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Physics: Post your doubts here!

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do we have to study the As part like waves etc for A2 ?
No this is not what it means.
You must have seen the past papers they had a section A which consisted of questions from the A2 syllabus except the applications which were in section B. Now they'll be merged together. There won't be any section B.
 
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someone help!!!!........
9702/s14/qp13/question 32

Thanks in advance.....
The electric field strength, E, is given by this formula:

E = F/Q (where F is force, Q is charge and E is electric field strength)

We know F = ma, so E becomes:
E = ma/q ----(i)

You're given that, K.E = k. Use the equation of K.E (1/2mv^2) to find the value of m and substitute this value of 'm' into (i)
m = 2K/v^2 ---- substituting in (i)

(i) will become:
E = ( (2K/v^2) * (a) )/ q ------(ii)

Now use the equation of motion: v^2 = u^2 + 2as
The charge starts moving from rest, so initial speed (u) is zero. And it covers a distance of 'd' for reaching plate Y, so s=d. So,
v^2 = 2ad ---- put this one into (ii)

E = ( (2K/2ad) * (a) )/ q

Simplify it, you'll get :

E = K/qd

So the answer is C.
 
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Using the formula you proved above, you can find E, the Kinetic Energy required to statisfy their condition.
Next, using the relation E = VQ, you can find V. This is because the kinetic energy is being provided by the electric potential energy. You already know the charge and mass of an electron from the data provided at the beginning of paper.
Thanks legend
 
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The electric field strength, E, is given by this formula:

E = F/Q (where F is force, Q is charge and E is electric field strength)

We know F = ma, so E becomes:
E = ma/q ----(i)

You're given that, K.E = k. Use the equation of K.E (1/2mv^2) to find the value of m and substitute this value of 'm' into (i)
m = 2K/v^2 ---- substituting in (i)

(i) will become:
E = ( (2K/v^2) * (a) )/ q ------(ii)

Now use the equation of motion: v^2 = u^2 + 2as
The charge starts moving from rest, so initial speed (u) is zero. And it covers a distance of 'd' for reaching plate Y, so s=d. So,
v^2 = 2ad ---- put this one into (ii)

E = ( (2K/2ad) * (a) )/ q

Simplify it, you'll get :

E = K/qd

So the answer is C.
thanks alot!
 
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This question is from a mock exam and I do not know its answer. I coudn't find the paper. And answer with the explanation would be appreciated.
upload_2016-4-4_0-38-7.png
upload_2016-4-4_0-38-37.png
 
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This question is from a mock exam and I do not know its answer. I coudn't find the paper. And answer with the explanation would be appreciated.
View attachment 60005
View attachment 60006
I think it's C. Because in a stationary wave, the points in adjacent loops are out phase with each other. So if the displacement of a point in one loop is S then in the adjacent loop, it should be -S.
 
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Hello,
Guys please solve these qs
21 C
23 C
29 D
Thanks
 

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You're asked to find the feedback resistances.

V out / V in = Feedback resistance / Input Resistance

They said for each switch position, V out = 1.0 V, so just plug that in for each part.

For R(A),

1.0 / 100x10^-3 = Feedback resistance of A / 1000
Feedback resistance = 10,000 ohms

For R(B),

1.0 / 10x10^-3 = Feedback resistance of B / 1000
Feedback resistance = 100,000 ohms

For V in of C, work backwards,

1.0 / V in = 1000 / 1000

V in = 1.0 Volts which is the same as 1000 mV

Hope that helped!
 
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You're asked to find the feedback resistances.

V out / V in = Feedback resistance / Input Resistance

They said for each switch position, V out = 1.0 V, so just plug that in for each part.

For R(A),

1.0 / 100x10^-3 = Feedback resistance of A / 1000
Feedback resistance = 10,000 ohms

For R(B),

1.0 / 10x10^-3 = Feedback resistance of B / 1000
Feedback resistance = 100,000 ohms

For V in of C, work backwards,

1.0 / V in = 1000 / 1000

V in = 1.0 Volts which is the same as 1000 mV

Hope that helped!
pretty much thanks :)
 
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