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AS Physics P1 MCQs Preparation Thread.

umarashraf

umarashraf

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MysteRyGiRl said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
i need help here plzzzzzz
in Q no 11,12,13,20,22,25....
Click to expand...
12 first... since applied force is constant, the torque (F*d) will be proportional to d.... torque for 100mm is 3.. so for 150mm , it will be 4.5Nm... so we have to look for the tensions now..torque is the product of force and moment arm.. and the only force present here is the tension in the belt so 3=F*.1 F=30N
for the other wheel, 4.5=F*.15 F=30... so total force is equal to 30N to 30N = 60N...
 
Casablanca

Casablanca

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bloody_mary said:
frequency = 1/T

and definition of frequency is how many cycles per second and we got 50revolutions per second so i found the time period which is 0.02 so 1 wavelength takes 0.02s to complete so i we use 0.01 we can see half the wave
Click to expand...

Oh, alright. Thanks. :D

And if it's not too much trouble, could you explain question 9 from the same paper? The one about the train.
 
umarashraf

umarashraf

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umarashraf

umarashraf

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MysteRyGiRl said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
i need help here plzzzzzz
in Q no 11,12,13,20,22,25....
Click to expand...
20.. was so simple yaar... find area under the curve.. ok.. draw a straight line from 0,0 to P... and use 0.5* base * height.. this wil give u 0.1 as answer.. now it is clear that the area you have found is less than the original one but double to it.. so option just little greater than 0.1 isC...
 
bloody_mary

bloody_mary

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Casablanca said:
Oh, alright. Thanks. :D

And if it's not too much trouble, could you explain question 9 from the same paper? The one about the train.
Click to expand...

ok first say intial speed is u
decelration is a
and distance given is x

using equations of motion with only data given we use v^2=u^2 +2as

rearrange it to make s the subject s= v^2 - u^2/2a | since v is zero i will omit it

so -u^2/2a = x

for an increase by 20% it is like mulitplying by 1.20 so new intial velocity is 1.20u
so -(1.20u)^2/2a gives -1.44u^2/2a = x'

divide x' by x to get ratio so:
x' /x = -1.44u^2/2a * 2a/u^2

the 2a and u^2 cancel to give 1.44x
 
MysteRyGiRl

MysteRyGiRl

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umarashraf said:
20.. was so simple yaar... find area under the curve.. ok.. draw a straight line from 0,0 to P... and use 0.5* base * height.. this wil give u 0.1 as answer.. now it is clear that the area you have found is less than the original one but double to it.. so option just little greater than 0.1 isC...
Click to expand...
haan i was confused wd C and B i fnk :p jo bhi options thay wd 0.1 lolz :D
 
MysteRyGiRl

MysteRyGiRl

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umarashraf said:
what what what...??? 37../??? isotopes have same number of protons mate... so D is the answer... whats so special with this one..???
Click to expand...
but shouldnt just da proton number increase????? why is da nucleon nmbr increasing if its beta emission?
 
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