• We need your support!

    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)

Physics: Post your doubts here!

G

geek101

Messages
398
Reaction score
233
Points
53
leosco1995 said:
The question says "describe what is seen..", yes you do talk about the smoke particles, but since they appear as tiny, white specks of light you should mention that in the answer to that question since it's an observation. Likewise you would mention that the motion of these specks of light is haphazard since that is also an observation.
Click to expand...

thnx got it...and i also read that the smoke particles are not what you see but the light reflected from them....thnx sista!
 
iKhaled

iKhaled

Messages
869
Reaction score
374
Points
73
geek101 said:
first here is how you convert the degrees and radians...
radians to degrees : multiply by 180/π
degrees to radians: multiply by π/180

Phase difference is simply is the fraction of the cycle (the angle) by which one wave moves behind the other.
you can calculate is by using this equation: (with degrees)
360 (x/λ) where x is the difference in the distance between the two waves...
or 360(t/T) where the waves are out of step by time t.
use 2π for radians, since 360 = 2π

For coherent waves....they have a constant phase difference...this is usually the case with waves fron the same source. It does not matter whether the waves are in phase ....but coherent waves must have a constant phase difference...which means that the difference between them is the same throughout. They must also have the same wavelength and the same frequency.
this is quite brief..but hope it helps.
Click to expand...
after readin this i felt i know nothing about waves LOL!
 
hmlahori

hmlahori

Messages
318
Reaction score
328
Points
73
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_21.pdf
Q 2(b) and Q 6(a ii) Can someone show exactly how these both are drawn? i have read the mark scheme and examiner report but i might be thinking it is drawn some other way entirely so just to verify my own drawings.

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_21.pdf
Q 7(b) in this i also want to confirm if my drawing is correct so if someone else can show the correct one..
 
Saad (سعد)

Saad (سعد)

Messages
224
Reaction score
183
Points
53
Okay...

In the syllabus, we have:

(e) describe the effect of a uniform electric field on the motion of charged particles.

And in the notes, I have things like Equipotential Surface, Potential Gradient, and 'Electric field lines must meet the surface at right angles'.

... What in the realm of Physics is this?
 
G

geek101

Messages
398
Reaction score
233
Points
53
iKhaled said:
sorry that's the correct paper

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_2.pdf
Click to expand...
well 3 b is :
1) the sum of the forces in any direction must be zero
2) the sum of the moments of the force about a point must be zero
and if you mean 3 c ii....
1) quite simple...
M abut a point = force x perpendicular distance
= (30/100) x 150
= 45 Nm
2) it must be equal to the moments so it is also 45N
3) since the torque is 45Nm
and torque = one force x perpendicular distance
and we know that the perpendicular distance is 12/100 = 0.12 m
and the value of total torque is 45 Nm
so just substitute...
therefore Force = torque / perpendicular distance
= 45 / 0.12 = 375 N
The T is the mark scheme is tension which is the force caused by the string.
 
iKhaled

iKhaled

Messages
869
Reaction score
374
Points
73
geek101 said:
well 3 b is :
1) the sum of the forces in any direction must be zero
2) the sum of the moments of the force about a point must be zero
and if you mean 3 c ii....
1) quite simple...
M abut a point = force x perpendicular distance
= (30/100) x 150
= 45 Nm
2) it must be equal to the moments so it is also 45N
3) since the torque is 45Nm
and torque = one force x perpendicular distance
and we know that the perpendicular distance is 12/100 = 0.12 m
and the value of total torque is 45 Nm
so just substitute...
therefore Force = torque / perpendicular distance
= 45 / 0.12 = 375 N
The T is the mark scheme is tension which is the force caused by the string.
Click to expand...
sorry for wasting ur time, u explained the wrong question i meant 4 b iii..i am so sorry bro :/ thank u though .
 
You must log in or register to reply here.
Top