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answering you in 2012:-their frequencies n wavelenght plz
path difference = 28 cmhttp://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
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.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 ??^_^
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 Dhttp://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
28) its d because decreasing the frequency means increasing the wavelength. v=f into lambda.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 11http://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 8 was already answered above.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
thanks a bunch =Dpath 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
Assalamoalaikum wr wb!Any Genious Here Giving A2
PLZ Help Me
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 ?- 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.
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.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 ?
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