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Yayyyyy!YES.
LET'S KNOCK SOME PINS DOWN IN ANGER.
*rage*
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Yayyyyy!YES.
LET'S KNOCK SOME PINS DOWN IN ANGER.
*rage*
yay *freefood* xDI think I could marry you. <3
where is Q7? and can you upload other Qs as well?
Yes yes, I'm a bit busy with other subjects, but yeahwhere is Q7? and can you upload other Qs as well?
i will be waitingYes yes, I'm a bit busy with other subjects, but yeah
Shortly inshaaAllah
bloody hell ! i took height of water 0.6 instead of 0.8
ik as i said there are two ways your way summing them individually without finding the constant and that way.i used that way to try to make him understandWow, you guys.
You created such a big fuss, even a Moderator walked in.
that equation only is applicable for t=8.finding the constant makes it as if the motion started from t=0 instead of t=8.the particle/car only shares the same distance,speed,acceleration at t=8,and not at any other time so if it shares the same distance of the particle at t=8 having constant acceleration,then u can directly sub t=27 in the equation as long as u found c.there hundreds of questions like this one i did everything from 2001-2013 the way of the answers are in the 2 ways ive listed.Okay so let's suppose that the equation you ended up with 's=3/10 T^5/3 - 1.6' is correct for the entire motion of the body, because at t=8 you get the same value of displacement i.e 8m. That means that this equation is applicable to the entire motion. So try it with a different time e.g t=4s. In the linear motion if you substitute the value of t=4s you will end up with a displacement of s=2m (s= 1/2 (0.25) (4^2). Substitute this value into your 'correct' expression of s and you end up with s=1.42m, clearly showing that this is not a universal expression for the entire motion. The motion had to be treated as two separate parts, from t=0 to t=8 and then t=8 to t=27. That is why when the second motion starts at t=8, which is equal to t=0 for motion 2.
yes the answer is 71.3m Well good luck for your next papersthat equation only is applicable for t=8.finding the constant makes it as if the motion started from t=0 instead of t=8.the particle/car only shares the same distance,speed,acceleration at t=8,and not at any other time so if it shares the same distance of the particle at t=8 having constant acceleration,then u can directly sub t=27 in the equation as long as u found c.there hundreds of questions like this one i did everything from 2001-2013 the way of the answers are in the 2 ways ive listed.
Just one mark!bloody hell ! i took height of water 0.6 instead of 0.8
how much is that gonna cost me
I've been posting them all throughout the thread. Look through the pages.Can someone tell the correct answers!nFor personal satisfaction maybe
9.how many marks was q6
5 & 6, will do 7
Haha. That's good, mashaaAllah. And don't worry, everyone goofed up in some parts. One of the smartest people I know messed up and thinks he's only gonna get 33. :3JazakAllah Khair! I got question 5 correct, and question 6 part a till the deceleration, didn't have time for part b. I guess the exam wasn't bad afterall!
ya........dont worry threshold will be under 10 marks.....for A gadwell , i think this paper was harder than any other m1 paper i have done...
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