Physics: Post your doubts here!

[sadiaali]

look u just have to use the formula
n*(lamda)=d*sin(theta)
where n is the order (its told the second order that means 2)
d is the spacing of lines (which we have to calculate)
lamda is the wavelength
put the values in the equation
2*(590*10^-9)=d*sin(43)
solve it u wil get 1.7

Oh, thank you

 

[ZaqZainab]

I need the mark scheme of physics for 2001 and 2002!

http://xtremepapers.com/community/attachments/marksc-1-pdf.12307/ 2001 winter
http://papers.xtremepapers.com/CIE/...A and AS Level/Physics (9702)/9702_w02_er.pdf this is the examiner report but it contain answers for MCQ 2001 winter

 

[Thought blocker]

Well Guys, kitkat <3 :P mehria
Mohammed salik


I am sorry for yesterday night.. I got it, it was v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v. SIMPLE I made it complicated, Abb shamajh aya hadu ne a ko 1 q liya tha...
I was so stupid, sorry for wasting your imp hours on this topic.

 

[kitkat <3 :P]

Well Guys, kitkat <3 :P mehria
Mohammed salik


I am sorry for yesterday night.. I got it, it was v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v. SIMPLE I made it complicated, Abb shamajh aya hadu ne a ko 1 q liya tha...
I was so stupid, sorry for wasting your imp hours on this topic.

its ok

 

[Thought blocker]

its ok

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf
Q 9
why not C?
it is t^2 so here t=(t2-t1)
Q10 why not D?
Q11 i thought it is C
Q14
Q21
Q32 R=V/I
so here I is infinite wont R be o??
because I=V/R and if I is infinite I=V/0??
I know there are so many but even if you help me out with one of this i will be great full

 

[kitkat <3 :P]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_1.pdf
Q 9
why not C?
it is t^2 so here t=(t2-t1)
Q10 why not D?
Q11 i thought it is C
Q14
Q21
Q32 R=V/I
so here I is infinite wont R be o??
because I=V/R and if I is infinite I=V/0??
I know there are so many but even if you help me out with one of this i will be great full

Q9 the answer is D
firstly u have to find x
s=u*t+0.5a*t^2
X=0.5a*t1^2(u=o)

for the accelaration we are going to use complete hieght i.e.
x+h so
s=ut+0.5at^2 (u=0)
x+h=0.5*a*T2^2
put value of x
o.5*a*t1^2 +h= 0.5*a*t2^2
h=0.5*a*t2^2- 0.5*a*t1^2
(o.5=1/2 so on the other side it will becme 2 so it wil be 2h)
now 2h= a*t2^2 -a*t1^2
take a common
2h=a(t^2 - t1^2)
a=2h/(t2^2-t1^2)
hope u got it :/

 

[kitkat <3 :P]

for 10 isnt D th answer :O??
for 11 when the direction is opposit we add the force when same we subtract so A is the answer
ZaqZainab

 

[ZaqZainab]

for 10 isnt D th answer :O??

A is the answer

Q9 the answer is D
firstly u have to find x
s=u*t+0.5a*t^2
X=0.5a*t1^2(u=o)

for the accelaration we are going to use complete hieght i.e.
x+h so
s=ut+0.5at^2 (u=0)
x+h=0.5*a*T2^2
put value of x
o.5*a*t1^2 +h= 0.5*a*t2^2
h=0.5*a*t2^2- 0.5*a*t1^2
(o.5=1/2 so on the other side it will becme 2 so it wil be 2h)
now 2h= a*t2^2 -a*t1^2
take a common
2h=a(t^2 - t1^2)
a=2h/(t2^2-t1^2)
hope u got it :/

thanks

 

[ZaqZainab]

for 10 isnt D th answer :O??
for 11 when the direction is opposit we add the force when same we subtract so A is the answer
ZaqZainab

ua is going in opposite direction to ub
so ua-ub??
va is in the same direction to vb so va+vb?
why is this wrong

 

[Kamihus]

A is the answer


thanks

The answer is D

 

[ZaqZainab]

The answer is D

Opps my bad
My apologies

 

[kitkat <3 :P]

ua is going in opposite direction to ub
so ua-ub??
va is in the same direction to vb so va+vb?
why is this wrong

when the are in differentdirections one is postive the other one is negative while when the move in the same direction both are either positive or negative so when we put them in the formula we get that the velocities of the balls in different directio will be added while the ones in same direction will be subtracted

 

[ZaqZainab]

when the are in differentdirections one is postive the other one is negative while when the move in the same direction both are either positive or negative so when we put them in the formula we get that the velocities of the balls in different directio will be added while the ones in same direction will be subtracted

either positive or negative so why subtracted it can be added too? for v and subtract for u
I am not saying that A is wrong but why is C wrong?
why cant it be like the velocities of the balls in different direction will be subtracted while the ones in same direction will be added?

 

[sagar65265]

ua is going in opposite direction to ub
so ua-ub??
va is in the same direction to vb so va+vb?
why is this wrong

An important point concerning elastic collisions is that the relative speed (relative is a very important word here) is the same for the colliding objects both before and after the collision occurs. Interestingly, this is independent of mass.

Suppose you are seeing the entire situation from the point of view of ball A. You appear to be stationary (i.e. the velocity of A from the perspective of A is zero) and B is coming towards you with a particular speed. This speed is equal to u(a) + u(b) - suppose you are moving at 100 kmph, and another object is coming towards you at 100 kmph, it will look as if that object is coming is coming at 200 kmph.

This is the relative speed for both objects before the collision [even if you are looking from the perspective of ball B, the velocity at which A is coming toward you is
u(a) + u(b)].

After the collision, the balls are travelling in the same direction. Suppose you are watching from the perspective of ball A. The relative speed here is v(b) - v(a).

To explain, suppose the following - you are traveling at 100 kmph, and someone in front of you is also traveling at 100 kmph. The distance between you remains the same, and from your perspective you are stationary - if the distance doesn't change, that must mean he is also stationary from your point of view, right?

This situation remains exactly the same if you replace your speed with v(a) and the other person's speed with v(b), and write v(a) = v(b). Furthermore, for the relative velocity to be zero (as discussed above) we cannot have Relative Speed = v(a) + v(b) because it would give us Relative Speed = 2v(a), which is wrong. But if we write Relative Speed = v(a) - v(b), we have Relative speed = 0 when v(a) = v(b), which is right.

Since the Relative Speed before the collision is the same as the Relative Speed after the collision, we equate the equations we have derived to get

u(a) + u(b) = v(b) - v(a)

(We could have u(a) + u(b) = v(a) - v(b), but that option is not there, so no need to worry about that).

Hope this helped!
Good Luck with all your exams!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_1.pdf
Q 9
why not C?
it is t^2 so here t=(t2-t1)
Q10 why not D?
Q11 i thought it is C
Q14
Q21
Q32 R=V/I
so here I is infinite wont R be o??
because I=V/R and if I is infinite I=V/0??
I know there are so many but even if you help me out with one of this i will be great full

9)
kitkat <3 :P answered it.

10)
Answer is D only Do you have eyes ? Jk Zaqyzaq

11)
Answered ._.

14)
Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm

21)
Use the formula p = ρgh
Hence when you find P for 1st you'll get is 3528
and when u'll find P for 2nd you'll get is 7056
This means When you multiply 2 with P1 you'll get P2 that is 7056 so ans will be 2P

32)
For this, its simple thing, for +1 V we have value of I as 50 * 10 ^ (-3) that will give R as 20 Ohm
And for -1V we have I as zero, hence as R = V / I : I = zer0, so when there is zero in denominator that is I here, so automatically, R is infinite

 

[ZaqZainab]

An important point concerning elastic collisions is that the relative speed (relative is a very important word here) is the same for the colliding objects both before and after the collision occurs. Interestingly, this is independent of mass.

Suppose you are seeing the entire situation from the point of view of ball A. You appear to be stationary (i.e. the velocity of A from the perspective of A is zero) and B is coming towards you with a particular speed. This speed is equal to u(a) + u(b) - suppose you are moving at 100 kmph, and another object is coming towards you at 100 kmph, it will look as if that object is coming is coming at 200 kmph.

This is the relative speed for both objects before the collision [even if you are looking from the perspective of ball B, the velocity at which A is coming toward you is
u(a) + u(b)].

After the collision, the balls are travelling in the same direction. Suppose you are watching from the perspective of ball A. The relative speed here is v(b) - v(a).

To explain, suppose the following - you are traveling at 100 kmph, and someone in front of you is also traveling at 100 kmph. The distance between you remains the same, and from your perspective you are stationary - if the distance doesn't change, that must mean he is also stationary from your point of view, right?

This situation remains exactly the same if you replace your speed with v(a) and the other person's speed with v(b), and write v(a) = v(b). Furthermore, for the relative velocity to be zero (as discussed above) we cannot have Relative Speed = v(a) + v(b) because it would give us Relative Speed = 2v(a), which is wrong. But if we write Relative Speed = v(a) - v(b), we have Relative speed = 0 when v(a) = v(b), which is right.

Since the Relative Speed before the collision is the same as the Relative Speed after the collision, we equate the equations we have derived to get

u(a) + u(b) = v(b) - v(a)

(We could have u(a) + u(b) = v(a) - v(b), but that option is not there, so no need to worry about that).

Hope this helped!
Good Luck with all your exams!

There i got it
Thanks you so much

 

[ZaqZainab]

9)
kitkat <3 :P answered it.

10)
Answer is D only Do you have eyes ? Jk Zaqyzaq

11)
Answered ._.

14)
Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm

21)
Use the formula p = ρgh
Hence when you find P for 1st you'll get is 3528
and when u'll find P for 2nd you'll get is 7056
This means When you multiply 2 with P1 you'll get P2 that is 7056 so ans will be 2P

32)
For this, its simple thing, for +1 V we have value of I as 50 * 10 ^ (-3) that will give R as 20 Ohm
And for -1V we have I as zero, hence as R = V / I : I = zer0, so when there is zero in denominator that is I here, so automatically, R is infinite

Q14 and why 100*10?? why 10?
Q 32 I is not 0 :O its infinite from the graph
Thanks for the rest you make it seem so simple I have just gone 'dumb'

 

[Thought blocker]

I is not 0 :O its infinite from the graph
Thanks for the rest you make it seem so simple I have just gone 'dumb'

Check this image and help me ._.
Welcome, I am even dumb, yesterday I waste my 2 hours in 1 question, read previous 2 - 3 pages

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31