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Biology; Chemistry; Physics: Post your doubts here!

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heyheyheyyyy, what is the 6th paper of physics? is it the practical one? :O
and can i use the notes for this paper to prepare for my ATP?
 
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if it is the extended syllabus what should i do to avoid over cramming? cuz i havent got much time left :|
P.S. IM SORRY FOR BEING SO PAKAOO :$
Our syllabus is the same as the extended syllabus for IGCSE, so you have to do everything.
Cramming doesn't help at all in the sciences subjects. Read the book/notes once. Then do ATP papers and MCQ's, they build up your theories and concepts.
 
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Our syllabus is the same as the extended syllabus for IGCSE, so you have to do everything.
Cramming doesn't help at all in the sciences subjects. Read the book/notes once. Then do ATP papers and MCQ's, they build up your theories and concepts.
Okayy.. THANKS A LOTTT!! :)
 
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Help, :( How do we determine the exact height of an object after it's light rays pass through a lense. I know when it enlarges and when does it increases in size, but how do we calculate the exact height? For reference, http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_s12_qp_21.pdf question 10 part b sub part 2

pass a line from the top of the object through the mid of the lens, and another from the top that is parralel to the ground and is refracted through the focal length (3cm) Where the first and the second ray meet, the image is formed. Set a ruler from the ground level to the image, and find the height
 
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Help, :( How do we determine the exact height of an object after it's light rays pass through a lense. I know when it enlarges and when does it increases in size, but how do we calculate the exact height? For reference, http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_s12_qp_21.pdf question 10 part b sub part 2
The distance from lens to object = 8 cm.
Focal length = 3 cm. (2F = 6cm).
The object is even further than 2F hence the image will be formed in between F and 2F (real, inverted and diminished), if you've correctly done part (i) then you'll get the height by measuring the height of the image. I reckon it'll be around 1-1.5.
 
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pass a line from the top of the object through the mid of the lens, and another from the topic that is parralel to the ground and is refracted through the focal length (3cm) Where the first and the second ray meet, the image is formed. Set a ruler from the ground level to the image, and find the height
The distance from lens to object = 8 cm.
Focal length = 3 cm. (2F = 6cm).
The object is even further than 2F hence the image will be formed in between F and 2F (real, inverted and diminished), if you've correctly done part (i) then you'll get the height by measuring the height of the image. I reckon it'll be around 1-1.5.
ah rite, as simple as it gets. use a ruler. i was looking fr some theory behind it. Thanku!
 
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ah rite, as simple as it gets. use a ruler. i was looking fr some theory behind it. Thanku!

furthermore, noting the length is all the more easier due to the fact that they have made a clear graph. So, i don't think it should be difficult at all.
 
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If we use cross multiplication method in moles do we get method marks for it? because in the marking schemes there are individual marks for taking out moles and masses.
 
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1 mol Mg-----1 mol H2
(24 g) -----24 dm3 H2
0.036 g --------------x
x=24*0.036/24=0.036dm3 H2 is produced with 0.036 g Mg
 
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