Physics: Post your doubts here!

[goodluckayesha]

my labeling is correct for showing which one is the width?
becuase if the its correct then if u increase the slit seperation than the distance between the slits( slit width) decreases isnt it? so

yh label is correct n no width of slit n slit separation have no connection
but ur using rong words distance between the slits is slit separation

 

[1357913579]

yh label is correct n no width of slit n slit separation have no connection
but ur using rong words distance between the slits is slit separation

ok thanks

 

[1357913579]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_23.pdf
can someone please sketch this
6(b)

 

[cyclone94]

gary221 or Dug........any1 plzzz help wid dis asap...
m/j-11-43 qs 12

 

[goodluckayesha]

ok thanks

ur welcum did u understand??

 

[Dug]

gary221 or Dug........any1 plzzz help wid dis asap...
m/j-11-43 qs 12

a)
i) Signal-to-noise ratio = 10lg(Signal power/Noise power)
Let minimum signal power = P
24 = 10lg(P/(5.6x10^-19)
P = 1.4x10^-16 W

ii) Attenuation per length = (1/L) 10lg(P-in/P-out)
1.9 = 1/L 10lg(3.5x10^-3/1.4x10^-6)
L = 71 km

 

[Warrior66]

This type of question has always been a disaster for me
Can you write all the possible effects of changes to the apparatus? (The ones you said, and also other changes like moving the screen further, changing wavelength, changing slit separation, etc.)

Well for the most basic relationships..you can use the formula pi = ax/D.
like for example, increasing the distance 'D' between the screen & the double slits whilst keeping everything same..the fringe separation 'x' increases.

 

[Sapphires]

Please help me with these questions - 6c (ii) Please draw the graph. and question 5b (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_4.pdf gary221 or Dug please. Thank you!

 

[goodluckayesha]

Please help me with these questions - 6c (ii) Please draw the graph. and question 5b (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_4.pdf gary221 or Dug please. Thank you!

u dint lyk my explanation??

 

[Sapphires]

u dint lyk my explanation??

Nooo, That's so not true!
I thought you only answer AS questions, not A2.
You can answer it, if you don't mind!

 

[goodluckayesha]

Nooo, That's so not true!
I thought you only answer AS questions, not A2.
You can answer it, if you don't mind!

oops!! srry I dint notice u asked an A2 question
I m doing AS only for now A2 later

 

[gary221]

Please help me with these questions - 6c (ii) Please draw the graph. and question 5b (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_4.pdf gary221 or Dug please. Thank you!

5 b ii) 1 In the equation, 1/2mv^2 = qV
mass n charge are constant, so velocity is directly proportional to p.d.
Reducing the pd ---> velocity of the particle will also decrease.
So, the deflection will b larger.

2 the magnetic field is providing the centripetal force in this case. So, when mag. field strength increases, the force exerted by it also increases ( F = BIl)
So, since the centripetal force now increases, the deflection will again b larger!

Hope u gt it!

 

[Sapphires]

5 b ii) 1 In the equation, 1/2mv^2 = qV
mass n charge are constant, so velocity is directly proportional to p.d.
Reducing the pd ---> velocity of the particle will also decrease.
So, the deflection will b larger.

2 the magnetic field is providing the centripetal force in this case. So, when mag. field strength increases, the force exerted by it also increases ( F = BIl)
So, since the centripetal force now increases, the deflection will again b larger!

Hope u gt it!

Thank you so much. Yes, I get it.
I needed help with 6c ii graph as well, please?

 

[gary221]

Please help me with these questions - 6c (ii) Please draw the graph. and question 5b (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_4.pdf gary221 or Dug please. Thank you!

 

[Sapphires]

View attachment 25122

Thank you, thank you so much.

 

[SalmanslK]

HELP !!

 

[gary221]

HELP !!

paper link??

 

[SalmanslK]

paper link??

MJ 2010 42 Q1

 

[gary221]

MJ 2010 42 Q1

2R above earth's surface ---->3R from the centre of the earth.
So, potential at 3R from center of the earth = 2.1 * 10^7 J/kg

v knw tht change in kinetic energy = change in potential energy
so, 1/2 * m * v^2 = mgh ----> φ * m
1/2 v^2 = φ
v^2 = (2.1 * 10^7) * 2
v = 6.5 * 10^3 m/s

Hope u gt it!

 

[SalmanslK]

2R above earth's surface ---->3R from the centre of the earth.
So, potential at 3R from center of the earth = 2.1 * 10^7 J/kg

v knw tht change in kinetic energy = change in potential energy
so, 1/2 * m * v^2 = mgh ----> φ * m
1/2 v^2 = φ
v^2 = (2.1 * 10^7) * 2
v = 6.5 * 10^3 m/s

Hope u gt it!

oh !! just got confused with the 2R and 3R thing..Tricky CIE...silly ME