Physics: Post your doubts here!

[SilverCrest]

monochtomatic light is incident on diffraction grating and a diffraction pattern is observed.
What is the effect of replacing the grating with more line sper metre?
The answer is the number of orders of diffraction visible decreases and the angle between the first and second orders o f diffraction increases?


can someone explain how?
thanks

 

[daredevil]

what should be the formula and numerical value for 2(e)(ii) http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_51.pdf

post a question once and if anyone knows the answer they will answer it
i don't know the answer becz i'm still in A1 but if anyone knows it they'll tell u. just wait for a bit
we have been asking and answering questions through the whole thread so we will definitely be doing it again if we can don't fret

 

[Kumkum]

question number 2(b) for the uncertainties in the last two data can we write 8 and 7 or should we write 8.4 and 7.5 respectively. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_51.pdf

the uncertainties should be to atleast one significant figure and the number of significant figures for this part will depend on all the previous uncertainties. if u put them to one sig. figure then u should do the same for this one as well. i think it should be 8 and 7.
hope i've helped

 

[selena]

The current in a component is reduced uniformly from 100 mA to 20 mA over a period of 8.0 s.
What is the charge that flows during this time?
A 160 mC B 320 mC C 480 mC D 640 mC

 

[Tkp]

cn any1 help me in mj12 qp12-30

 

[Tkp]

The current in a component is reduced uniformly from 100 mA to 20 mA over a period of 8.0 s.
What is the charge that flows during this time?
A 160 mC B 320 mC C 480 mC D 640 mC

u need to find the average of current as it says the current is reduced uniformly.so current is 60A and t is 8.so q=it and 480C

 

[unseen95]

the uncertainties should be to atleast one significant figure and the number of significant figures for this part will depend on all the previous uncertainties. if u put them to one sig. figure then u should do the same for this one as well. i think it should be 8 and 7.
hope i've helped

ya it helped

 

[bogus]

for the sound at m to be zero that means destructive interference.so for destructive interference the path difference must be be a odd number of multiple wavelength
and for the second conditon if the amplitudes are same so there will be destructive interference as the two waves cancel out each other
so for the first 1 it would be the path difference must be be a odd number of multiple wavelength and 2 nd one amplitude amplitude of both the waves are same at m.
b)v=f lambda.so u put these values and u get 2 value of l(convert the unit in cm).that means the minima would be within the range.now find the path differnce.find the length of s2m by pythagoras theorem which is 128 cm.so the path difference for s2m-s1m is 28 cm.now apply the formula of path difference for destructive interference
28=(n+.5)l put the value of n to o,1,2,3,4.now u will get 2 values which are within the range.so 2 minima would be detected

Thanks!

 

[yousef]

CAN SOME ONE PLZ SHOW ME ,, HOW TO DO ... MAY/JUNE 2005 Q 6 (bi) (biii) >>>>>http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_2.pdf

 

[Ahmad Zaeem]

M/J 2012 P11 Qs 12 ....concept pls ???..........

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

 

[Light Yagami]

Why is the answer C:-

 

[Malik777]

Any One Plz Upload Physics Revision Guide

 

[Ahmed Ali Akbar]

Why is the answer C:-
View attachment 22713

 

[syed1995]

M/J 2012 P11 Qs 12 ....concept pls ???..........

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

First See the the two 30N forces (They are a couple since they are parallel and in opposite direction with a distance between them.)

Torque is produced by a couple. Since they are in opposite directions .. there will be a torque .. so resultant torque is non-zero.

Now look at the 20N forces .. since they are NOT in opposite directions .. They are NOT forming a couple.. so there won't be torque. since both are in the same direction and the forces are not cancelling each other out .. the resultant force will be non-zero.

To understand this one .. forget that those 30N forces are even there.. What will happen if you pull a tyre from the top and bottom via a rope in the same direction.. the tyre will move .. meaning the resultant force is non-zero.

Answer: D

 

[Tkp]

Why is the answer C:-
View attachment 22713

no need to think of the other things.they are saying they are using the same material.so the young modulus should be same as the 1st 1.so ans ic c

 

[snowbrood]

i can solve this but i dont get one thing is the force exerted on X by Y or force exerted by Y on X the same?

 

[Ahmad Zaeem]

May/June 2012 P11 Q 12....Concept with complete solution plsss!!!!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

 

[Ahmad Zaeem]

First See the the two 30N forces (They are a couple since they are parallel and in opposite direction with a distance between them.)

Torque is produced by a couple. Since they are in opposite directions .. there will be a torque .. so resultant torque is non-zero.

Now look at the 20N forces .. since they are NOT in opposite directions .. They are NOT forming a couple.. so there won't be torque. since both are in the same direction and the forces are not cancelling each other out .. the resultant force will be non-zero.

To understand this one .. forget that those 30N forces are even there.. What will happen if you pull a tyre from the top and bottom via a rope in the same direction.. the tyre will move .. meaning the resultant force is non-zero.

Answer: D

thanks..buddy!

 

[Malik777]

View attachment 22749
i can solve this but i dont get one thing is the force exerted on X by Y or force exerted by Y on X the same?

i Think B :S

 

[Ahmedraza73]

May/June 2012 P11 Q 12....Concept with complete solution plsss!!!!

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

you can look at this as a pully attached with 2 mass, the man is the lighter object while the barrel is heavier object

so here you need the acceleration since its the same rope the tension and acceleration would same for both cases ie for the man and the barrel

first you need to make simultaneous equations relating the man and the the barrels mass with T and a

T-m1g=m1a this is for a mass which is moving upwards
so we use it for the man
T-800=80a ------ eq1

m2g-T=m2a this is the vice versa of the last one
so we use ir for the barrel
1200-T=120a

solving them simultaneously u will get the acceleration
a= 2
now we use kinamatics here
since it is said that "when the head of the man and the bottom of the barrel are at equal level" it is obvious that when they r equal the must have covered half the distance
so the distance we take here is s=9
likely u= 0 , s= 9 , a=2

v^2 -u^2= 2as
solve it and u get v= 6 m/s