Physics: Post your doubts here!

[ashcull14]

 

[Xylferion]

View attachment 52039

Energy ( E ) = Power ( P ) * Time ( t )
Power = I^2 * R -----> ( 40 x 10^-3 )^2 * 60 = 24 J

e.m.f = useful volts + lost volts
Since 6.0 J is lost, we need to find how many volts are lost.
Voltage ( V ) = Energy ( W ) / Charge ( Q )
Charge ---> Q = It = 40 x 10^-3 * 60 = 2.4 coulombs

Voltage = 6 J / 2.4 C = 2.5 volts
This is the number of volts lost.

Emf = Useful volts + 2.5 volts
Useful volts = Those used up by the resistor
Voltage at resistor = IR = 40 x 10^-3 * 250 = 10 volts

Emf = 10 volts + 2.5 volts = 12.5 volts

So the answer is D.

Hope that helped!

 

[zain ul abidin]

Can someone please explain 0/n 11 p43 q7 b ii 2
m/j 12 p42 q1 c ii
and o/n 12 p43 q5 c ii

 

[TheJDOG]

Can someone please explain 0/n 11 p43 q7 b ii 2
m/j 12 p42 q1 c ii
and o/n 12 p43 q5 c ii

Hello zain,

o/n 11 p43 q7 bii 2

Cannot find this question

m/j 12 p42 q1 cii
c) ii) First off, were told that the radius of the orbit of the moon increases by 4 cm each year.
We need the energy change in one year
We also know that Energy change= m.Change in phi ,where m is the mass and change in phi is change in potential.
Moon mass: 7.4 x 10^22 Kg
Change in potential= GM(1/R1 - 1/R2)
Now just substitue: Change in potential= (6.67 x 10^-11)(6 x 10^24) (1/3.84x10^8) - 1/3.84x10^8 + 0.04)= 1.09 x 10^-4
Now Energy change = moon mass x change in potential = (7.4 x 10^22)(1.09 x 10^-4)=8 x 10^18 J

o/n 12 p43 q5 c ii

Cannot find this question

 

[amal sharkawi]

Can anyone answer this question plz??

 

[amal sharkawi]

Can anyone answer plz??

 

[amal sharkawi]

Can anyone answer plz??

 

[zain ul abidin]

Hello zain,

o/n 11 p43 q7 bii 2

Cannot find this question

m/j 12 p42 q1 cii
c) ii) First off, were told that the radius of the orbit of the moon increases by 4 cm each year.
We need the energy change in one year
We also know that Energy change= m.Change in phi ,where m is the mass and change in phi is change in potential.
Moon mass: 7.4 x 10^22 Kg
Change in potential= GM(1/R1 - 1/R2)
Now just substitue: Change in potential= (6.67 x 10^-11)(6 x 10^24) (1/3.84x10^8) - 1/3.84x10^8 + 0.04)= 1.09 x 10^-4
Now Energy change = moon mass x change in potential = (7.4 x 10^22)(1.09 x 10^-4)=8 x 10^18 J

o/n 12 p43 q5 c ii

Cannot find this question

o/n 11 p43 q7 bii 2
http://studentbounty.com/pastpapers...el/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf

o/n 12 p43 q5 c ii
http://studentbounty.com/pastpapers...el/Physics (9702)/2012 Nov/9702_w12_qp_43.pdf

 

[Muskan Achhpilia]

Hey,

Can someone please help me out with the following mcq and structured questions?

http://www.docdroid.net/xk0o/2002-onwards-physics-structure-compiled-all-doubts.docx.html

http://www.docdroid.net/xjzg/mcq-all-combined-physics.docx.html

For the structured questions I have also mentioned the paper

Thanks a lot,

 

[TheJDOG]

o/n 11 p43 q7 bii 2
http://studentbounty.com/pastpapers/Cambridge International Examinations (CIE)/International AS and A Level/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf

o/n 12 p43 q5 c ii
http://studentbounty.com/pastpapers/Cambridge International Examinations (CIE)/International AS and A Level/Physics (9702)/2012 Nov/9702_w12_qp_43.pdf

O/n 11 p43 q7 bii 2

We need a value for plancks constant, we have to find the gradient of the graph fig. 7.1.
I will take these two points on the graph (1.3x10^-19,2.6x10^6) and (3.7x10^-19,3.8x10^6)
Gradient = y2-y1/x2-x1= (3.8-2.6)x10^6/(3.7-1.3)x10^-19= 5x10^24
But we know that the MaxEnergy of a photoelectron= hf=hc/lambda - phi

Notice that normally we plot max energy on y axis and frequency or 1/lambda on x axis. With gradient=hc

But here, they're swapped max energy of photoelectron on x axis and 1/lambda on y axis and so the gradient must equal 1/hc.

We can now continue calculating gradient =1/hc and h=1/gradient.c
So h= 1/(5x10^24)(3x10^8)= 6.67 x10^-34 Js

O/n 12 p43 q5 cii
Hmm I'm honestly not sure about how to sketch this one I don't want to give you a wrong answer

 

[amal sharkawi]

Can anyone answer plz??

 

[Physux]

View attachment 52103
Can anyone answer plz??

Is the answer A because for an elastic collision relative speed of approach= relative speed of separation?

 

[Physux]

Can anyone explain how to solve the last part of paper 3 question 1? For example part f of variant 34 M/J 2010.

 

[zain ul abidin]

check question 576 at
http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-113.html

E=-(d phi/d t) so shouldn't it be in opposite direction i.e first pulse in negative side and second at positive?

 

[zain ul abidin]

O/n 11 p43 q7 bii 2

We need a value for plancks constant, we have to find the gradient of the graph fig. 7.1.
I will take these two points on the graph (1.3x10^-19,2.6x10^6) and (3.7x10^-19,3.8x10^6)
Gradient = y2-y1/x2-x1= (3.8-2.6)x10^6/(3.7-1.3)x10^-19= 5x10^24
But we know that the MaxEnergy of a photoelectron= hf=hc/lambda - phi

Notice that normally we plot max energy on y axis and frequency or 1/lambda on x axis. With gradient=hc

But here, they're swapped max energy of photoelectron on x axis and 1/lambda on y axis and so the gradient must equal 1/hc.

We can now continue calculating gradient =1/hc and h=1/gradient.c
So h= 1/(5x10^24)(3x10^8)= 6.67 x10^-34 Js

O/n 12 p43 q5 cii
Hmm I'm honestly not sure about how to sketch this one I don't want to give you a wrong answer

http://studentbounty.com/pastpapers...el/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf
can u explain q7 b ii 1 also?

 

[TheJDOG]

http://studentbounty.com/pastpapers/Cambridge International Examinations (CIE)/International AS and A Level/Physics (9702)/2011 Nov/9702_w11_qp_43.pdf
can u explain q7 b ii 1 also?

In this question were asked to calculate phi the work function energy using the eqaution we wrote in (i) and Fig 7.1
The maximum kinetic energy of an emitted photoelectron is Emax= hc/lambda - phi
Now go back to Fig 7.1 and pick any point on the line
I'll pick (1.5 x 10^-19,2.7x10^6)
h the Planck constant= 6.63 x 10^-34 Js
Just substitue and work for phi
1.5x10^-19=(6.63x10^-34)(3x10^8)(2.7x10^6) - phi
So work function of metal =Phi= 3.87 x 10^-19 J

 

[zain ul abidin]

In this question were asked to calculate phi the work function energy using the eqaution we wrote in (i) and Fig 7.1
The maximum kinetic energy of an emitted photoelectron is Emax= hc/lambda - phi
Now go back to Fig 7.1 and pick any point on the line
I'll pick (1.5 x 10^-19,2.7x10^6)
h the Planck constant= 6.63 x 10^-34 Js
Just substitue and work for phi
1.5x10^-19=(6.63x10^-34)(3x10^8)(2.7x10^6) - phi
So work function of metal =Phi= 3.87 x 10^-19 J

but as we have to calculate 'h' in next part i don't think that we can use it in this part!

 

[TheJDOG]

but as we have to calculate 'h' in next part i don't think that we can use it in this part!

No bro I understand what you're trying to say but we can use h Planck's constant because we know it, and it's given in the formula sheet h= 6.63 x 10^-34 Js
In part (ii) we're asked to find a value for it using Fig 7.1 and our equation. we cool?

 

[Muskan Achhpilia]

most of these are already solved at
http://physics-ref.blogspot.com/

check for the corresponding years there

Okay thanks

 

[zain ul abidin]

No bro I understand what you're trying to say but we can use h Planck's constant because we know it, and it's given in the formula sheet h= 6.63 x 10^-34 Js
In part (ii) we're asked to find a value for it using Fig 7.1 and our equation. we cool?

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