Author Topic: CIE P3 Doubt!  (Read 1957 times)

Offline HUSH1994

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CIE P3 Doubt!
« on: March 26, 2011, 08:06:50 am »
i need some help in P3 CIE,every single paper Q1 the last part,after when you draw the graph and find the equation of the line,which you put the formula they give you with the (y=mx+c).
Anyone can help me with that,

Thanks,
HUSH

Offline sabbath_92

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Re: CIE P3 Doubt!
« Reply #1 on: March 26, 2011, 08:49:48 am »
Huh? I don't get your question?

Offline Arthur Bon Zavi

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Re: CIE P3 Doubt!
« Reply #2 on: March 26, 2011, 09:26:56 am »
i need some help in P3 CIE,every single paper Q1 the last part,after when you draw the graph and find the equation of the line,which you put the formula they give you with the (y=mx+c).
Anyone can help me with that,

Thanks,
HUSH

Like if the equation is : 9 - 55x = 3y, then rearrange in the form y = mx + c.
So 3y = -55x + 9; or : y = (-55x + 9)/3

Continuous efforts matter more than the outcome.
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Offline HUSH1994

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Re: CIE P3 Doubt!
« Reply #3 on: March 31, 2011, 10:11:47 am »
no man thats simple algebra,take a look at any P3,question one the last part,thats what i mean,when they give you a new formula and you have to find some values

elemis

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Re: CIE P3 Doubt!
« Reply #4 on: March 31, 2011, 10:50:03 am »
no man thats simple algebra,take a look at any P3,question one the last part,thats what i mean,when they give you a new formula and you have to find some values

Give us an example question.

Offline HUSH1994

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Re: CIE P3 Doubt!
« Reply #5 on: March 31, 2011, 05:06:04 pm »
MJ 2008 P31,Q1 part g please?

elemis

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Re: CIE P3 Doubt!
« Reply #6 on: March 31, 2011, 05:11:58 pm »
MJ 2008 P31,Q1 part g please?

There is no part (g) in Question 1 Summer 2008 :S

Offline HUSH1994

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elemis

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Re: CIE P3 Doubt!
« Reply #8 on: April 01, 2011, 04:21:50 am »
there is,its on page 6
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_31.pdf

Look at your graph again.

On the  Y-axis you plotted cos(\frac{\theta}{2}) and on your x axis you plotted n

Looking at the equation :
<br />cos(\frac{\theta}{2}) = \frac{mgn}{2T} + k

Hence, substitute cos(\frac{\theta}{2}) for Y and n for X

You will get :

y = \frac{mg}{2T}x +k

Hence, \frac{mg}{2T} is the gradient of the graph and k is your y-intercept.