Physics: Post your doubts here!

[mahabaloch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q 5 part c and d

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2
Q5 part a

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q3 part b
Q7 part b2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 part c
Q5 part a 2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Q4partb 2
Q6

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q3 part c

http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_qp_22.pdf
Q5 partc

I know its Aaaa lot but i really need help
Thank you


Some body please help

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q 5 part c and d

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2
Q5 part a

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q3 part b
Q7 part b2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 part c
Q5 part a 2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Q4partb 2
Q6

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q3 part c

http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_qp_22.pdf
Q5 partc

I know its Aaaa lot but i really need help
Thank you


Some body please help

i need to sleep. i will answer in the noon today. if i forget please remind me.

 

[Thought blocker]

i need to sleep. i will answer in the noon today. if i forget please remind me.

I'll remind you

 

[Suchal Riaz]

I'll remind you

you still alive?

 

[Thought blocker]

you still alive?

On the urge of death

 

[Suchal Riaz]

On the urge of death

*verge

 

[Thought blocker]

*verge

i AM WAITING to hear your accent. Good night :')

 

[Hijab]

Well, i only answered the second part For the first part, it would
1/12 + 1/6 + 1/12 = 3

I am sorry... I misread ur statement..... Thankyou

 

[Hijab]

For the first part, the circuit is:
the upper two resistors are in series
The lower two resistors are in series
And the the one in the middle is well.. alone xD
And all the above are in parallel to each other
I made that diagram just for the second part, it is not applicable to part i

Thankyou.....

 

[Hijab]

Is the answer C?
Note: I am not complete sure about this question
extension = F * l/A*young modulus

Now in the model, all the linear dimensions are one tenth.

Now, the length, width and height of the load are 10 times the one in the model. Thus 10*10*10 = 1000 times the volume of the load in the figure. As volume and mass are directly proportional, the mass should also increase 1000 times, and so should the weight.

Now, the radius of the wire in the model is times the one in the model (linear dimensions) Thus the area should be 100 times more than the one in the model due to the square in the area of a wire

The length of the wire would just be more by a factor of 10.

The young modulus should not change, as it is the same for a material

(1000 * 10/100)/1 = 100

Kindly confirm about this one.

Can u solve this question as well?

 

[Hijab]

I have one more question.... If someone could answer it... I would be really thankful....

 

[Thought blocker]

Can u solve this question as well?

Wall area = A m²
Volume of air colliding with wall per sec .. V/s(m³/s) = u(m/s) x A(m²) = uA
Mass colliding per sec .. m/t (kg/s) = V/s(m³/s) x D(kg/m³) = uA D

Force = rate of change of momentum ..
F = (mv - mu) / t = m/t (v-u) .. .. assuming the air comes to rest, then v = 0 and ..
F = -(m/t)u .. .. (against the motion)

F = (uAD)u .. .. u²AD .. .. ►F = 33² x A x 1.20 .. (N)

 

[usama321]

Can u solve this question as well?

Learn the formula as it is used in some other mcqs too. Rate of flow Kg/s = density * area* velocity...
Thus we get 33 * 1.2 * 12 = 475.2 kg/s
Thought blocker has explained it quite well

 

[WaqarAhmedJamali]

please send me ir of chemistry and physics for pakistan

 

[Hijab]

Wall area = A m²
Volume of air colliding with wall per sec .. V/s(m³/s) = u(m/s) x A(m²) = uA
Mass colliding per sec .. m/t (kg/s) = V/s(m³/s) x D(kg/m³) = uA D

Force = rate of change of momentum ..
F = (mv - mu) / t = m/t (v-u) .. .. assuming the air comes to rest, then v = 0 and ..
F = -(m/t)u .. .. (against the motion)

F = (uAD)u .. .. u²AD .. .. ►F = 33² x A x 1.20 .. (N)

Thankyou

 

[Hijab]

Learn the formula as it is used in some other mcqs too. Rate of flow Kg/s = density * area* velocity...
Thus we get 33 * 1.2 * 12 = 475.2 kg/s
Thought blocker has explained it quite well

Thankyou

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(a)(i)", I didn't get the marking scheme on this one but is the answer something like this that resistance decreases with increasing p.d. since its an increasing curve (current is increasing)

 

[Hijab]

Wall area = A m²
Volume of air colliding with wall per sec .. V/s(m³/s) = u(m/s) x A(m²) = uA
Mass colliding per sec .. m/t (kg/s) = V/s(m³/s) x D(kg/m³) = uA D

Force = rate of change of momentum ..
F = (mv - mu) / t = m/t (v-u) .. .. assuming the air comes to rest, then v = 0 and ..
F = -(m/t)u .. .. (against the motion)

F = (uAD)u .. .. u²AD .. .. ►F = 33² x A x 1.20 .. (N)

Can u solve this question as well

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
"Q6(a)(i)", I didn't get the marking scheme on this one but is the answer something like this that resistance decreases with increasing p.d. since its an increasing curve (current is increasing)

It can be seen from the upward bending curve that the current does not increase proportionally to the voltage. It increases more than the voltage, which hints that with increasing voltage, the resistnace of C decreases.

 

[usama321]

Can u solve this question as well

Not sure about this one. Here's what the examiner report says:
The answer to this question cannot reliably be obtained by guesswork. Eliminating the pressure at P gives
(h1–h2)ρg = 8000 Pa, so (h1–h2) = 0.060 m. D is the only answer that fits.