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LooolOfc I am. I am so good at it.
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LooolOfc I am. I am so good at it.
Hehhe....Loool
Yes.Like I said around 17 marks... The momentum thing... Then... Some parts of electricity....
Today u did 22 right??
So the wabout the gt??Yes.
I guess it will be around 40 - 42So the wabout the gt??
I guess it will be around 40 - 42So the wabout the gt??
no way ... it cant be in that range .... i cudnt even complete the paper ... it was a disaster for meI guess it will be around 40 - 42
How much did u left?no way ... it cant be in that range .... i cudnt even complete the paper ... it was a disaster for me
Even tho all the questions were pretty simple .. it was very lenghty and that was the only problem i faced :/
half of a questionHow much did u left?
What was change in KE?I did so bad in this paper T_T. I didn't even have time to recheck my answers and at end of exams I noticed I forgot to convert 450g to kilograms for the Kinetic change in energy( the slope diagram question)
And the waves A, B explanation the questions were like chinese to me, had it ask in a more direct way I would have understood it was asking for where the nodes and antinodes were situated
And that final question, I completely forgot to consider mass of atoms ;_;
Dont worry, GT will b low.. 35 - 36half of a question
Did you appear paper 22?Dont worry, GT will b low.. 35 - 36
Paper 22 wasnt that bad it was medium not hard not easyDid you appear paper 22?
Paper 22 was quite lengthy and i guess gt would be 30-32.
What was change in KE?Paper 22 wasnt that bad it was medium not hard not easy
I think it was 2.2** I got 22** (forgot the 2 numbers after) Because I forgot to divide by 1000What was change in KE?
Same that question was weird.
mass would be cancled dont worry.
Though I used, 1/4/3Pi(d/2)^3I think it was 2.2** I got 22** (forgot the 2 numbers after) Because I forgot to divide by 1000
And thank you for telling me about the mass, yeah coincidentally a friend of mine just told me the same thing before I went for mechanics paper lol
The mass actually is so small that you consider the mass of both atoms to be similar( so using only 1/r^3 is right)
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