x a
19.62 - 6 = (8+2) x a
a = 1.362. Approximately = 1.4 m/s^2
Now the main doubt here is why are we doing 8+2. It would help if you've studied P4 mathematics, but if you haven't, then this is why:
T = tension, F = frictional force
Forces of box: T-F = 8a
Forces on 2kg mass = 2g - T = 2a (Note: we're taking -T here because in the previous equation we took T as positive when it was away from the box, not T is towards the box and away from the 2kg mass)
Substitute the first equation in the second one. 2x9.8 - 8a - F = 2a
10a = 13.6
a = 13.6/10 = 1.36 m/s^2. Approximately = 1.4 m/s^2

- Question 28 was already explained above.

- Question 32
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

thanx very much jaf....for question 11 i did it the same way i just wanted to check if my way was correct....and for question 32 how did u no they where in parrallel ?

 

[Jaf]

thanx very much jaf....for question 11 i did it the same way i just wanted to check if my way was correct....and for question 32 how did u no they where in parrallel ?

Think about it... how else could they have been connected? There are clearly spaces between the individual wires. This shows that they are infact 7 distinct wires running alongside each other. The only way they could have been in series would be if they were connected end to end.
(PS: Real Madrid FTW!! )

 

[A.ELWY 7]

well thanx

 

[A.ELWY 7]

so how does everyone do in paper 1....i wish the curve become low this year!!