Physics: Post your doubts here!

[Jaf]

thank you but, ALL of this i do know jaf...but i cant understand why 25cos(55)??

Whenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.

 

[A.ELWY 7]

Whenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.

jaf i really dont no how to thank you...and this note about the given force being the hypotinse it really save me thank you

 

[Jaf]

jaf i really dont no how to thank you

Oh but I do... kind duas and prayers for me would suffice.

 

[A.ELWY 7]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37

Q13:CLOCK WISE moment = anticlockwise moment so FxD=FxD and the rule have a mass of 100g in the centre at 50cm mark so 10 cm from the 40 mark so it is (20g x 60)+(100g x 10)/50 = 44 which is C
Q15: i spend days trying to do and i couldn't do it
Q25:also i didnt know it
Q27: find d which is (1/500) time ten to the power of minus 3 = 2 x 10^-6...then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D
Q37: he want the ratio of the V1/V2 so we need the distance of R1 from x divided by the distance of R2 from x...the distance from R1 from X is x and the distance of R2 from X is the total wire length - the distace of R1 from X so it is L-x so the answer is D x/L-x

 

[sadman]

I have a question from June 2007 paper 4...........
this question is number 7 part (c).........why will increasing the load of the resistor reduces damping of the oscillation?

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37

Q25)
Let's say the length of the trumpet is L and the speed of sound in the horn is c.
At the first frequency, only one note and one antinode are produced. This is 1/4 the wavelength of the sound wave. So L = 1/4 lambda and hence lambda is 4L. c = f x lambda = 75 x 4L = 300L

In the next frequency, there HAS to be a node at the mouthpiece and an antinode at the bell. This means there's another node inside the horn. So now L = 3/4 lambda and lambda = 4/3 L. We also know the speed of the wave remains constant.
c = f x lambda = f x 4L/3
300L/(4L/3) = 225. There's only one such answer with this as the first frequency and that's D.

I'll do 15 later. I'm kind of in a hurry.

 

[JD REBORN]

Q 10 and Q31 with proper explanation.

 

[JD REBORN]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37

For q15 if u have done A level M1 then it will be no problem however this is the solution:
Gain in Kinetic Energy= 1/2 x 2.0 x(v^2 -0)
=v^2
Loss in Potential Energy= 2.0 x 9.81 x 3=58.86 J

Work done against friction= 7.0 x 5.0=35 J

Loss in potential energy -work done against friction=v^2

58.86 -35=v^2
v=4.9 m/s

 

[smzimran]

waalaikumassalam wr wb
haha...nice one

ok and for ques 6..i wanna know what's the difference b/w flux and flux linkage..see ms...it says smthng like that...

and jazakAlllahhhhhhhhhhhhh for the help...
many duassss 4 u

Magnetic flux linkage means No of magnetic field lines (Symbol: Φ )
Magnetic flux density is no. of field lines per unit area (Symbol:B )
Φ = BA

B = Φ / A

And thanks 4 the prayers, i really need them

 

[smzimran]

...some more doubts in June 2006
· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?

Aoa,
Q2:
In (i) you have found the amount of total hydrogen which was found out to be 415 mol H2
Now in (ii), they are asking about how many balloons may be filled by that 415 mol of H2
The problem is that they have given the amount of H2 in one balloon not at a pressure of 2.50 × 10^7 Pa but they have given for 1.85 × 10^5 Pa.
So, we need to find volume at pressure of gas at 1.85 × 10^5 Pa.
Volume of gas at 1.85 × 10^5 Pa = V2
The formula is p1V1 = p2V2
V2 = p1V1 / p2
V2 = (2.5 × 10^7× 4.00 × 10^4) / (1.85 × 10^5)
V2 = 5.41 × 10^6 cm3

Now, the gas that remains in the cylinder is :
V = (5.41 × 10^6) - ( 4.00 × 10^4) = 5.37 × 10^6 cm3
1 balloon contains (7.24 × 10^3) cm3 of H2
1 cm3 H2 fills 1 / (7.24 × 10^3) balloons
(5.37 × 10^6) cm3 fills (5.37 × 10^6) / (7.24 × 10^3) balloons
= 741 balloons

If we had solved this question by dividing no of moles of H2 in cylinder by no of moles of H2 in one balloon, the answer would be a little high value because we would have ignored the amount of gas that remains in the cylinder.

I hope u got it!

 

[smzimran]

...some more doubts in June 2006
· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?

Q4
Attachment

 

[smzimran]

...some more doubts in June 2006

· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?
· Q:6 a what I understand is the field should be circle..right? I don’t get this :s
· Q:6 c what values do I use? :s
· Q:7

Q6(a)
The field due to current in wire PQ will be circular at point Q
But they are asking about field due to WIRE XY AT POINT Q
I think you did not read the question correctly...


Q6(c)
Any values eg I = 10, d= 0.1
I = 11, d= 0.1
etc

 

[SkyPilotage]

Magnetic flux linkage means No of magnetic field lines (Symbol: Φ )
Magnetic flux density is no. of field lines per unit area (Symbol:B )
Φ = BA

B = Φ / A

And thanks 4 the prayers, i really need them

I just wanted to point out,
Magnetic Flux is the number of magnetic field lines which is Phi.
Magnetic Flux Linkage is the product of the magnetic flux and the number of turns of the coil. which is N x Phi.
"Magnetic flux density is no. of field lines per unit area (Symbol:B )" like you said.
Thank you

 

[smzimran]

I just wanted to point out,
Magnetic Flux is the number of magnetic field lines which is Phi.
Magnetic Flux Linkage is the product of the magnetic flux and the number of turns of the coil. which is N x Phi.
"Magnetic flux density is no. of field lines per unit area (Symbol:B )" like you said.
Thank you

Yes, i admit a little mistake of mine!

 

[Unicorn]

Can someone please explain to me how to do question 15, 18 and 20?
I kinda understand 18 my reasoning is that the pressure already in the flask is already hdg so in the increase would also be hdg giving 2hdg (correct me if i am wrong)

 

[SkyPilotage]

Can someone please explain to me how to do question 18 and 20?
I kinda understand 18 my reasoning is that the pressure already in the flask is already hdg so in the increase would also be hdg giving 2hdg (correct me if i am wrong)

Question 18:-
Your reasoning for question 18 is wrong :S The change in pressure is denstiy (rho) x gravity (g) x DIFFERENCE IN HEIGHT (delta h)
If it increase by "h" on one side, it should go down by "h" on the other side so the change in height is 2h. Therefore, the pressure is 2hdg. i.e D

 

[SkyPilotage]

Can someone please explain to me how to do question 15, 18 and 20?
I kinda understand 18 my reasoning is that the pressure already in the flask is already hdg so in the increase would also be hdg giving 2hdg (correct me if i am wrong)

Question 20:-
Tension in P:- is :"T"
Tension in Q: - 2l means it is easier to pull so there is less tension ---> 1/2T
A/2 means its is thinner so its also easier to pull ------> 1/2T/2 = 1/4T
Therefore:- T/1/4T = 4/1 i.e D!

Hope it helped Feel free to post more doubts

 

[Unicorn]

Question 18:-
Your reasoning for question 18 is wrong :S The change in pressure is denstiy (rho) x gravity (g) x DIFFERENCE IN HEIGHT (delta h)
If it increase by "h" on one side, it should go down by "h" on the other side so the change in height is 2h. Therefore, the pressure is 2hdg. i.e D

Ohhh that is why thnx ^^

 

[Unicorn]

Question 20:-
Tension in P:- is :"T"
Tension in Q: - 2l means it is easier to pull so there is less tension ---> 1/2T
A/2 means its is thinner so its also easier to pull ------> 1/2T/2 = 1/4T
Therefore:- T/1/4T = 4/1 i.e D!

Hope it helped Feel free to post more doubts

which formula are you using here??

 

[SkyPilotage]

Proportionality formulas
I used Young's Modulus where it is equals Stress/Strain= (F/A)/(x/L) so F = xY(A)/L so F is proportional to A and inversely proportional to L.
So if l doubles , F halves. If Area doubles, F doubles
You dont need to learn the derivation i showed you. This is just to make you understand it better but If otherwise, neglect it hehe
--Glad to help---