Physics: Post your doubts here!

[omg]

any vid animation for rectification??

 

[Oliveme]

14) rebounding height is half of initial means potential energy is half after rebounding.
p.e = k.e therefore k.e is also half after the rebound
final k.e = 1/2 of initial k.e
1/2*m*v^2 = 1/2 (1/2*m*u^2)
v^2 = 1/2* u^2
taking square root on both sides
v = u/sqr rt of 2
v/u = (u/sqr rt of 2) / u
u cancels out
ratio is 1/sqr rt of 2

Thank you very much. You're amazing.

 

[Oliveme]

question 15
look the box at rest having potential energr mgh = 9.81*3*2=58.9J.


now this energy will be converted to kinetic energy at the bottom of the slope.
means at bottom 1/2mv^2 = 58.9J now subtract the work done against friction while coming down! W=f.d



means W= 5*7 = 35J now subtract it from the original ans. u will get the net kinetic energy at the bottom of the slope which is 58.9-35=23.9J



now look 1/2 mv^2 =23.9 find v??

v= 4.88 round off u will get 4.9ms-1Ans

thank you. stuck on this question for ages and you solved it in seconds.

 

[Oliveme]

35) total resistance is
1/r = 1/1 + 1/2 + 1/5
r= 0.588 ohms
total currrent is 5A
v = IR
v = 5*0.588
= 2.94
voltage is parallel does not divide
so voltage in 2ohms resistor will be 2.94 volts
v=ir
2.94 = i * 2
i = 1.5

thank you

 

[Oliveme]

we know that current entering circuit is equal to leaving the circuit according to khirchoff's law

now I =v/r thsi is the equation for current
now current in resistor of 1 ohm = I1=v/1
2 ohm = I2=v/2
5 ohm = I3 v/5 ..... remember V being same as voltage never divides in paralel circuit!

now total curren I is 5 amperes we have the eq. I1+ I2=I3 = 5

substitute values of I1 , I2 ,I3 and find the unknown voltage! which will be V = 50/17. = 2.94volts

now fr finding current in resistor of 2 ohms! V=IR I=V/R 2.94/2 = 1.47 round of to get 1.5amperes Ans.



hope u get it well???? let me know abt it???

yes, i get it. thank you. thumbs up.

 

[Oliveme]

Asalam-o-alaikum
can somebody please help me with these questions?
questions 10, 27, 30, 34. thank you very much.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

 

[rz123]

Asalam-o-alaikum
can somebody please help me with these questions?
questions 10, 27, 30, 34. thank you very much.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

for 30: u need to calculate the electric field strengt first by using E= V/d i.e 12k / 25mm = 480000 v/m.

use this to find the force: E= F/q 480000 into 1.6* 10^-19 this will give u the answer 7.7 into 10^-14 Newton uncle.

 

[Mustehssun Iqbal]

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??

 

[leadingguy]

Asalam-o-alaikum
can somebody please help me with these questions?
questions 10, 27, 30, 34. thank you very much.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

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

 

[unique840]

Thank you very much. You're amazing.

ur welcum

 

[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