do we have to study the As part like waves etc for A2 ?The applications will be merged with the questions. There will not be a separate section B for applications.
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do we have to study the As part like waves etc for A2 ?The applications will be merged with the questions. There will not be a separate section B for applications.
No this is not what it means.do we have to study the As part like waves etc for A2 ?
The electric field strength, E, is given by this formula:someone help!!!!........
9702/s14/qp13/question 32
Thanks in advance.....
Thanks legendUsing 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 alot!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.
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.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
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf
can someone please help me with 5 b?
thanks!
See solution 186Hello,
Guys please solve these qs
21 C
23 C
29 D
Thanks
pretty much thanksYou'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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