Author Topic: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE  (Read 18109 times)

Offline thecandydoll

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #60 on: June 06, 2010, 12:33:43 pm »
Anyone solve this for me  ? :D
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!

Offline ruby92

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #61 on: June 06, 2010, 02:02:43 pm »
m/j 2004 paper 2
3 bii????

Offline halosh92

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #62 on: June 06, 2010, 02:44:20 pm »
m/j 2004 paper 2
3 bii????

3bi) the gradient is d/t^2
3bii) s= ut + 1/2at^2
u=0
s=1/2at^2
s/t^2  * 2 = a          (making a as the subject)
substitute the gradient instead of s/t^2 and multiply by 2  thats all


voila   ! :)
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Meticulous

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #63 on: June 06, 2010, 03:22:13 pm »
3bi) the gradient is d/t^2
3bii) s= ut + 1/2at^2
u=0
s=1/2at^2
s/t^2  * 2 = a          (making a as the subject)
substitute the gradient instead of s/t^2 and multiply by 2  thats all


voila   ! :)

WOW! finally halosh answering and not asking :P jk :P

+rep.!

Offline halosh92

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #64 on: June 06, 2010, 03:26:12 pm »
WOW! finally halosh answering and not asking :P jk :P

+rep.!

OUCH that hit me right in the heart  :P :P
haha i used to help alot before the exams started now am just blank ::)
thx! ;)
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Phosu

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #65 on: June 06, 2010, 04:03:29 pm »
jun 09 var 1
question 5b

Offline aldehyde1612

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #66 on: June 06, 2010, 04:26:35 pm »
Anyone solve this for me  ? :D
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!

i need help with them too... can someone please answer!

and oct/nov 2007... q6. b ... how did u get the pd to be 5.0V ???
.We know a little about a lot of things; Just enough to make us dangerous.

Offline poisonedrose

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #67 on: June 06, 2010, 05:09:25 pm »
i need help with them too... can someone please answer!

and oct/nov 2007... q6. b ... how did u get the pd to be 5.0V ???
The power supply gives 230 volts, but the shower unit only needs 225 volts( it says there in the question) so if you subtract 230 and 225, you get 5 volts.


Offline halosh92

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #68 on: June 06, 2010, 05:14:36 pm »
in series:
extensions are added
in parallel :
extensions divided by number is springs

in parallel:
spring constants are added
in series:
i have no idea wat happens  :)

could someone check if this is correct or wrong?
« Last Edit: June 06, 2010, 06:09:02 pm by halosh92 »
everyday we wake up is a miracle, then how do we say miracles dont happen?????

nid404

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #69 on: June 06, 2010, 05:14:45 pm »
jun 09 var 1
question 5b


phase difference=( 2pi/ lamda)  X path difference

phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi  X path difference.
path difference is S2M - S1M 
find S2M using Pythagoras = root of 802 + 1002
                                                   =128
path difference= 128-100=28
wavelength= 2pi/pi X 28
               =56cm                 
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz   wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas

Offline poisonedrose

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #70 on: June 06, 2010, 05:24:03 pm »
Anyone solve this for me  ? :D
M/J 2007-----Physics Q4d (i) and (ii)
And M/J/2007----PHYSICS q1 the whole thing!
I dont get it =(
thank youuuu!
Using an appropriate scale, draw a horizontal line representing horizontal velocity of 18m/s.I used 2cm (10 boxes)=5 m/s. And it is a straight, horizontal line because horizontal velocity does not change,.

then calculate vertical velocity at point of impact with ground by using the equations of motion with the formula v2=u2+2as. U being 0. a being 9.8 and s being 16. The vertical velocity gottn is approximately 17.7m/s which we can draw using our scale, as a vertical straight line.

We join the two lines- making a triangle and we can find the resultant velocity by the using Pythagoras theorem .. (which would be the hypotenuse) a2+b2=c2.

Once you are drawn with the vector diagram, measure the asked angle with a protractor or with trigonometry

Offline zxcvbnm

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #71 on: June 06, 2010, 06:07:24 pm »
may 06-question 5, part (c)

Offline sizbeauty

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #72 on: June 06, 2010, 07:16:59 pm »
m/june02
Q4(b)(part 1)
oct/nov02
Q5(b)(part1) how do u calculate the phase difference in this one  ??? ??? ??? ???
plz help me!!!!!!

nid404

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #73 on: June 06, 2010, 07:17:42 pm »
may 06-question 5, part (c)

check the attachment.

nid404

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Re: P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
« Reply #74 on: June 06, 2010, 07:23:39 pm »
m/june02
Q4(b)(part 1)


a) v2=u2+2as
v2=2X9.8X1.6
v=5.6m/s

b) p.e lost=k.e gained

p.e lost=mgh=73/1000 X 9.8 X 1.6=1.14
k.e gained=90% of p.e lost= 1.03J

1.03=mgh
1.03=73/1000 X 9.8 X h
h=1.44 m