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Why not 7?
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AS Physics P1 MCQs Preparation Thread.
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use the equation nλ=dSinθ where n(number of order), λ(wave lengeth), d(distance between each grating) and θ being the angle.
since he's looking at the maximum number of images so Sinθ= 1. and nλ=d.
to find d, you divide 1mm/500 so you get 2x10^-6/
n=d/λ=2x10^-6/(600x10^-9)= 3.33 on each side of the zero order
so you add 3+3 to find images on both sides of Y and add the zero order line which also forms an image. so 3+3+1=7
oops somebody already answered
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Haha - That's a relief. I got mega-confused just then.yeah dude you're right.im mistaken
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Haha - That's a relief. I got mega-confused just then.
lol sorry for that. Good luck for the exam today hope you do very well
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Haha no harm done.lol sorry for that. Good luck for the exam today hope you do very well
U admitted ur mistaken-ness and that, my friend, means u've got guts.
Technnically speaking, everyone here HAS guts... but yeah u get the point.
U too!!!
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answer is D, because only points within the overlap are in phase and so A and C are wrong. also B is wrong because stationary waves have varying amplitudes.
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27 shd b A.
No. 1. Nodes are always formed at closed ends so discard D.
No.2 In one loop, ie between two nodes, particles are in phase. WITHIN one segment. (Or loop)
So particles r in phase, one direction, in bottom segment.
Particles are in phase in second (top) segment. BUT particles in bottom and top segments ARE NOT in phase with EACH other.
I hope u get what I mean.
Amplitude of particles equidistant from a node is same.
So A.
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only the loop i mean so points on N1 to N2 are in phase and N2 and N3 as well, however points on N1 arent in phase with points on N3.
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yes, i understood. thank you.
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#32
Power = I^2 R
so power in Y = (I/2)^2 x 2R = I^2/2R
power in X = I^2 R
Y/X = 1/2 so answer is B
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When a battery is charged, does it receive enrgy or power??
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Energy is always conserved so
Initial energy = final energy
so Ek = new Ek + change in Ep
(mu^2)/2 = (mv^2)/2 + mgh
cancel out all the m ( masses)
(u^2)/2 = (v^2)/2 + gh
28^2 /2 = v^2 / 2+ 9.81x22
392 = v^2 /2 + 215.82
v^2 /2 = 176.18
v^2 = 352.36
v = 18.8
answer is B
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q.9,Q.27,Q.29,Q.33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf Q.18,Q.34 why arent the wires in the cable considered to be parallel i.e shouldnt we divide the R by 2 not multiply it by 2.
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf Q.18,Q.34 why arent the wires in the cable considered to be parallel i.e shouldnt we divide the R by 2 not multiply it by 2.