Qualification > Sciences
P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE
Phosu:
jun 09 var 1
question 5b
aldehyde1612:
--- Quote from: thecandydoll 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!
--- End quote ---
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 ???
poisonedrose:
--- Quote from: aldehyde1612 on June 06, 2010, 04:26:35 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 ???
--- End quote ---
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.
halosh92:
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?
nid404:
--- Quote from: Phosu on June 06, 2010, 04:03:29 pm ---jun 09 var 1
question 5b
--- End quote ---
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
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