Physics: Post your doubts here!

[unique111]

I seriously need help in P4 physics! any smarties out there? Please help me.

 

[Thought blocker]

I seriously need help inP4 physics! any smarties out there? Please help me.

Which type of help ?
I have formulas , may that help you!

 

[Thought blocker]

8000 Post complete

 

[unique111]

Which type of help ?
I have formulas , may that help you!

Thank you, bro! You giving AS only?

 

[Thought blocker]

Thank you, bro! You giving AS only?

Yup

 

[unique111]

Yup

I want help in Paper 4. Hope to find someone as helpful as you over here to help me in a-level too

 

[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

w07:
q5 a

if you want to know why 0.7 a here is why:
I is directly proportional to A²
I = k A²
we want to make I half times. multiply both sides by ½
½ I = ½(k A²) =k ½(A)²
we want to make that coefficient go inside that square
square of 1/√2 is ½ so
½ I = k( 1/√2 A)²
as 1/√2 ≈ 0.7
½ I = k( 0.7 A)²

and about 60 degree:
the phase difference is sixty degree which is 1/3 of 180. so one big box is equal to sixty degrees. so start the new waves sixty degree left or right from where the previous waves started and the amplitude will be 0.7a and rest of it will be the same i mean the frequency and wavelength.
more detailed description:
i drew the line at 0.7a and -0.7a which will be the point where the crests and trough of the graph are going to be. as the amplitube is 1/sqrt(2) * a
and i dragged every point right by one big box marked by 60 degree, 120 degree and 180 degree(interval of 60) because the complete wave took 6 squares and complete wave is 360 so each box becomes of 60 degree.
drag every point to the right/or left one box.
see this i hope it is clear. i tried my best.

 

[Suchal Riaz]

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

s08:
Q2 (b)
from graph, there is force of 28 N for extension of 3.5 cm
E=1/2 eF = 1/2 * 3.5X10^-2 m * 28 N
=0.49 J
Q4 (b)
P/2.2X10^9 = change in V /V
as 2.2X10^9 is a very big number, the change in v/V is very large number, the change in V will be very small.
Q5(c)
first find λ:
difference between two consecutive nodes(or antinodes)= 1/2 λ so λ = 2*17.8 cm = 35.6 cm .
convert λ into m: λ = 0.356 m
v=fλ so v= 125 Hz * 0.356m = 44.5 m/s
v=√(T/m) so v²=T/m so m=T/v²
m=4 N/ (44.5 m/s )² = 2.0 X 10^-3 kg/m
Q6 (b)
there is a long way this question can be done. but i have a shortcut which always works. i don't have any proof for that. if you want to use it then here is how it works:
power add up in parallel and add like this in series 1/Pc = 1/P1 + 1/P2
(1) circuit is not complete so there will be no current at all. so power = 0
(2) as there is shortcircuit no current will pass through resistor B. only current will pass through B. so P=1.5 KW
(3) again resistor B is short-circuit so no current through it. A and C are in parallel. by the formula i gave you, the power will add up in parallel. so P=3 KW
(4) now there is no current through C but A and B are in series. so power will be calculated through the formula i gave you:
1/P = 1/1.5 + 1/1.5 ==> 1/P = 2/1.5 ==> p=1.5/2 = 0.75 KW
(5) now power of A and B are in parallel with C. so power of A and B will add up with C. 0.75 KW + 1.5 = 2.25 KW

 

[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

now i am tired please remind me to solve rest of them in evening.

 

[Thought blocker]

s08:
Q2 (b)
from graph, there is force of 28 N for extension of 3.5 cm
E=1/2 eF = 1/2 * 3.5X10^-2 m * 28 N
=0.49 J
Q4 (b)
P/2.2X10^9 = change in V /V
as 2.2X10^9 is a very big number, the change in v/V is very large number, the change in V will be very small.
Q5(c)
first find λ:
difference between two consecutive nodes(or antinodes)= 1/2 λ so λ = 2*17.8 cm = 35.6 cm .
convert λ into m: λ = 0.356 m
v=fλ so v= 125 Hz * 0.356m = 44.5 m/s
v=√(T/m) so v²=T/m so m=T/v²
m=4 N/ (44.5 m/s )² = 2.0 X 10^-3 kg/m
Q6 (b)
there is a long way this question can be done. but i have a shortcut which always works. i don't have any proof for that. if you want to use it then here is how it works:
power add up in parallel and add like this in series 1/Pc = 1/P1 + 1/P2
(1) circuit is not complete so there will be no current at all. so power = 0
(2) as there is shortcircuit no current will pass through resistor B. only current will pass through B. so P=1.5 KW
(3) again resistor B is short-circuit so no current through it. A and C are in parallel. by the formula i gave you, the power will add up in parallel. so P=3 KW
(4) now there is no current through C but A and B are in series. so power will be calculated through the formula i gave you:
1/P = 1/1.5 + 1/1.5 ==> 1/P = 2/1.5 ==> p=1.5/2 = 0.75 KW
(5) now power of A and B are in parallel with C. so power of A and B will add up with C. 0.75 KW + 1.5 = 2.25 KW

Question 6 part b Explain it again please.

 

[sweetjinnah]

That's what I meant. Why is the collision inelastic when an atom collides with a moving piston?

Energy is not conserved.. so inelastic

 

[sweetjinnah]

sweetjinnah sitooon

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c


Plz help someone?

Question 6ci and ii)
ci) 1. A= decay constant * N
=> (3.76*10^6) = (0.025)*N... solving for N = 4.67*10^15
2. moles= mass/Mr
moles = no. of paticles /Avogadros constant.
Mass= (4.76*10^15) / (6.02*10^23) = ans * 90/1000 solving mass = 6.98*10^-10 kg

 

[sweetjinnah]

sweetjinnah sitooon

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c


Plz help someone?

 

[sweetjinnah]

sweetjinnah sitooon

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c


Plz help someone?

 

[sweetjinnah]

sweetjinnah sitooon

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c


Plz help someone?

Q7c) N/No = 1/10
N/No = e^(-decayconstant * t)
putting the values nd solve for t = 3.11*10^4 sec convert into hrs dividing by 3600 = 8.63 hrs.

 

[Musab100]

can anyone do this question for me ?
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Q 5

 

[Suchal Riaz]

Question 6 part b Explain it again please.

 

[Mohammed salik]

View attachment 40175

Amazing !! i wasted 15 mins on this Quesstion !

 

[unique111]

Thank you so much! you've come as an angel in need

View attachment 40162

Could you explain this graph? And this too https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-401#post-751864

 

[unique111]

Question 6ci and ii)
ci) 1. A= decay constant * N
=> (3.76*10^6) = (0.025)*N... solving for N = 4.67*10^15
2. moles= mass/Mr
moles = no. of paticles /Avogadros constant.
Mass= (4.76*10^15) / (6.02*10^23) = ans * 90/1000 solving mass = 6.98*10^-10 kg

The answer for N is correct, but the equation does not solve it. Could you see this too?