A2 Physics | Post your doubts here

[angelgirl:)]

im doing them from my sir's compiled booklet . he has removed all the questions which arnt in our sylabus now.

ok...

 

[leadingguy]

oh God u r doing those years...now so many changes r there in our syllabus...some topics r removed and some r added...so dont depent so much in these years...

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_43.pdf

question 3 part bi. 2 )) why is the direction upwards?????

question 7 part b )
can u please help ?????

 

[wahab305]

assalamoalaikum wr wb!

i need help with the virtual earth part in the inverting amplifier...

ok dude simple things
in non inverting and inverting amp one of the output is linked with earth line hence one of them has potential zero.now according to conservation of energy the the output must be less than or equal to the voltage supplied if not saturation occurs and same voltage comes and as the supplied voltage.as the open loop gain 10^5 is very high hence for the other output corresponding to the one at zero potential(virtual earth) the value of potential must be small hence saturation is avoided as it is so small it is called virtual earth

 

[leadingguy]

ok dude simple things
in non inverting and inverting amp one of the output is linked with earth line hence one of them has potential zero.now according to conservation of energy the the output must be less than or equal to the voltage supplied if not saturation occurs and same voltage comes and as the supplied voltage.as the open loop gain 10^5 is very high hence for the other output corresponding to the one at zero potential(virtual earth) the value of potential must be small hence saturation is avoided as it is so small it is called virtual earth

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_43.pdf

question 3 part bi. 2 )) why is the direction upwards?????

question 7 part b )
can u please?????

 

[wahab305]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_43.pdf

question 3 part bi. 2 )) why is the direction upwards?????

question 7 part b )
can u please help ?????

direction is upward as the phase angle is below 180 degree hence direction is the same as if piston moves upwards
yar for question 7 simple thing electric field force and magnetic field force oppose each other okay...and relation is v=e/b doesnot include mass and charge only on velocity as magnetic force increase with velocity...and yes alpha particles are positive keep in mind now index finger will be towards the direction of velocity

 

[angelicsuccubus]

iF u checked and it does not show then thats great apart there are some questions in papst paers regarding this.. but yes since last few years there are no questions... so pray fr the tradition to be maintained

trust me... they will not give anything that will ask us to calculate the bandwidth, if they do .. just blame it on candidate number 0609, okay?

 

[angelgirl:)]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_43.pdf

question 3 part bi. 2 )) why is the direction upwards?????

question 7 part b )
can u please help ?????

Q3...direction is upword becouse the piston was initiallly at AB position. which is -4cm above the equlibrium position..and we got the value 2cm from the calculation...
-4=-4cos220t
since we r using cos it means it started from the max...amplitude..so its moving 120' phase difference towords equlibrium...
this is how i assume it...but not sure of the answer...

for Q7...since the particle has the same speed as the e- and moving in the same path as the e- ...it means its mocving through the equal magnetic and electric fields...so it experiances the upord force by magnetic field and a downword force be electric field...direction of fields u have to find by felimings left hand rule...

hope u get it...

 

[leadingguy]

direction is upward as the phase angle is below 180 degree hence direction is the same as if piston moves upwards
yar for question 7 simple thing electric field force and magnetic field force oppose each other okay...and relation is v=e/b doesnot include mass and charge only on velocity as magnetic force increase with velocity...and yes alpha particles are positive keep in mind now index finger will be towards the direction of velocity

thanx alot fr question 7 but fr question 3 we always see in the graph that til 90 amplitude increases, means it wil move upwards then aftr 90 til 180 amplitude decreases so wil move downwards???? i am a little confused at this can U explain with elaboration ??

 

[leadingguy]

Q3...direction is upword becouse the piston was initiallly at AB position. which is -4cm above the equlibrium position..and we got the value 2cm from the calculation...
-4=-4cos220t
since we r using cos it means it started from the max...amplitude..so its moving 120' phase difference towords equlibrium...
this is how i assume it...but not sure of the answer...

for Q7...since the particle has the same speed as the e- and moving in the same path as the e- ...it means its mocving through the equal magnetic and electric fields...so it experiances the upord force by magnetic field and a downword force be electric field...direction of fields u have to find by felimings left hand rule...

hope u get it...

thanx alot fr Ur response i understood question 7. thanx alot.
fr question 3 u reached the final answer correctly, yes it is upwards but the method u used is too time consuming and alittle bit tricky too.


isn't it the simple use of sinosidal graph??? to predict the direction wid use of angle??

 

[angelgirl:)]

thanx alot fr Ur response i understood question 7. thanx alot.
fr question 3 u reached the final answer correctly, yes it is upwards but the method u used is too time consuming and alittle bit tricky too.


isn't it the simple use of sinosidal graph??? to predict the direction wid use of angle??

ya u can do that as well ... i guess dats the real method...but i was showing u how i arrieved at my answer...

 

[Jonathan127]

Hey you guys ! Can anyone please summarize the hand rules used to determine the field direction , current direction and motion using the hand rules :/ I am confusedddd ! Please include the rules for current carrying conductor , electromagnetism for charge moving in a uniform feild .. Basically all the situations where we need to use the hand rules Thanks alottttttttt !

Sorry for the late reply!
This is a really big question. Like, there are two hand rules.
The left hand rule is to predict the direction of a force on a current carrying conductor in a magnetic field. In this one, when the conductor carrying a current is placed in a uniform magnetic field, a field is also created from the current carrying wire, and so, there is a resultant force on the wire.

The right hand rule is to predict the direction of induced emfs and currents in a straight conductor. So, this is the one where the conductor is cutting the flux in the conductor, and a emf is generated. It is where the formula E= -d(flux) / dt is used and proves Faraday's Law.

Sorry for the IGCSE type explanation, but the question is very general.

 

[angelgirl:)]

m/j 2009 Q4b...? help needed?

 

[angelgirl:)]

m/j 2009 Q4b...? help needed?

ppl plz help for this Q?

 

[leadingguy]

m/j 2009 Q4b...? help needed?

yah I too ponderd over this question yesterday.. DID not get a satisfactry answer but wat I get was;

we have already find out that distance r = r.sinwt

now fr finding velocity we can convert the expresion into this V = r.w.coswt (by differentiating above eq.)

and fr finding acceleration we can convert it into a = - r.w^2.sinwt (again by differentiating the above one)


now max. accelleration is at 90 degrees. so a = - r.w^2.sin(90 = a = - r.w^2 proven


in paer there is no need to show the whole procedure jxt state that

r = r.sinwt

a = - r.w^2.sinwt

acceleration max. so sin90 and therefore a = - r.w^2 proven

what do u say??

 

[grumpy]

what is the relationship between attenuation and frequency of the electromagnetic wave
qualitatively

 

[Jonathan127]

yah I too ponderd over this question yesterday.. DID not get a satisfactry answer but wat I get was;

we have already find out that distance r = r.sinwt

now fr finding velocity we can convert the expresion into this V = r.w.coswt (by differentiating above eq.)

and fr finding acceleration we can convert it into a = - r.w^2.sinwt (again by differentiating the above one)


now max. accelleration is at 90 degrees. so a = - r.w^2.sin(90 = a = - r.w^2 proven


in paer there is no need to show the whole procedure jxt state that

r = r.sinwt

a = - r.w^2.sinwt

acceleration max. so sin90 and therefore a = - r.w^2 proven

what do u say??

Wow, such a complicated answer but its the only one that makes sense so far. I'm sure there is a more simple answer since its only 2 marks. I did that paper a few days ago, and then gave up on it and hoped it wouldn't be in the exam .

 

[ABDSyed]

URGENT
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_4.pdf

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Q 3(ii) Y Does Amplitude is 11mm Plz Help!​

 

[angelgirl:)]

yah I too ponderd over this question yesterday.. DID not get a satisfactry answer but wat I get was;

we have already find out that distance r = r.sinwt

now fr finding velocity we can convert the expresion into this V = r.w.coswt (by differentiating above eq.)

and fr finding acceleration we can convert it into a = - r.w^2.sinwt (again by differentiating the above one)


now max. accelleration is at 90 degrees. so a = - r.w^2.sin(90 = a = - r.w^2 proven


in paer there is no need to show the whole procedure jxt state that

r = r.sinwt

a = - r.w^2.sinwt

acceleration max. so sin90 and therefore a = - r.w^2 proven

what do u say??

ho i did is....SQrsinwt...from the aii answer..
which can be written as x=x.sinwt....
v=dx/dt=x.wcoswt
a=-xw^2
a is proportional to -x...means S.H.M....by differentiation...i guess i copied from the ms...i didnt remember...
a=d^2x/dt^2 = -x.w^2sinwt.....

 

[Mobeen]

URGENT
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf

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Q 3(ii) Y Does Amplitude is 11mm Plz Help!​

in the first line the question says that the needle travels a total distance of 22mm . amplitude is half the distance traveled in one cycle . = 22/2 = 11mm

 

[angelgirl:)]

URGENT
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf

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Q 3(ii) Y Does Amplitude is 11mm Plz Help!​

the needle oscillate through a total distance of 22cm...means at equalibrium it will be at 22/2=11cm