Physics: Post your doubts here!

[hassam]

AND we DO Not have the picture for spectrum of FM in any of our books.....

 

[Anonymous']

Assalamualaikum, and hey guys! hmm.. I need help on question 3 (b) november 2001 -------> http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w01_qp_2.pdf

 

[DragonCub]

Assalamualaikum, and hey guys! hmm.. I need help on question 3 (b) november 2001 -------> http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w01_qp_2.pdf

Via intuition you can deduce that the extension in S3 is also x.
The springs are identical, so their constant of elasticity can all be seen as k.
Force on S1 (call it F1) is kx, so is force on S2 (call it F2).
The resultant force of F1 and F2 is to the left, since their vertical components cancel out each other.
Horizontal resultant force = F1 cos60 + F2 cos60 = kx/2 + kx/2 = kx
Due to equilibrium, this is the force on S3. Extension of S3 = kx/k = x

 

[Anonymous']

Via intuition you can deduce that the extension in S3 is also x.
The springs are identical, so their constant of elasticity can all be seen as k.
Force on S1 (call it F1) is kx, so is force on S2 (call it F2).
The resultant force of F1 and F2 is to the left, since their vertical components cancel out each other.
Horizontal resultant force = F1 cos60 + F2 cos60 = kx/2 + kx/2 = kx
Due to equilibrium, this is the force on S3. Extension of S3 = kx/k = x

hmm.. I don't really get this -------> Horizontal resultant force = F1 cos60 + F2 cos60 = kx/2 + kx/2 = kx
Due to equilibrium, this is the force on S3. Extension of S3 = kx/k = x

can you explain why did you put: kx/2 + kx/2 = kx and why is it: Extension of S3 = kx/k = x ?

 

[Anonymous']

Via intuition you can deduce that the extension in S3 is also x.
The springs are identical, so their constant of elasticity can all be seen as k.
Force on S1 (call it F1) is kx, so is force on S2 (call it F2).
The resultant force of F1 and F2 is to the left, since their vertical components cancel out each other.
Horizontal resultant force = F1 cos60 + F2 cos60 = kx/2 + kx/2 = kx
Due to equilibrium, this is the force on S3. Extension of S3 = kx/k = x

Wait, I think I already figured it out! heheh.. Thank you for your explanations! heheh..

 

[tom ed]

can someone plz tell me why was the average of current taken to calculate the amount of charge flowing....in (question no 30..paper 1 june 2003).the question is:.
30 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 160mC B 320mC C 480mC D 640mC

the answer is C

 

[DragonCub]

can someone plz tell me why was the average of current taken to calculate the amount of charge flowing....in (question no 30..paper 1 june 2003).the question is:.
30 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 160mC B 320mC C 480mC D 640mC

the answer is C

You can assimilate this with linear motion
charge = current × time
displacement = velocity × time
You must be familiar with calculation involving acceleration, aren't you?
It is quite similar for charge. Here the "initial current" 100 mA, "final" 20 mA, "current acceleration" is constant (uniform).
So charge (displacement) = [initial current (velocity) + final current (velocity) ]/2 × time
= (100 mA + 20 mA)/2 × 8.0 s = 480 mC

 

[tom ed]

You can assimilate this with linear motion
charge = current × time
displacement = velocity × time
You must be familiar with calculation involving acceleration, aren't you?
It is quite similar for charge. Here the "initial current" 100 mA, "final" 20 mA, "current acceleration" is constant (uniform).
So charge (displacement) = [initial current (velocity) + final current (velocity) ]/2 × time
= (100 mA + 20 mA)/2 × 8.0 s = 480 mC[/quote

yeah rite.....yes i am........thanku very much.......for telling....u have certainly mastered ths part...XD

 

[MEGUSTA_xD]

An electric railway locomotive has a maximum mechanical output power of 4.0 MW. Electrical power is delivered at 25 kV from overhead wires. The overall efficiency of the locomotive in converting electrical power to mechanical power is 80 %.
What is the current from the overhead wires when the locomotive is operating at its maximum power?
A 130A B 160A C 200A D 250A



Answer: C

 

[unique840]

output power given is 4MW. efficiency is 80%.
efficiency = output power/input power
input power = output / efficiency
input = 4MW/80%
input = 5MW
P=VI
5MW/25kV = I
I= 200A

 

[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