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Physics: Post your doubts here!

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the distance of block from the level of head of the man is 18m and when to level the head with the block man travels 9m upwards and block 9m downwards.
tension acts towards the pulley and the weight acts opposite to tension i.e downwards.
force acting in the direction of acceleration is considered more in magnitude than the force in opposite direction

for the man
T-80g=80a equation 1
120g-T=120a equation2
adding both equations we get
40g=200a
a=0.2g (given g as 10ms^-2)
a=2ms^-2
v^2=2as (as initial velocity of both the objects is zero )
v^2=2*2*9
v^2=36
v=6ms^-1

for question 24
stress=F/A
F=stress*A
F=(stress/4)*4A (area would be 4A as (2D)^2 is four times greater than (d)^2 for e.g (2*2)^2=16 x(2)^2=4 see
stress should be divided by 4 as the examiner states that load is the same which means the force applied is the same
so to get F 4 must be divided by 4 to get 1 ok!!

stress is directly proportional to strain
so if stress is stress/4 .strain would also be strain/4
stress/4=x/3l.
stress=x/l main formula
coefficient of x would be such that it equals 3/4 because only then we would get our formula u see coefficients will change but the formula wont ok
that mean x=3/4
 
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snowbrood
the distance of block from the level of head of the man is 18m and when to level the head with the block man travels 9m upwards and block 9m downwards.
tension acts towards the pulley and the weight acts opposite to tension i.e downwards.
force acting in the direction of acceleration is considered more in magnitude than the force in opposite direction

for the man
T-80g=80a equation 1
120g-T=120a equation2
adding both equations we get
40g=200a
a=0.2g (given g as 10ms^-2)
a=2ms^-2
v^2=2as (as initial velocity of both the objects is zero )
v^2=2*2*9
v^2=36
v=6ms^-1

i understood qs 24 very well ....but still haven't got an idea about this qs 12...plsss...label the words "a"...etc...and write eqs first ....like mg=ma ...what ever is the equation.... i am sorry for the inconvinience but i am a medical student ....and mathematics has always been my enemy .... :p...lol
 
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May june 2012.. Paper 11 Q34 ....i really don't have the ideo hoe to solve these kind ov qs!!!....need help with a concept pls. (-_*)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

For these questions it's easier if you compare ratio of the two resistance. The resistance of the first wire is R, let the resistance of the second be R2.
Now, you know that resistance = (resistivity* length)/ cross-sectional area.
So, R/ R2 = [(Resistivity* L1)/ A1]/[(Resistivity* L2)/A2]
for the same material the resistivity is the constant. So, cancel resistivity and cross-multiple and you get R/ R2= L1*A2/ L2*A1
once you understand this ratio you can start from here for all such question in future. Just put the values in now and remember to convert all the units for radius to meter. Area of cross-section Pi r^2 but again, you can cancel out pi for both. So it's just length*radius^2
you get, R/R2 = 8* (5*10^-4)^2/ 2* (2.5*10^-4)^2
that is, R/R2 = 16
so R2 = R/16

It might seem a little confusing but just write it out as you go and it gets simpler. Hope this helps!
 
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For these questions it's easier if you compare ratio of the two resistance. The resistance of the first wire is R, let the resistance of the second be R2.
Now, you know that resistance = (resistivity* length)/ cross-sectional area.
So, R/ R2 = [(Resistivity* L1)/ A1]/[(Resistivity* L2)/A2]
for the same material the resistivity is the constant. So, cancel resistivity and cross-multiple and you get R/ R2= L1*A2/ L2*A1
once you understand this ratio you can start from here for all such question in future. Just put the values in now and remember to convert all the units for radius to meter. Area of cross-section Pi r^2 but again, you can cancel out pi for both. So it's just length*radius^2
you get, R/R2 = 8* (5*10^-4)^2/ 2* (2.5*10^-4)^2
that is, R/R2 = 16
so R2 = R/16

It might seem a little confusing but just write it out as you go and it gets simpler. Hope this helps!

Got it very well!!!!....thanks alot!!!!!!......wasn't confusing even a bit ....!!!
 
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9702/12_2011 Oct\Nov paper 12
Can Somebody PLEASE Give Reasons and concepts to the answers....................pls..pls..pls...
I'm in a hurry...............Thanks
 

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Tkp

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the distance of block from the level of head of the man is 18m and when to level the head with the block man travels 9m upwards and block 9m downwards.
tension acts towards the pulley and the weight acts opposite to tension i.e downwards.
force acting in the direction of acceleration is considered more in magnitude than the force in opposite direction

for the man
T-80g=80a equation 1
120g-T=120a equation2
adding both equations we get
40g=200a
a=0.2g (given g as 10ms^-2)
a=2ms^-2
v^2=2as (as initial velocity of both the objects is zero )
v^2=2*2*9
v^2=36
v=6ms^-1

for question 24
stress=F/A
F=stress*A
F=(stress/4)*4A (area would be 4A as (2D)^2 is four times greater than (d)^2 for e.g (2*2)^2=16 x(2)^2=4 see
stress should be divided by 4 as the examiner states that load is the same which means the force applied is the same
so to get F 4 must be divided by 4 to get 1 ok!!

stress is directly proportional to strain
so if stress is stress/4 .strain would also be strain/4
stress/4=x/3l.
stress=x/l main formula
coefficient of x would be such that it equals 3/4 because only then we would get our formula u see coefficients will change but the formula wont ok
dnt we need to find the radius here?
 

Tkp

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guys please help me to solve this questions 10,19,21,23,26,28,29,31,34,36,37,38
 

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Hiii can some body please help me
In paper 5 phy ! For drawing graph in question it is said to include error bars what does that mean???
We calculate the uncertanties in part 1 !!
What to do for value we plot in the graph?
 
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