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Mathematics: Post your doubts here!

blabla

blabla

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Nibz said:
Q.7
(ii)
When t = 2.5
d = 2.6 x 2.5 = 6.5

sin theta = 1.6 / d = 1.6 / 6.5 = 0.246

since a = 10 sin theta
a = 10 x 0.246 = 2.46 m/s'2
Click to expand...

could you please explain why you did the a = 10 sin theta bit? thanks :D
 
donor

donor

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donor

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reina81 said:
why can't we use the polar from to solve it? and also how did you solve for y^2?
Click to expand...
we could use the polar form but my answer is not the one given in the markscheme. for y^2, i made a quadratic equation which gave the roots for y^2. To make things easier you could replace the y^2 with any variable such as 'a'
 
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reina81

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donor said:
we could use the polar form but my answer is not the one given in the markscheme. for y^2, i made a quadratic equation which gave the roots for y^2. To make things easier you could replace the y^2 with any variable such as 'a'
Click to expand...
oh ok thanks alot
 
Unicorn

Unicorn

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1357913579 said:
please can any solve this problem tomarrow is my mecanics exam please reply as soon as possible. number 5 first part where we have to show tht tention is 4N
[URL='http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s06_qp_4.pdf[/quote']http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s06_qp_4.pdf[/URL]
Click to expand...

dasasas.pngOk if you look in the picture you can see the forces already present

the 4root2 force is the normal to the force applied ON the pulley so if you rearrange the forces then use pythogras to get 2T^2 =32 and solve u get T= 4
 
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SUMMERLOV35

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Ashique said:
Again, I request someone to help me out with the last part-
Can someone PLEASE have a look at questin number 7 iii part, please, I realllu don't get it-
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf
Click to expand...

this question is repeated , you will find it in 2003 or 2004 paper too...
anyways for part i
r= 4500
f= 3150 so coefficient is 0.7

for second part they want the acceleration as in
between the blocks so obviously top one would the one to slip away...
and i figured out that whenever they show that a would not be greater than a value for the forces you only
take frictional force
0.2x2000=200a
a=2

to find the max value of P
there is frictional force so
P-F=450X2

REMEMBER FOR NET FORCE=MA
you always take net force

hope this helps
 
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