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ok here is what i mean. thats how i'd solve it
OMG it was that easyy
Man M1 is getting to my head Thanks soo much Pie-man
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ok here is what i mean. thats how i'd solve it
Youu are great mash'allah thank youuu
sinx/cosx =wg/5.5
tanx * 5.5 = wg
Substitute this in the second equation
5.5cosx + sinx( tanx * 5.5) = 7.3
after simplifying the brackets, you'll get
5.5cos^2 x + 5.5 sin^2 x = 7.3 cosx
5.5( cos^2 x + sin^2 x) =7.3 cosx
5.5/7.3 = cosx
x= 41.4
now wg= tan41.4 * 5.5
you're welcome but try to complete it first cause i didn't check how it turned out though.OMG it was that easyy
Man M1 is getting to my head Thanks soo much Pie-man
where did the 2nd equation come from? :/Oh that's easy just make 2 simultaneous equations.
Use s=ut+1/2at^2
So your first equation is: 55=5u+12.5a
And second one is: 120=10u+50a
Solve them simultaneously, you should get a=0.4 and u=10 m.s-1
where did the 2nd equation come from? :/
You're welcome x)
Thank you, my brain stopped e_e
may june 2009 question 1 c)ii) my answer is different than the mark scheme and i can't get the write answe
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s09_qp_4.pdf
Oh righttt.. i was using B to C... -_-
Anyways q4(ii) same paper?
By what time will u be here to help? im still doing papers and problems are arising.. :/Just add up your values of the forces in the y-direction , and they've already given you the ones in the x-direction.
Use Pythagoras to find resultant, and then use whatever to find the angle .
Hope you got it x)
Oh righttt.. i was using B to C... -_-
Anyways q4(ii) same paper?
for keeping the block equilbrium a frictional force must act on the block in the right directionJune011 v.43 no.4 and no.5 please ,, Honestly they are the worst questions ever
By what time will u be here to help? im still doing papers and problems are arising.. :/
and will u also be available in the morning?
Which variant is tougher, 41 or 43 ?
Lemme know ASAP! Thanks a lot.
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