Physics: Post your doubts here!

[Soldier313]

10 b)

View attachment 24228
since we only want the current to flow from positive to negative therefore we use the diode to forward bias he circuit

Thanx bro
But um the ms says that there should be a second diode in the circuit, i am confused as to where do i have to place it.

 

[Soldier313]

Aoa wr wb

Can someone please draw for me the diagrams for 7b i and 7b ii ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_41.pdf

Thank you

 

[Tabi Sheikh]

Thanx bro
But um the ms says that there should be a second diode in the circuit, i am confused as to where do i have to place it.

u can put another diode to protect the relay its always parallel to the relay and is reverse biased always it used for protection only .... but not sure about it whether ms talk abut the same BTW Ur welcome

 

[Tabi Sheikh]

i

Aoa wr wb

Can someone please draw for me the diagrams for 7b i and 7b ii ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_41.pdf

Thank you

Diode A will act as half wave rectifier as it will on be active while the voltage is positives when voltage is negative it will be zero
second part will be the other way round when negative voltage then positive voltage but when positive i will be zero
in diagram i have just pointed out the points where it will be zero the others on both graphs will be positive Curve .
BTW u can switch the graphs .....

 

[snowbrood]

wooaah fancy working out to be done in a paper where u have got just one hour -_-
thanks btw,... i do get it now but how r v supposed to think it over in the paper is what bugs me :O

yea u are right lets hope the best remember dont panic i got a grade B just because of panicking

 

[daredevil]

yea u are right lets hope the best remember dont panic i got a grade B just because of panicking

oh okay thanks for the heads up ! 0-0
Best of Luck -.-

 

[Tabi Sheikh]

yea u are right lets hope the best remember dont panic i got a grade B just because of panicking

and don't be too slow like me

 

[daredevil]

please explain Q4b all parts from this paper:
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

 

[Tabi Sheikh]

please explain Q4b all parts from this paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf

q4 b i) The area between both the graphs..... is the energy Es

 

[Tabi Sheikh]

reason that is the energy lost and is clearly visible

 

[Soldier313]

u can put another diode to protect the relay its always parallel to the relay and is reverse biased always it used for protection only .... but not sure about it whether ms talk abut the same BTW Ur welcome

i
Diode A will act as half wave rectifier as it will on be active while the voltage is positives when voltage is negative it will be zero
second part will be the other way round when negative voltage then positive voltage but when positive i will be zero
in diagram i have just pointed out the points where it will be zero the others on both graphs will be positive Curve .
BTW u can switch the graphs .....

Thank you so so much bro!
just a clarification, for the second half wave rectification diagram for diode B, the diagram will be like this yeah?
sorry it's kinda rough :/

 

[sma786]

AS physics formulae pleasE?!

 

[Tabi Sheikh]

Thank you so so much bro!
just a clarification, for the second half wave rectification diagram for diode B, the diagram will be like this yeah?
sorry it's kinda rough :/

View attachment 24265

yeah its true

 

[snowbrood]

can anyone solve q4b(iii) of november 2012 paper 21 calculate the wavelength of another part of visible spectrum that gives a maximum for a different order at the same as in (ii)
wavelength 625nm in part (ii) order is 2 it is second order maximum at an angle 61 to the incident direction number of lines per metre are 7*10^5 can anyone solve the question using this data??

 

[snowbrood]

yeah its true

can u solve ?

 

[Tabi Sheikh]

can u solve ?

i will give it a try

 

[Shreeram]

since he measured the distance from the top f ball but when the ball hit the table only the bottom of the ball hits the table therefore the y intercept of graph is the diameter of the ball that is approximately 1.75cm according to my line.

but how do u calculate the acceleration

 

[Tabi Sheikh]

can u solve ?

bro its just trial and error put in the different whole number values for n and use d*sin(61)=n*wavelength where d is carried forward from previous part.

 

[snowbrood]

bro its just trial and error put in the different whole number values for n and use d*sin(61)=n*wavelength where d is carried forward from previous part.

why 1250 and 312.5 are out of visible i dont get it

 

[Tabi Sheikh]

Shreeram the graph is distance - time^2 graph there fore the gradient of the graph will be the acceleration i.e. m/s^2