- Messages
- 532
- Reaction score
- 151
- Points
- 53
ok...im doing them from my sir's compiled booklet . he has removed all the questions which arnt in our sylabus now.
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
ok...im doing them from my sir's compiled booklet . he has removed all the questions which arnt in our sylabus now.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdfoh 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...
ok dude simple thingsassalamoalaikum wr wb!
i need help with the virtual earth part in the inverting amplifier...
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdfok 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
direction is upward as the phase angle is below 180 degree hence direction is the same as if piston moves upwardshttp://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 ?????
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?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
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...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
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...
ya u can do that as well ... i guess dats the real method...but i was showing u how i arrieved at my answer...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??
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 !
ppl plz help for this Q?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;m/j 2009 Q4b...? help needed?
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 .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..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??
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 = 11mmURGENT
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
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=11cmURGENT
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
Q 3(ii) Y Does Amplitude is 11mm Plz Help!
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now