• 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!

Messages
258
Reaction score
99
Points
28
Messages
373
Reaction score
33
Points
28
honey are you ok? there is still no 4b(iv) :confused:

yeh honey im ok. the problem here is that even though i changed the ppr from w10 to w11 it was still opening the previous paper. just go to the adressbar and change it to w11
 
Messages
224
Reaction score
116
Points
53
for question 2 you have
accn = 2.5 x10^15
time = 2.4 x10^-9
intial speed= 5 x10^7

and u know that the electron is launched halfway between the plates so the distance between it and the plate is 1.5/2 =0.75

using equation of motion= s= ut +(at^2)/2 you get s to be 0.1272 which is much less than 0.75 so no it does not hit it

for question 4 ur diagram is probably wrong draw it again
I drew again. didn't get anywhere close to 41-48 :s cud u draw it for me?
 
Messages
83
Reaction score
7
Points
18
Ok View attachment 10429

as you can see from the diagram I rearranged the forces but didn't change the direction
the angle between T1 and T2 is 90 because T1 is at an angle of 50 to horizontal and T2 at 40 to horizontal
and the angle between 7.5N and t=T1 is 40 because it is vertically downwards so is perpindicular to horizontal

to find T1 and T2 use trignometery

hope it helped


Yes i got it now :)
Tysm :D
 
Messages
258
Reaction score
99
Points
28
Jeez
i am so sorrry!
q2 it is!

ok so using the equation v^2=u^2 +2as

for intial state (v would be zero and zero squared is zero so i will leave it out to make it easier to calculate)
so distance will be -u^2/2a =s

for half speed u= 0.5u therefore -> -(0.5u)^2/2a =s' (s' is the unknown distance)
simplyfy equation u get -1/4 *u^2 which is basically -u^2/4/2a which gives -u^2/8a =s'

when ever you have a situation like the above divide the thing you are trying to find by original
ie s'/s = (-u^2/2a) /(-u^2/8a) which gives s'/s =2/8 so s'= 0.25 s

hope it helped
 
Messages
55
Reaction score
32
Points
28
Can someone please explain how to do q30
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_13.pdf

because that is not on the AS syllabus

I have found a site to help me but I don't get anything from it
http://www.physicsclassroom.com/Class/sound/u11l5d.cfm equation for closed end tube: f=v/4l
http://www.physicsclassroom.com/class/sound/u11l5c.cfm equation for open end tubes: f=v/2l

does the frequency have to be like greater than 0.5 for stationary waves??

They wouldn't set a question outside of the syllabus. You just need to apply your knowledge to the situation.
What you need to do is find out if you can draw a stationary wave in both the tubes P and Q. If you are able to draw a stationary wave in a tube, then a stationary wave can be formed in it.
Closed end of a pipe is always a node, and the open end is always an anti-node.
For P you have an anti-node at the open end, and a node at the closed end. You are able to draw a stationary wave 35cm long with 4 nodes, and 4 anti-nodes so P works.
For Q, both ends have an anti-node. You are able to draw a stationary wave 50cm long with 5 nodes, and 6 anti-nodes, so Q works too.
Final answer is A!
 
Messages
84
Reaction score
12
Points
8
Hey people i have an important question.. Does number of significant figures matter in phy paper 2? If answer scheme gives 2sf for final answer but i gave in 3sf, do i lose any mark?
 
Messages
960
Reaction score
3,500
Points
253
i just read something about simple harmonic motion and was wondering is that covered in our AS Physics syllabus?
 
Top