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http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf
Q.4, 15, 27, 28, 34 please?
Q.4, 15, 27, 28, 34 please?
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thanks alot but there are few aswell pls resolve them too tey are on page 38Q15)
At terminal velocity, weight = air resistance
so mg = kv
And v = mg/k
Now substitute v with mg/k in 1/2 mv^2 formula to get the answer.
Q22) You want high sensitivity for small masses, i.e. a big change in extension for smaller masses. This means a lower value of 'k' is needed for the smaller masses. The rigid box is there to prevent external forces like wind from pushing the masses on the spring. Also, when you have larger masses on the spring, then you need a higher value of 'k', i.e. a small extension for a large load. So the answer would be A.
Q32)
A and B are wrong because the e.m.f. can't decrease, its the energy being provided by the battery.
Also, VT = E - Ir (terminal voltage i.e. voltage across the battery is equal to e.m.f. - internal resistance * current).
From this equation, we can confirm that answer C is correct.
S11/12
Q16) Assume the length of the rod to be 1m. You know that the weight lies exactly in the middle, so each side is 0.5m apart. If you take the wall to be the pivot, then you can form an equation,
F sin 30 * 1 = 0.5 * 10
F = 10.0 N
Q19)
Work = force * distance in direction of force
10 = 0.005 * force
force = 2000 N
Q34) The answer to this one is a fact, you have to learn it. The e.m.f. depends on both r and R.
A is wrong because using the equation VT = E - Ir, the terminal voltage will change.
B is wrong because the current would decrease, not increase.
C is wrong because the e.m.f. of the battery can't change.
D is a correct statement. You can verify this yourself using P = I^2 * R
buddy kindly resolve my doubts on page 38 pls27.. electric field is away from the observer and into the page.. elecrtic field is always from positive to ngative... this means that the observer is facing positive plate directly... electrons would be repelled by the negative plate and will move towards the positive plate... behind where the observer is standing... i think...
for q.23 young modulus is a ratio and it is always constant for the same material regardless of dimenionsCan someone explain q16, q23 and q33 ? :S Confusing
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Oh yes. My bad ! Can you explain the other questions as well ?for q.23 young modulus is a ratio and it is always constant for the same material regardless of dimenions
q16 i wuld recommend u take values of ur ownCan someone explain q16, q23 and q33 ? :S Confusing
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
for q.16, kinetic energy before the collision = 1/2mv^2 + 1/2mv^2 => mv^2Oh yes. My bad ! Can you explain the other questions as well ?
17.. efficiency=output/input * 100 if input is 100 output is 8... so 8/100 * 100 =8...
31... electron would be attracted by the positive charge, so B and D are fundamentally wrong... now since the electron is moving against the electric field, it will be deflected out of the field by the negative charge... line of field represents the direction of the field ...
yea lol, it is A :/whaaaat
wat is da explanationyea lol, it is A :/
thnx i didnt get it eitheruse the equation P=hdg since the liquids are in equilibrium the pressure of air (since it is open) on both of them is equal
so P/hg = d
for P p/2xg
for Q p/xg
now divide P by Q to get 0.5
thnx i didnt get it either
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