ALL CIE PHYSICS DOUBTS HERE !!!

[Ghost Of Highbury]

Winter 2009, paper 22, Q1)c)ii)
 (CIE AS level Physics)

Please explain the steps to determine the no. of s.f in 'g' and the uncertainty clearly.
If any rules are to be followed to determine them, please state them too. Thanks a lot

[polltery]

Draw tangent to the curve at 0.4s, calculate the gradient of the tangent.

THANKS ALOT  MATE!

[elemis]

Winter 2009, paper 22, Q1)c)ii)
 (CIE AS level Physics)

Please explain the steps to determine the no. of s.f in 'g' and the uncertainty clearly.
If any rules are to be followed to determine them, please state them too. Thanks a lot

Hang on.

[Ghost Of Highbury]

Hang on.

Sure thing

[elemis]

First calculate the value of g using the equation, which is 9.75064.... The equate the fractional uncertainties of the left hand side of the equation to the right hand side of the equation.

Since we are calculating the absolute uncertainty of g we let it be delta x :





Looking at the initial data we can see that the absolute uncertainties we used were accurate to only 1 s.f. Hence, we express our calculated absolute uncertainty (delta x) to a maximum of 1 s.f.

Look at our value of g and comparing it to the original data we can see each value was accurate to a minimum of 2 s.f. Hence we express g to a max of 2 s.f.

[$H00t!N& $t@r]

need help with  O/N 2010 variant 23 q6b (ii) 

[Ghost Of Highbury]

First calculate the value of g using the equation, which is 9.75064.... The equate the fractional uncertainties of the left hand side of the equation to the right hand side of the equation.

Since we are calculating the absolute uncertainty of g we let it be delta x :





Looking at the initial data we can see that the absolute uncertainties we used were accurate to only 1 s.f. Hence, we express our calculated absolute uncertainty (delta x) to a maximum of 1 s.f.

Look at our value of g and comparing it to the original data we can see each value was accurate to a minimum of 2 s.f. Hence we express g to a max of 2 s.f.

Raw data, ofcourse! DARN!

Thanks Ari. Been a long time man, mind catching up? yahoo?

[elemis]

Raw data, ofcourse! DARN!

Thanks Ari. Been a long time man, mind catching up? yahoo?

No worries.

Hahaha !! It has been very long. Unfortunately, I'm neck deep in revision, maybe after the exams ?

[Ghost Of Highbury]

need help with  O/N 2010 variant 23 q6b (ii) 

Dark fringe, destructive interference, out of phase, 180^o

2. Superposition can cause the amplitudes to add up,( Crest Crest), or get subtracted.
  Max amplitude = 2+1.4 , min = 2-1.4
   3.4, 0.6

   I = kA^2
   Ratio = k*3.4^2 / k*0.6^2 = 3.4^2/0.6^2 = 32

[Ghost Of Highbury]

No worries.

Hahaha !! It has been very long. Unfortunately, I'm neck deep in revision, maybe after the exams ?

Ah, same here man. Past papers -_-
You're giving physics paper 2 on Monday too right?

After exams, done. Good luck

[elemis]

need help with  O/N 2010 variant 23 q6b (ii)  

To be out of phase the phase diff. has to be 180 degrees.... ALWAYS.

A bright fringe is formed when two crests meet.

Amplitude of crest A = 2

Amplitude of Crest B = 1.4

Sum = 3.4

Intensity = k3.4²

Can you figure out the rest ?

[$H00t!N& $t@r]

but i thought you have to find the phase difference using 360(x/lambda) ... when do we use this?

[elemis]

Ah, same here man. Past papers -_-
You're giving physics paper 2 on Monday too right?

After exams, done. Good luck

Yup, though I've finished all of the papers.

What variant are you doing ?

Same to you.

[Ghost Of Highbury]

Yup, though I've finished all of the papers.

What variant are you doing ?

Same to you.

Paper 22

[$H00t!N& $t@r]

but i thought you have to find the phase difference using 360(x/lambda) ... when do we use this?

hello?