A2 Physics | Post your doubts here

[angelgirl:)]

please tell how to draw the diodes in june 2010 paper 41 Q6

i no i didnt drew it nicely...but hope it helps u...

 

[miss irfan]

i no i didnt drew it nicely...but hope it helps u...

thankyou, i just had to confirm

 

[SkyPilotage]

apart from bridges where else is strain gauge used?

walls, aircrafts, buildings, wings, doors, anything which has a small change in length over a unit of time..to be observed!

 

[smzimran]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_4.pdf
Q1 (b) (i)
Anyone ?

 

[ousamah112]

i knw its a silly question
answers should be in two or three significant figures??? bcoz some answers in ms are in two sf while some are in three.....!!

 

[SkyPilotage]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
Q1 (b) (i)
Anyone ?

I talk about this in details with alot of websites in previous pages but you werent here I guess
When its on the axis of rotation i.e at poles , radius is zero so centriopetal force is zero, so the normal reaction is equal to weight.
** Extra information:-
When its at the equator, the centripetal force which is the resultant of the gravitational and normal force , is towards the centre.
Meaning gravitational force is greater than normal reaction force.

 

[SkyPilotage]

i knw its a silly question
answers should be in two or three significant figures??? bcoz some answers in ms are in two sf while some are in three.....!!

You use 3s.f if not stated in the question, but if values used in the question are 2 s.f then use 2 s.f in ur answer!

 

[ousamah112]

I talk about this in details with alot of websites in previous pages but you werent here I guess
When its on the axis of rotation i.e at poles , radius is zero so centriopetal force is zero, so the normal reaction is equal to weight.
** Extra information:-
When its at the equator, the centripetal force which is the resultant of the gravitational and normal force is towards the centre.
Meaning gravitational force is greater than normal reaction force.

how radius is zero at centre???

 

[smzimran]

I talk about this in details with alot of websites in previous pages but you werent here I guess
When its on the axis of rotation i.e at poles , radius is zero so centriopetal force is zero, so the normal reaction is equal to weight.
** Extra information:-
When its at the equator, the centripetal force which is the resultant of the gravitational and normal force is towards the centre.
Meaning gravitational force is greater than normal reaction force.

At the poles, radius is zero ???

 

[SkyPilotage]

how radius is zero at centre???

Radius of rotation, not radius of the planet!
AT the poles, you are the centre of rotation i.e radius is zero!

 

[ousamah112]

You use 3s.f if not stated in the question, but if values used in the question are 2 s.f then use 2 s.f in ur answer!

but in some questions values are to 3 sf but still the ans is in 2sf

 

[SkyPilotage]

but in some questions values are to 3 sf but still the ans is in 2sf

no, there should be atleast one value with 2 s.f , you sue the smallest available!

 

[SkyPilotage]

At the poles, radius is zero ???

Did you get it?
If you are the centre of rotation, how far are you from the centre? zero! so centripetal force is zero hence weight = normal force.

 

[aliya_zad]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s05_qp_4.pdf

Q6 part b) how do u draw the graph for emf versus time!

 

[angelgirl:)]

m/j 2010 ppr42 Q no.4c...? plz answer with explaination...

 

[ousamah112]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s05_qp_4.pdf

Q6 part b) how do u draw the graph for emf versus time!

see post #678 on page 34

 

[smzimran]

Did you get it?
If you are the centre of rotation, how far are you from the centre? zero! so centripetal force is zero hence weight = normal force.

Thanks

 

[OrlandoBloom<3]

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 !

 

[ousamah112]

m/j 2010 ppr42 Q no.4c...? plz answer with explaination...

first find the total potential at midpoint due to both charges (A n B) i.e at 6micrometre.
then find potential due to both charges at P
so change in potential= potential at P -potential at midpoint
and work done= change in potential x charge of electron...

 

[leadingguy]

m/j 2010 ppr42 Q no.4c...? plz answer with explaination...

well always remember that when ever we are asked to calculate wrk done

we need to calculate it with this formula ..... w = ∆Vq
(∆v is the change in pottential which the charge faces in traveling frm point to another point )
whereas the q represnts the charge (1.60*10^-19)

so frst calculate the pottential "V" at the mid point of line AB where the charge is currentlY situated.

the formula fr calculating V is V = KQ/r remember the charge at mid point is experiencing to pottentials 1 frm rite hand side charge 2nd frm left hand size charge.

we know that the charge is of same coloumbs and is 6.0μm away frm each point A and B


so we need to twice the pottential i.e V = (KQ/r) ×2 frst part completed

Now the pottential where the charge is to be taken (at point P )

point p is experiencing pottentail frm point A calculate it by V1 = KQ/r

point P is also experiencing pottential frm Point B calculate it by V2 = KQ/r

remember r represents the distance from A to P (3.0μm)

r represents the distance frm B to P (9.0μm) \


now aftr calculating these voltages add them to get V at point P 2nd part completed

now that we have to subtract these two potentials at point p and potential at midpoint to obtain ∆V
do that and the ans ∆V wil be malltiplied by "q" to get wrk done


HOPE U gt what I said????