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P1 CIE physics doubts

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astarmathsandphysics:
year? I suppose p1

sweetie:
its nov08 ( i've attached the file) :)

astarmathsandphysics:
10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions  D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B

dodo:
hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b) :)

nid404:

--- Quote from: dodo on April 26, 2010, 03:58:10 pm ---hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b) :)

--- End quote ---

here u go

there is a formula u need to know

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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