ALL CIE PHYSICS DOUBTS HERE !!!

[TJ-56]

I drew a light source with a cardboard with a slit in it to act as the place where the diffraction will occur. Then a screen where the diffracted light will be seen.
In your case I dont think we can use an eye as a suitable detection media. We will need to use some sort of equipment in order for us to interpret/observe the diffracted light.

For the second one i simply drew a sound source with a small obstacle and had a microphone and cathode ray oscilloscope on the other end to see the wave pattern on the CRO.
I think your second one is fine.

You may want to check the mark scheme to see what cambridge are looking for exactly in your answers

ok thanx, just wanted to make sure

[lana]

may june 03 question 3 b(i) please
thanks =]

[$!$RatJumper$!$]

may june 03 question 3 b(i) please
thanks =]

The circumference of a circle is known to be 2*pie*r OR d*pie

In this case we will use d*pie since they have given us the diameter as 3.
So the actual circumference is 3*pie. But then it says that it is turned by 6.5 degrees. Now to find the amount it moved in terms of its circumference, we use: (6.5/360)*(3*pie) = 0.17 cm

P.S we divided it by 360 because we need to express 6.5 in terms of its angle of deflection.

Hope that helped

[lana]

thank youu for your help =]
+rep

[$!$RatJumper$!$]

could someone please help me with S09 P21 Q5b and S08 P2 Q6b

thankx

[astarmathsandphysics]

s08 p2

a) s1 closed so no current or power

b) s1 and s2 closed so voltage acros A is 240 so power is 1.5KW

c)1.5W+1.5KW since voltage across ABC is 240V, 0V and 240V resp

d)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W

e)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W +1.5KW in C=2.25KW

Can you upload S09 P21 Q5b cos I don't have it

[$!$RatJumper$!$]

s09 p2
a)s1 closed no no current or power
b) s1 and s2 closed so voltage acros A is 240 so power is 1,5KW
c)1.5W+1.5KW since voltage across ABC is 240V, 0V and 240V resp
d)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W
e)voltage across A and B is 120V so using P=V^2/R power in each element is 1/4 of 1.5KW=375 W and total power is 750W +1.5KW in C=2.25KW

Can you upload S09 P21 Q5b cos I dont have it

thankx for that

ok iv attached s09 here..

and could you also help me with s10 P22 Q4c

[ashish]

NOV 2008 p1 no.11

i have done it but a bit like mechanic question
here it is

8.0a = T
2.0a=20-T-6

after solving it i got 1.4ms^-2

there is another method of working it! can anyone please help me with the other method !

[$!$RatJumper$!$]

NOV 2008 p1 no.11

i have done it but a bit like mechanic question
here it is

8.0a = T
2.0a=20-T-6

after solving it i got 1.4ms^-2

there is another method of working it! can anyone please help me with the other method !

This is how i would have done it. But i do not get the answer as 1.4ms^-2. Can someone tell me what im doing wrong. Thankx

The 2kg mass hanging would be the force that pulls the 8kg mass. Since it is hanging gravity would act on it making it have a force of 2*9.81 = 19.62N

Since frictional forces are 6N, we do, 19.62-6 = 13.62N

Thats the resultant force. Now we use F=ma to get its acceleration.

13.62/8 = 1.7 ms^-2

[Deadly_king]

This is how i would have done it. But i do not get the answer as 1.4ms^-2. Can someone tell me what im doing wrong. Thankx

The 2kg mass hanging would be the force that pulls the 8kg mass. Since it is hanging gravity would act on it making it have a force of 2*9.81 = 19.62N

Since frictional forces are 6N, we do, 19.62-6 = 13.62N

Thats the resultant force. Now we use F=ma to get its acceleration.

13.62/8 = 1.7 ms^-2

You cannot take the acceleration to be 9.81 since that would not be the case due to the tension in the string which is linked to the other mass.

Using Newton's 2nd law of motion for the 2kg mass => W - T = 2a
W : weight of the object.(2 x 9.81)

NOTE : Newton's 2nd law applies only to the resultant force. Here the resultant is W - T.

So you have to do it the way ashish did it. You need to find two equations involving the two variables you're to find; acceleration and tension.

Then you need to solve them simultaneously to get the required solution which is a = 1.4ms^-2.

[Deadly_king]

and could you also help me with s10 P22 Q4c

Jun 10 p22

4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.

Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4

Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m

V = f x lambda

Therefore f = v/lambda = 330/1.8 = 180 Hz

Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)

Hope it helps

[Arthur Bon Zavi]

+rep Deadly_king.

[Deadly_king]

+rep Deadly_king.

Hehe...........thank you Mr the Ancestor

[Dania]

Refer to the attachments
Look at the paper, for the full question. [Question no. 4]
Thanks in advance.

[$!$RatJumper$!$]

Jun 10 p22

4.(c) Sound with lowest frequency is produced => An antinode is found at the end of the tube and it is the only one.

Hence there should be a node at the piston. Distance between a node and the consecutive antinode is given to be lamda/4

Lambda/4 = 45 cm -----> lambda is given to be 180cm = 1.8m

V = f x lambda

Therefore f = v/lambda = 330/1.8 = 180 Hz

Frequency is given to the nearest tenth since v = 330 and lambda=1.8 (nearest 10)

Hope it helps

Thank you DK

Another one, please explain
S09 P21 Q5b
W09 P22 Q5b

Thank you