Physics: Post your doubts here!

[omarjaved619]

never mind i got it
ok it is D bcz c in a closed pipe the smalest frequency possible is f. note that on an open end there is always an antinode and on a cosed one there is always a node.
so if there is an antinode on the open end of X and a subsequent node on the closed end the frequency is f
In Y however there should be an antinode on both the ends and for the smallest frequency only one node in between. so the pipe will have an antinode, a node and another antinode which will make the distance 2f. (antinode to node is f as in X)
i hope u got it

Yes I did! Got some more, if you dont mind

Q35(same paper)

Q15, Q18 from http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf Thanks a tonnn!

 

[zackle09]

For q3. I'll tell you a method if they ask for resultant, choose an axis, let's say y vertical and x is horizontal. I hope till that part you know how to resolve the forces. Then if you know, get the resultant of the forces that is parallel to the y axis and then the same for the x axis. Now use pythagoras, so that Fresultant=Square root of [(resultant force parallel to y)^2 + (resultant force parallel to x)^2]

For q30.If you are asked to AVERAGE a value what you do? sum of all the values/number of values right? So stick to it.

ohhhhh.. i did what you said..i got 11 and answer is 10..are you supposed to get an exact answer or is this fine
and for question 30..they didnt tell us to find the avg though :/

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf Q15 any one

In this ques, we consider the moment of the forces.
http://www.s-cool.co.uk/gcse/physics/forces-moments-and-pressure/revise-it/moments
The moment of L is in the clockwise direction, whereas the moment of W = anticlockwise.
They balance each other → system is in equilibrium.
When, the dist of L from the pivot increases → the moment of L now increases.
[Since moment = force * dist from pivot]
To balance this, we need an increase in moment in the ANTICLOCKWISE direction.
Note → the force needs to be perpendicular to the pivot.
When R is shifted to the left, it produces a moment in the clockwise direction.
and towards the left, it produces a moment anticlockwise.
So, ans = B

Hope u get it

 

[TaffsAsLevel]

ohhhhh.. i did what you said..i got 11 and answer is 10..are you supposed to get an exact answer or is this fine
and for question 30..they didnt tell us to find the avg though :/

It is 10 I just did that few days ago, ok I will give you my working and you try see your problem. square root of[(10cos30)^2 + (10-10sin30)^2)] check you get 10 N, if you don't understand my working then you don't know resolving forces or maybe you didn't recognize the angle just because 120 degrees is given. Sorry because it is hard to type out compared to writing down on paper

 

[zackle09]

It is 10 I just did that few days ago, ok I will give you my working and you try see your problem. square root of[(10cos30)^2 + (10-10sin30)^2)] check you get 10 N, if you don't understand my working then you don't know resolving forces or maybe you didn't recognize the angle just because 120 degrees is given. Sorry because it is hard to type out compared to writing down on paper

ohhh noo i got it! was making a stupid mistake..thanks alot! )

 

[ahmed abdulla]

It is 10 I just did that few days ago, ok I will give you my working and you try see your problem. square root of[(10cos30)^2 + (10-10sin30)^2)] check you get 10 N, if you don't understand my working then you don't know resolving forces or maybe you didn't recognize the angle just because 120 degrees is given. Sorry because it is hard to type out compared to writing down on paper

>>>>???

 

[A star]

In this ques, we consider the moment of the forces.
http://www.s-cool.co.uk/gcse/physics/forces-moments-and-pressure/revise-it/moments
The moment of L is in the clockwise direction, whereas the moment of W = anticlockwise.
They balance each other → system is in equilibrium.
When, the dist of L from the pivot increases → the moment of L now increases.
[Since moment = force * dist from pivot]
To balance this, we need an increase in moment in the ANTICLOCKWISE direction.
Note → the force needs to be perpendicular to the pivot.
When R is shifted to the left, it produces a moment in the clockwise direction.
and towards the left, it produces a moment anticlockwise.
So, ans = B

Hope u get it

ans D :/

 

[ShaanSiddiq090909]

Can someone please explain q 34 of june 2008 pleaaaaaseee!!!

 

[white rose]

Did u get how we calculated n?
Formula → nλ = dsinΘ
(1*10^6) * λ = 1 * sin 35
[d = 1, as it is a 1st order diffraction]
So, rearranging the eqn abv,
λ = (1*sin35)/(1*10^6)
λ = sin 35/(1*10^6)

Hope u get it!

d is 1.0*10^-6 den y r u taking 1
and n is the no.of orders which is 2 here

 

[gary221]

ans D :/

sorry, m bad.
I meant D.
It cannot be A or B, as the force H will not affect the movement of the pivot.

 

[TaffsAsLevel]

>>>>???

Okay, you should know what voltage stands for it is potential DIFFERENCE, so..

You have a 12 V from P to Q, and you know that you are seeing a parallel circuit so voltage is same. Now we need to know the p.d at Q, we can use a voltage division? so that you will get 4V, so what you have left? 8 V, you are left with 8V you can see at the intervals if you jot down the values 12 to 8 to 0, that is why it is potential difference, now go to Y do the same you should get 8V and remaining of 4 V. so you got the remaining potential at P and Q now find their difference, I'll let you do that one

 

[ahmed abdulla]

d is 1.0*10^-6 den y r u taking 1
and n is the no.of orders which is 2 here

can u help

 

[ShaanSiddiq090909]

can u help

June 2008 q 34 and 37 please?

 

[Maira Siddique]

Can anyone explain Q 28 of the followin paper:

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

and Q6 of :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf

PLEASE

 

[white rose]

can u help

here v need to find the voltage og 500ohm & 2000ohm.
how?.....use this formula Vtotal/Rtotal=V1/R1
for 500ohm....12/1500=V1/500...cross multiply ull get 4V
for 2000ohm....12/3000=V1/2000...ull get 8v
now subtract both the values 8V-4V=4V
hope u get it

 

[gary221]

i think u frgot q. 12...

12) We know that loss in P.E. = gain in K.E.
For the man's head to be level with the bottom of the barrel, the barrel must fall halfway → 9 m.
So, loss of P.E. of the barrel = mgh = 120 * 10 * 9 → 10800 J
Now, gain in K.E. = ½mv² = 10800 J
So, velocity → ½ * 80 * v² = 10800 ------> v = 16 m/s

Since the gravity also acts on the man in the OPPOSITE direction,
the resultant velocity = 16 - 10 → 6 m/s
Ans = A

Hope u get it!

 

[Champ101]

Can anyone explain Q 28 of the followin paper:

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

and Q6 of :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf

PLEASE

hi maira, i can help you with q6.
horizontal component has constant velocity, therefore : s= v*t
for vertical component initial velocity is 0, therefore: s=1/2gt^2
now divide the two and you will get : 2v/gt

 

[white rose]

12) We know that loss in P.E. = gain in K.E.
For the man's head to be level with the bottom of the barrel, the barrel must fall halfway → 9 m.
So, loss of P.E. of the barrel = mgh = 120 * 10 * 9 → 10800 J
Now, gain in K.E. = ½mv² = 10800 J
So, velocity → ½ * 80 * v² = 10800 ------> v = 16 m/s

Since the gravity also acts on the man in the OPPOSITE direction,
the resultant velocity = 16 - 10 → 6 m/s
Ans = A

Hope u get it!

i got wht u r sayin but again...wht i think is that the barrel can fall 9m only when the man is also 120 kg to balance the rope
on both the sides.
do u get my point?

 

[Maira Siddique]

hi maira, i can help you with q6.
horizontal component has constant velocity, therefore : s= v*t
for vertical component initial velocity is 0, therefore: s=1/2gt^2
now divide the two and you will get : 2v/gt

Got it,thanks!

 

[gary221]

i got wht u r sayin but again...wht i think is that the barrel can fall 9m only when the man is also 120 kg to balance the rope
on both the sides.
do u get my point?

No, the support/equilibrium is not what we're looking for here.
Here we just want the dist at which the man's head will be level with the barrel's bottom.
That will only happen when the barrel is at 9m.