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change in gravitational potential energy = 3.8*1.5*10^-2(because extension is in centimetre)but I am not getting the answer
But the question didn't ask for drawing such diagrams ! I did this way and got the answer ! but I'm not sure did this answer 1 e (i) or not ! please someone correct me If I'm wrong !
now use the cosine rule. angle is 90+45. scale drawing will not give accurate result because the velocity of wind gives 1.8cm only if i took 1 cm = 20 m/s
What about the second part (2.)change in gravitational potential energy = 3.8*1.5
can u plz explainSnow Angel
question number 4 (b) (iv)
current = 0.24 A
(a) p.d = IR = 0.24*5.5=1.32V
(b) terminal p.d = emf - lost volts = 4.4 V - (0.24 A * 2.3 ohm) = 0.552 V
(c) terminal p.d = emf - lost volts = 2.1 V - (1.8 ohm * 0.24 A) = 1.67 V
acceleration = gsin15http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_23.pdf
Q 1 e both i and ii
and Question 2 a ii
yesIS THIS OK?
if you see the the fig 5.2 you will realize the total wave produce are 3/2
use the formula E=1/2*k*(x^2)What about the second part (2.)
new wave will be straight line.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_22.pdf
guys how are we suppose to do q 5part b iii i dont get how we r suppose to draw the new wave
I just did fag -_-http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf
Can anyone PLEASE explain Q4 c ii) full
sorry dude if I am bothering you too muchuse the formula E=1/2*k*(x^2)
you do know the formula for elastic potential energy, right?sorry dude if I am bothering you too much
can you please explain it
Change in U = mgh = 2.8N * 0.015 m = 0.057Jhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf
Can anyone PLEASE explain Q4 c ii) full
Idon't get how does this represent diffractionHere you need to draw labelled diagram of source of light, apparatus to diffract/interfere and how to observe it.
To observe the diffraction the light must be perpendicular to the slit and the screen must be long distance from slit
to observe. we will decrease slit separation(in (ii) ) and slit width in (i) part.
(i) when we will decrease it, the the spot will start to spread into geometric shadows. along the bright big spot a few fringes will also be observed but they will be dim.
(ii) the fringe seperation will increase. alternate bright and dark fringes will be produced.
correct. but i drew it to make it easy for you to understand. i always use cosine rule or Pythagoras theorem unless examiner himself says to draw scale drawing.But the question didn't ask for drawing such diagrams ! I did this way and got the answer ! but I'm not sure did this answer 1 e (i) or not ! please someone correct me If I'm wrong !
Suchal Riaz
the laser is normal to a slit and that difracted light goes to screen. right click that image and open it in new tab because the image is rotated in this website.I
Idon't get how does this represent diffraction
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