Physics: Post your doubts here!

[unkidd]

im losing atleast 50 marks

 

[abdul rehman 123]

Can we discuss the questions

 

[Champ101]

not yet

 

[Alool]

physics paper 42 was very difficult

 

[Champ101]

physics paper 42 was very difficult

Keep smiling ?

 

[roxy roro]

paper 42 was very very difficult !!

 

[NinjaInPyjamas]

It is given that Lambda=630nm=630 x 10^-9 m
It is given that the grating is 450 Lines per 1mm
First, we need a, which is distance between 2 lines only.
To get a, we need to use a=d/n, where d is the whole length of grating and n is number of lines.
We need to change 1 mm into m, which is 0.001 m.
Now, a= 0.001/450=2.2 x 10^-6m
We now need to use, asintheta=nLambda where a is separation between 2 lines.
We need to find n, we put theta=90, so (2.2 x 10^-6)sin90=n(630 x 10^-9), notice that all values should be the same unit. I prefer changing them all to meters.
Notice that our answer of n=3.52, but n has to be a whole number, so we round it down to 3.
I hope I helped you, i tried to be as clear as possible.

thank you so much! it was really good!

 

[Alool]

Keep smiling ?

lol not after physics

 

[Champ101]

lol not after physics

hmm its true but there's nothing we all can do now! Except doing good in paper 5! R u maths givin tomorrow?

 

[Alool]

hmm its true but there's nothing we all can do now! Except doing good in paper 5! R u maths givin tomorrow?

Inshallah we do good in paper 5..
and no i don't take maths.

 

[Champ101]

Inshallah we do good in paper 5..
and no i don't take maths.

Inshallah!!

 

[salvatore]

How was paper 41 guys? I found it okayish

 

[Mohamed1212]

How was paper 41 guys? I found it okayish

I hated it I made lots and lots of mistakes and got the CT scanner question all wrong. If you remember any answers please tell me.

 

[Snow Angel]

Hi please check and reply Suchal Riaz
Thnx in advance

 

[Suchal Riaz]

Hi please check and reply Suchal Riaz
Thnx in advance

i can't see the images. they are broken -_-

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_31.pdf

Q1d please tell me how to find the units for such questions and what makes it different from other questions !

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_31.pdf

Q1d please tell me how to find the units for such questions and what makes it different from other questions !

In Physics, whenever you are adding two or more quantities, their units have to be the same; adding meters to seconds makes no sense, so we cannot perform that kind of operation. Therefore, we know that the units of R/P are equal to the units of Q/P. Since P is common to both, with common units, we can say that the units of R are equal to the units of Q.

A further observation is that since their sum is equal to the left-hand-side of the equation given, the units of R/P (or the units of Q/P) should be equal to the units of 1/I.

Since we know that current (in amperes, A) is a base unit, we can write that the units of the left side are A^-1, since we are taking the reciprocal of the current.

Furthermore, since we know that the Resistance can be defined as the ratio of V/I and since V = W/Q = Joules / (Ampere * Second), we can write that the units of R = Joules/ (Ampere^2 * Second). Further simplifying this, we get ( using Joules = 1 kg * m^2 * s^-2) the units of R as 1 kg·m2·s-3·A-2 .

Now, we know that the units of Q are equal to the units of R, so the units of Q = 1 kg·m2·s-3·A-2.

However, another requirement for the equation to be valid is that the units of all terms on the right are equal to the units for all the terms on the left. Since the units of the quantities on the left side are equal to A^-1, we can write:

(Units of R)/(Units of P) = A^-1

So, solving this, (Units of P) = A * (Units of R) = 1 kg·m2·s-3·A-1.

In terms of known units the derivation is simpler.

(Units of R)/(Units of P) = 1/A

So (Units of P) = ΩA.

Hope this helped!
Good Luck for all your exams!

 

[Snow Angel]

i can't see the images. they are broken -_-

ok dats fyn

 

[Snackbox86]

7a(ii):
1.
(I'm pretty sure the E.g thing is a typing error, since there's no "g" variable in the question - the answer should be E.q or qE).
Since the moving particle is positively charged and the upper plate is negatively charged (not to forget, the lower plate is positively charged), the moving particle will be attracted upwards (by the negative plate) and repelled in the same direction (by the lower plate).
This force does not depend on the position of the particle or it's velocity, and it is constant all throughout.
The magnitude of the force exerted on a particle with charge q when the strength of the electric field at it's location is E, is given by:

F = q * E

So, since the electric field strength is constant and directed upwards (between the plates) all along the length of the plates, this is the force exerted by the plates on the particle. Therefore, F = qE.

2.
Since the force exerted on the particle is always in the vertical plane, the horizontal component of the particle's velocity will remain constant throughout it's journey, because no force component (neither the force of gravity or the electric force have a horizontal component) will influence it's velocity.
So, since the horizontal displacement of the particle as it travels from one side of the plates to the other is equal to L and this is related to the velocity by

v = s / t = L / t

we can write

t = L / v.

7b(i)

The law of conservation of momentum says that "the momentum - or quantity of motion - of a particle or the center of mass of a system of particles will not change unless a net external force acts on that particle or system of particles".
In other words, "the momentum of a closed system will not change unless a net external force acts on that system of particles".
It can be extended to the situation where two or more objects, taken as components of a complete system, interact by collision or action-at-a-distance forces; if no force exerted by a body from outside the system exists, then the momentum of the system (of all the components of the system) summed up can never change.

So, if no external forces act on a system, momentum before interaction = momentum after interaction.

7b(ii)

This question requires the use of the Impulse theorem, which can be written in symbols as

Δ(mv) = F(net) * t

And so, we can write

Δ(mv) = qE * L/v = qLE/v

7b(iii)

The charged particle is not an isolated system; since a net force exerted by a body other than itself acts on the particle, it cannot be considered a close system and as such it's momentum will change, so the conservation of momentum doesn't apply here.
However, if you take all the objects involved to be part of the system (i.e. if you say that the particle as well as the plates form the system) all the forces involved in the diagram are between the objects of the system, and no external force acts on it. Therefore, it's momentum is conserved.

(I'll try the first part after some time, the path thing).

Hope this helped!
Good Luck for all your exams!

Thank you so much!

 

[DeViL gURl B)]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s10_qp_11.pdf

Can someone please explain me question 5 and 7.. Detailed!!!
Please thank u