Physics: Post your doubts here!

[asd]

please help me with question number 7(b) i went through the marking scheme but couldn't understand it, please explain
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf


View attachment 24630

Use the Geiger Muller tube to check the count-rate. Now place a thin sheet of paper in front of the tube, if alpha is present, it will alter the rate. Note this down. Now place an aluminium sheet in front of the tube, if it stops the radiation by the same amount, It means beta is not present.

For b, Just put two magnets such that the field between them is perpendicular to the direction of radiation. Now if beta is present it should move out of the plane. Just use a detector to check if any radiation is detected.

 

[syed1995]

http://tinypic.com/view.php?pic=wslo45&s=5

<3 ya for that.. that's exactly what I was looking for!

Thanks for the diagram and that link! I will have a look at it.. Although that diagram did it for me.

 

[asd]

No problemo.

 

[syed1995]

physics

Why are you not easy like AS Maths?

 

[A star]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf 3ci and6bii anyone ?

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf 3ci and6bii anyone ?

in 3ci see that other than the weight the only external force that can act on the card is where its in contact with something and its DEFINITELY not the hand because they r asking about AFTER u release the card so it will be around the rod... that figured out u need to think of one force basically becuase one is a definite friction..... so think over it what is keeping the card on the rod?? there is a reactive force of the rod holding the card up ryt?? against its weight?? so there must be and upward force or there is another option of the rod's weight acting bcz that too is happening ryt?
tell me if I'm wrong
I'll c the other question and get back to u assap

 

[asd]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf 3ci and6bii anyone ?

From 6bi, you know the total resistance between A and B is 960
1/960=1/1600 + 1/x
1/960=(x+1600)/1600x
1600x/960=x+1600
gives you x =2400 ohms = 2.4 kohms
From the graph, we get 11 C.

 

[asd]

And yeah for the first one, daredevil has already answered.

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf 3ci and6bii anyone ?

ok i;m not solving this part but here's wat u need to do... I'll just run u through the method

ok have this covered that the voltage iin the voltmeter is the voltage of the thermistor too
calculate the total current in the circuit. calculate the current in the resistor having the 1.6kOhm resistance and thus calculate the current left to run through the thermistor part of the circuit.
once u have the current and the voltage of the thermistor calculate the resistance and then go over to ur graph with ur resistance value for the thermistor.... from there match the correct temperature (on the x-axis) for the attained resistance of the thermistor.
If u don't get it or if i am wrong in any of the two questions then tell me

 

[asd]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
6b (i) and (ii)
If for the first part we're considering that there will be a short circuit and therefore the ammeter reading is going to be 0, why doesnt the same reason go for the second part? (cause the question says the wire has a negligible resistance)

 

[abruzzi]

I is directly propotional to A^2

I = kA^2
I/A^2 = k

We always find the value of k and then find the twice the amplitude etc etc..

Like if I is 6 and A is 2 .. and another wave has Double the amplitude..

6= k(2^2)
6= 4k
k= 1.5
I=kA^2
I=1.5(2*2)^2
I=24 Answer..

similarly in terms of I and A..

I = kA^2
k = I/A^2

Doubling A means ..

I = kA^2
I = k (2A)^2
I = k*4A^2
I = 4A^2 * I/A^2
I = 4I

Thanks bruv

 

[daredevil]

okay guys plz explain this concept in Q6aii from this paper:
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_22.pdf
how do v get the phase difference? I got the one at P but even that was a guess... help plzz

 

[asd]

Koi Meri Problem to solve karde :'(

 

[asd]

okay guys plz explain this concept in Q6aii from this paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf
how do v get the phase difference? I got the one at P but even that was a guess... help plzz

when you get a bright fringe, the waves interfere constructively i.e the phase difference is either 0* or 360* or n(360)
when you get a dark fringe, the waves interfere destructively, the phase difference has to be 180* or n(180)
where n is the number of order.

 

[daredevil]

Koi Meri Problem to solve karde :'(

I looked over it but i couldn't explain it either.. sorry :/ and tag me wen u get the answer plz

 

[asd]

I looked over it but i couldn't explain it either.. sorry :/ and tag me wen u get the answer plz

Okaaay, I will.
Did you get the answer for your problem?

 

[hela]

can you plaese explain in details
c)i and c)ii
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

 

[littlecloud11]

can you plaese explain in details
c)i and c)ii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
View attachment 24641

ci)Two other forces would be-
1) the normal reaction force, which acts upwards from the rod.
2) Friction between the rod and the card as the card swings.

cii) the card comes to rest when it's center of gravity is directly along the vertical axis through the rod. This is because at this position the weight does not create any moment about the rod. No resultant moment so the card rests.

 

[hela]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf CAN YOU PLAESE EXPLAIN WITH IN DETAILS GTHIS QUESTION

 

[daredevil]

ci)Two other forces would be-
1) the normal reaction force, which acts upwards from the rod.
2) Friction between the rod and the card as the card swings.

cii) the card comes to rest when it's center of gravity is directly along the vertical axis through the rod. This is because at this position the weight does not create any moment about the rod. No resultant moment so the card rests.

woww i explained that question too earlier today but my exlanation seeems like layman language and kindergarten stuff when compared to urs ... hats off man #RESPECT!! O__O