Further Math 2

[ima_shah2002]

Okay so i noe there aint gonna be many ppl who have done further math.. but guys can we discuss

1. First Question regardin SHM... can form two equations and find unknown of angular frequence and amplitude.. pretty easy(5 marks)
2. Damn i had some prob here waddya guys do... ( i think 7 marks)
3. First part was easy in provin tan<0.5 , 2nd part no idea..(9 marks)
4. super easy... just gotta be careful wid equations(10 Maks)
5. again this was tricky how did u prove the angle.. speed is then easy to calculate.. and the next part.. luckily i somehow got the answer yaay.. find time for such horizontal distance to be coverd(acceleration is 0) and then in that time find vertical distance... add that to distance above 0 and u get 31/30 yesss (12 mark)

For Stats...hmm.. somethin i was unprepared

6. did u use the chi square test with a contingency table
7. i dont remember... was it regarding some t test... dunno.. m confused
8.confidence interval with t test i think....
9.i think we had to use a z test with difference between means with known variance rite..
10. This was the one wid product moment relation coefficient.. i think i got 0.75... i was confused why next was 4 marks....
11. 8/3 ma2 i got 4/3ma2.. damnn.. .then 11/3 i think... then u just use energies to find v think (14)

 

[supermagic]

for question 3

u taking the moments about the point where the hemisphere and the inclined wall meets.

then we have

a x sin(alpha) x M = m x a x (1 - sin(alpha))

and we can deduce that from part (i) sin (alpha) < 1/(root5)

we put sin(alpha) as 1/root5, the we can get the answer!

and the last Either question,

the sqaure ABCD forms with 4 rods which has negligible thinkness

the each rod inertia = 1/3 x M/4 x ( 2x (root2 a /2 )^2 ) = (Ma^2)/12

the about the center ( by parallel )
I = (Ma^2) /12 + M/4 x (root2 a /2 )^2 = (1/12 + 1/8) Ma^2 = (Ma^2)/6

so inertia of ABCD aobut the centre,

I = (Ma^2)/6 X 4 (cuz there are 4 rods) = (2ma^2)/3

inertia of ring = 2M x (a)^2 = 2Ma^2

therefore total inertia = (2ma^2)/3 + 2Ma^2 = ( 8Ma^2)/3

then use perpendicular theorem, and parallel about A, u get, (13Ma^2)/3