Physics: Post your doubts here!

[unique840]

thank you

welcum welcum

 

[leosco1995]

December 2003, Q3 (c)

What are the directions of S1 and S2?

 

[sweetiepie]

For Notes Check here

http://www.xtremepapers.com/community/threads/some-different-notes-website-are-available.10423/

 

[hassam]

This questions is about the hall effect in electromagnetism;
The free delocalised electrons are moving freely within a conductor, and they're doing so continuously. And they're moving within the conductor in random motion. If that's so, then the hall effect might just act on each of those free electrons. Some electrons might drift to the right, some to the left, etc and they're moving inside the conductor irrespective of the motion of conductor, to some extent. And the magnetic force, that acts on each of these electrons might vary in direction, from one electron to another. Keeping this into account, how is it known that the force on the electrons act in just one direction?? Or is it the net force on the electrons that is taken into consideration in the hall effect??

a current is passed through that piece of semiconductor.......it is placed perpendicularly to the unknown magnetic field....this field causes a bill force on positive charges and electrons so they collect on opposite sides giving rise to a p.d .....this is measured by the voltmeter as hall voltage......this hall voltage is directly proportional to magnetic field strength so calibrating Vh against known magnetic field will let us know the value for field stregth at each Vh

 

[omg]

a current is passed through that piece of semiconductor.......it is placed perpendicularly to the unknown magnetic field....this field causes a bill force on positive charges and electrons so they collect on opposite sides giving rise to a p.d .....this is measured by the voltmeter as hall voltage......this hall voltage is directly proportional to magnetic field strength so calibrating Vh against known magnetic field will let us know the value for field stregth at each Vh

do we need to know abt the hall effct in detail???????????? 0.O

 

[smzimran]

do we need to know abt the hall effct in detail???????????? 0.O

You should know how a hall probe works...
It is important for P5 also

 

[hassam]

yea.....there is one question tilll now in paper 4 and 2 in paper 5 for which u need to understand wat i wrote above

 

[omg]

yea.....there is one question tilll now in paper 4 and 2 in paper 5 for which u need to understand wat i wrote above

CAN U TELL ME THE YEARS OF THOSE PPRS PLS??

 

[Anonymous']

December 2003, Q3 (c)

What are the directions of S1 and S2?

The directions of S1 and S2 are to the left and to the right respectively..

 

[Oliveme]

Q.10) ball moving towards wall will have momentum= p= mu (consider the direction positive )

now aftr striking wall it rebounds back having same speed as stated. Now p= mv= m(-u) (as the direction of ball is now changed so itx velocity should also change).


now at frst the momentum was mv
aftr striking it was -mv


for change in two things simply subtract both ................... (-mv) - (mv)
change in p = -2mv Ans


q.27) well I am sorry but I am unable to recall formulas for waves at the moment.



q.34) to calculate resistance, R = pl/A so A= l^2


R = pl/l^2

R = p/l now convert l in terms of V l^3 = V so l = V^1/3

sustitute V for l then R = p/V^1/3 Ans

q.30) IS already answered

thank you very much

 

[Oliveme]

can someone pls help me with ques 27? thank you
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

 

[Anonymous']

can someone pls help me with ques 27? thank you
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

hmm.. let's see.. it is given : 500 lines in 1 mm, therefore, 1 line = 0.001/500 = 2 * 10^-6
the general equation to find the maximum no. of order, i.e the maximum no. of images = d/lamba = (2 * 10^-6)/(600*10^-9) = 3.33 so, the maximum no. or order is 3, but, you have to consider the other 3 images at position Z, hmm.. firstly, zero order is at position Y, so, that's one image, then there are three orders, i.e, three images at position X as well as at position Z, therefore, altogether there are 7 images, 3 images at position X, 1 image at position Y, and 3 images at position Z.. I hope you understand my explanations.. heheh..

 

[leadingguy]

thank you very much

welcome!

 

[Oliveme]

hmm.. let's see.. it is given : 500 lines in 1 mm, therefore, 1 line = 0.001/500 = 2 * 10^-6
the general equation to find the maximum no. of order, i.e the maximum no. of images = d/lamba = (2 * 10^-6)/(600*10^-9) = 3.33 so, the maximum no. or order is 3, but, you have to consider the other 3 images at position Z, hmm.. firstly, zero order is at position Y, so, that's one image, then there are three orders, i.e, three images at position X as well as at position Z, therefore, altogether there are 7 images, 3 images at position X, 1 image at position Y, and 3 images at position Z.. I hope you understand my explanations.. heheh..

umm...i'm sorry but i still don't get it. how do you know there is only one at Y and 3 at X and Z.

 

[Ashique]

^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf

 

[unique840]

^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf

8) u=o
s=40
for the first 30 m:
2as = v^2 - u^2
2(10)(30) = v^2 - 0
v = 24.5 m/s
this v will be initial speed for the last 10 m
so for the last 10 m:
s = ut + (1/2)at^2
10 = 24.5t + 5t^2
t = 0.38 or -5.3
so t is 0.38

 

[unique840]

^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf

13) torque is force*perpendicular distance
the perpendicular distance is not given. the distance given is hypotenuse so we will calculate the perpendicular distance by 0.6sin60
so torque will be 0.6sin60 * 8 = 4.2

 

[unique840]

^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf

24) option D. its theory u have to learn the spectrum
28) D. cox the point negative charge has its field lines directed inwards

 

[unique840]

^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf

33) at point s, voltage is v1. in the first branch the voltage is divided in 1:1 ratio. so v1 is 1 volts
in second branch the voltage on 3 ohm resistor is 3/5 * 2 = 1.2 volts
v1 - v2 = 1.2 - 1
= -0.2 volts

 

[Ashique]

Thanks, unique840. It's really simple once you break it down. Thank you once again.