Physics: Post your doubts here!

[Mayarzawaydeh]

can somone pleassseeee help me in this question??
this paper is summer 2011 paper 52 question 2
the question says that it is suggested that V and R are related by the equation

V = F E /R + E

Where F is the resistance of the fixed resistor in the circuit and E is the e.m.f of the cell
a graph is plotted of V/E on the Y axis against 1/R on the X axis. Express the gradient in terms of F

can someone explain why the gradient answer is F ?
Thanks in advance !!!

 

[muhammadali233]

Part b says to calculate the new height gained by the ball,its kinetic energy is 90% of what was before

I went through this working and was able to calculate the right answer but the ms says to just take the 90% of the height and not going through the long working,the right answer scores only 1/2 why is that?What concept am i missing?

 

[The Sarcastic Retard]

can somone pleassseeee help me in this question??
this paper is summer 2011 paper 52 question 2
the question says that it is suggested that V and R are related by the equation

V = F E /R + E

Where F is the resistance of the fixed resistor in the circuit and E is the e.m.f of the cell
a graph is plotted of V/E on the Y axis against 1/R on the X axis. Express the gradient in terms of F

can someone explain why the gradient answer is F ?
Thanks in advance !!!

Y = mx + c
Y = V/E
x = 1/R
V/E = m(1/R) + c
m = F(E/E)(1/R)
m = F

 

[Rizwan Javed]

Part b says to calculate the new height gained by the ball,its kinetic energy is 90% of what was before
View attachment 59692
I went through this working and was able to calculate the right answer but the ms says to just take the 90% of the height and not going through the long working,the right answer scores only 1/2 why is that?What concept am i missing?
View attachment 59694

Your method is correct, but it can be done in a simpler way. Make use of the concept of conservation of energy. The P.E at the h1 (the initial height from which the ball was dropped), is converted into K.E fully when it hits the plate. That means that

P.E at h1 = K.E

Now after leaving the plate, the K.E (which is 90% of the K.E before hitting Or 90% of P.E at h1) is converted into P.E as it reaches h2. So,

90/100 * P.E at h1 = P.E at h2
90/100 * m g h1 = m g h2
90/100 * h1 = h2

 

[cashew]

PLEASE tell me how to solve it.
correct answer is C.

 

[Saad the Paki]

PLEASE tell me how to solve it.
correct answer is C.

You need to divide it into two separate parts.

First the part where he accelerates. Find the distance covered while he accelerates : 2as=v^2-u^2 ; s = (10^2)/2×2.5 =20m. Now find the time during this acceleration : s = ut+1/2at^2 ; t^2 = 16 therefore t = 4s.

Now the second part. We need to find the time taken when we moves at a constant speed of 10. Distance is 100-20 = 80m ,
t = distance/speed ; 80/10 = 8s

Now find the total time 8+4 = 12s. The answer is C.

 

[cashew]

You need to divide it into two separate parts.

First the part where he accelerates. Find the distance covered while he accelerates : 2as=v^2-u^2 ; s = (10^2)/2×2.5 =20m. Now find the time during this acceleration : s = ut+1/2at^2 ; t^2 = 16 therefore t = 4s.

Now the second part. We need to find the time taken when we moves at a constant speed of 10. Distance is 100-20 = 80m ,
t = distance/speed ; 80/10 = 8s

Now find the total time 8+4 = 12s. The answer is C.

Jazak Allah!!! I was finding the solution since hourssss

 

[usman]

Here are some resources for A level Physics I used in my time (includes my lecture notes and the coursebook's CD content):

https://www.mediafire.com/folder/h9q8om4c2c9bl/A_Level_Physics

 

[cashew]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s15_qp_12.pdf

q12. answer with explanation?
correct answer is C but I don't understand why

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_12.pdf

q12. answer with explanation?
correct answer is C but I don't understand why

This question involves the concept of projectile motion. You know that whenever there's a force acting on a body, the body accelerates. For a projectile in the absense of air, the horizontal velocity is constant. Since the velocity is constant, it means that there is no acceleration. No acceleration means, no force. So there is no force horizontally on the projectile.

Now moving to the vertical motion. On moon, there's a force of gravity which pulls the projectile down. This force of gravity is constant.

So the answer is C.

 

[cashew]

This question involves the concept of projectile motion. You know that whenever there's a force acting on a body, the body accelerates. For a projectile in the absense of air, the horizontal velocity is constant. Since the velocity is constant, it means that there is no acceleration. No acceleration means, no force. So there is no force horizontally on the projectile.

Now moving to the vertical motion. On moon, there's a force of gravity which pulls the projectile down. This force of gravity is constant.

So the answer is C.

THANKYOU! unfortunately, I was not aware of this concept.

 

[Mayarzawaydeh]

does anyone know where I can find winter 2015 papers for cie physics 9702 ?

 

[qwertypoiu]

does anyone know where I can find winter 2015 papers for cie physics 9702 ?

Try at gcecuide

 

[hafsa saleem]

can anyone please help me with mcq 37
http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_13.pdf

 

[Eugene99]

I got that 2nd Edition of Physics book, but will it cover 2016 syllabus?

 

[My Name]

I got that 2nd Edition of Physics book, but will it cover 2016 syllabus?

Did you get an ebook? :3

 

[My Name]

Can anyone please give me simple notes for HALL PROBE and how the HALL VOLTAGE expression is derived.

 

[Rizwan Javed]

Did you get an ebook? :3

I got one :3

 

[Eugene99]

Did you get an ebook? :3

nope, I got the hard copy

 

[Eugene99]

I got one :3

where did you get that from?