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AS Physics P1 MCQs Preparation Thread.

lavanyamane

lavanyamane

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fatemakhan said:
33:
overall R increases as now one resistor is in series wid 2 resistors in parallel
therefore overall I decreases....now watever the current is,X gets the whole of it while Y and Z get a half-half share,Thus, (since I is directly propotional to power keeping other factors constant) power to X increases whlie that to Y and Z decreases.(C)
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Thank you :)
 
confused123

confused123

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SalmanslK said:
View attachment 12682

The answer is C. can anyone please tell me how...and whats the concept behind this question..?
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resultant force will be zero as same force is applied on both equal charges due to uniform electric field. the charges have opposite sign so the forces cancel out each other.
anticlockwise moment as the negative Q is attracted by positive charge on the plate, and the positive charge Q is move downwards as field lines coming from positive to negative. i.e downwards.
 
confused123

confused123

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KurayamiKimmi said:
can anyone PRETTY PLEASE! help me with q no 13 and 15 coz i spent at least 20 min thinkin bout these two but i couldnt solve them:mad:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
the ans for both of them is B
(curse those Bs!!)
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haha easy. spending time and short tempering will not help. its the game of concept :D

see in 13th mcq. you should recall that torque of a couple is one of the force into perpendicular distance between the two same forces. means F into 1.20 = 900 into .20 , this will give you force 150 N i.e the very dear and sweet B option!
for the minimum force to be applied we need to get same torque below and on the spindle in order to create rotation effect so that the poor weight can be lifted

15th: height decreases by 2 in both containers. pressure reduces by 4 times altogether...no convincing answer available at the moment.
 
applepie1996

applepie1996

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lavanyamane said:
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf
Help, please! Q. 5, 7, 9, 13, 18, 24, 31, 37. I know they are quite a few, but please?
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Q5)Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01

Q7)Q7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory

Q)9)Q9)use the formula v^2 = u^2 + 2as
where
positive direction = direction of train
u = original velocity (speed of train)
v = final velocity (zero)
a = acceleration of train (in this case, acceleration is negative)
s = distance (from point where velocity = u to where velocity = v)

The problem uses x for distance, not s, so the equation we'll use is v^2 = u^2 + 2ax

because v = 0, we can write
u^2 + 2ax = 0, rearranging as:
x = -(u^2) / 2a . . . . . . . . Note - a has a negative value, thus making x positive
x = Ku^2 where K = -1 / (2a)
ie. x varies as the square of u
so if u increases by 20% (ie changes by a factor of 1.2), then x will change by a factor of (1.2)^2
which is a factor of 1.44
thus the minimum distance between yellow and red must now be 1.44x
 
applepie1996

applepie1996

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S

Saloon

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bloody_mary said:
just one question how did you get x and z to be 45 degrees? :S
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sorry for the late reply... the working i showed first is wrong, I'm sorry if i made you waste your time (found out looking at http://www.xtremepapers.com/papers/...A and AS Level/Physics (9702)/9702_s11_er.pdf pages 5 and 6)
This is the correct working added a bit more detail, and also d=1/N you can look at this link http://www.s-cool.co.uk/a-level/physics/diffraction/revise-it/diffraction-from-a-diffraction-grating. sorry again :(
correct%20question%2026.jpg
 
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