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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
questions 4,26,29,30,32,37 really need help!
questions 4,26,29,30,32,37 really need help!
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for Q4 use trial and error method until the units on both sides of the equation are equalhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
questions 4,26,29,30,32,37 really need help!
I am sure after going thorough this site you will be able to answer that question.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
Can someone briefly explain question 6 please
for q37http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
questions 4,26,29,30,32,37 really need help!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
questions 4,26,29,30,32,37 really need help!
26)for Q4 use trial and error method until the units on both sides of the equation are equal
L and a has units (m )while T has unit (s)
by using option B we get units on the right hand side equal to the left hand side
View attachment 44333
for q29
For a stationary wave in P there should be a node at the closed end and an antinode at the open end.For a wave to be formed in Q there should be an antinode at both ends.
length of P=35 cm
wavelength of given wave = 20 cm
so 35/20 = 1.75 lambda can fit in P. if you draw a wave 1.75 lambda long starting with a node you will find that it ends with an antinode so this means that a stationary wave would be formed in P
for Q 50/20 = 2.5 lambda would fit in it.
start by drawing an antinode at one open end and continue until you draw 2.5 lambda .u will find that an antinode also forms at the other end.This satisfies the condition for a stationary wave .hence you will get a stationary wave in Q as well
We know that V is proportional to A so V will be constant as A is constantWhich graph represents the motion of a car that is travelling along a straight road with a speed
that increases uniformly with time?
View attachment 44342View attachment 44342
How the the hell A is correct. C should be correct it beacause Question say "moving in straight line" which mean positive s.Why the ** A is correct?
Thank you sooooo sooo much idk how to thank you enough you helped me aloooot26)
∆L is basically compression.
L = Original length
When compression increases, original length decreases - fact - hence, inverse proportionality.
You can also prove it like this:
Strain = extension/original length => e / L
Compression = 1/e
Therefore, strain = 1/eL
When you look at this equation, you'll see that 1/e (compression ∆L) and L are inversely proportional.
30)
Waves have to superpose for interference pattern to be produced at RS the waves have not been yet superposed.
If you dont get me, here is something elaboration :¬
At first the waves have to be produced, when two waves meet they superpose but at line XY there is maximum of interference pattern compared to RS beacause waves have to meet first, so answer is C
32)
resultant force is zero because it rotates at a constant speed because and anti clockwise because E is downward so downward is negative so it repels it and it became anti clockwise, C
37)
Above post
Thank me by having A in phyThank you sooooo sooo much idk how to thank you enough you helped me aloooot
for q15
Thank you sooooo sooo much that helped me aloot i swear to god the way u explainfor Q4 use trial and error method until the units on both sides of the equation are equal
L and a has units (m )while T has unit (s)
by using option B we get units on the right hand side equal to the left hand side
View attachment 44333
for q29
For a stationary wave in P there should be a node at the closed end and an antinode at the open end.For a wave to be formed in Q there should be an antinode at both ends.
length of P=35 cm
wavelength of given wave = 20 cm
so 35/20 = 1.75 lambda can fit in P. if you draw a wave 1.75 lambda long starting with a node you will find that it ends with an antinode so this means that a stationary wave would be formed in P
for Q 50/20 = 2.5 lambda would fit in it.
start by drawing an antinode at one open end and continue until you draw 2.5 lambda .u will find that an antinode also forms at the other end.This satisfies the condition for a stationary wave .hence you will get a stationary wave in Q as well
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
can anyone please explain q18 19 and 20
Thanks!
Wait, I am helping some other .. please.Please can anyone explain:
q8 and q 21
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
And also can anyone kindly help me out in this?
Also visit https://www.xtremepapers.com/community/threads/physics-practical-tips.6306/People i compiled some of the necessary stuff for the upcoming physics practical from this community and notes written by some members
I tried my best and i hope it helps u .
Best of Luck for the paper
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