P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE

[halosh92]

1st variant
Q5c
could someone give it in more details plz?
Q7b
is it going to pass straight? or is it going to deviate at a smaller angle?

one more question: 4 pie L/T^2
if we found the total percentage uncertainity of L/T^2
then dont we have to multiply by 4
thx

could anyone solve this???

[Meticulous]

could anyone solve this???

when i get up halla..in 2 hours

[sweetie]

cud sum1 plz solve Q5b of the 1st variant

[nid404]

cud sum1 plz solve Q5b of the 1st variant

https://studentforums.biz/index.php/topic,8636.60.html

[AN10]

Q4c: stress= force/area
first find the minimum area of the tube from the graph , the highest stress is 9.5 * 10^8
area= (1.9 × 103) / (9.5 × 108)
= 2 * 10^-6
maximum area is given in the question
so subtract the minimum area from the maximum

i dnt get da last step...y do we subtract the areas

[The SMA]

Q4a
brittle: when the ultimate tensile stress occurs at the elastic limit ( thats when it breaks)
ductile: it can stretch after the elastic limit ( it becomes plastically deformed) then it some point it breaks
plastic: they usually have a low stress value and a large area under the graph , in the begiinng they are ductile but as stress increases they become plastically deformed

Q4c: stress= force/area
first find the minimum area of the tube from the graph , the highest stress is 9.5 * 10^8
area= (1.9 × 103) / (9.5 × 108)
= 2 * 10^-6
maximum area is given in the question
so subtract the minimum area from the maximum

Thanks! I got it now ^^

[The SMA]

i dnt get da last step...y do we subtract the areas

we need to minus bcoz we want to find the max area which is (3.2*10^-6)-(2.0*10^-6). not the min area we solved earlier.

[sweetie]

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

Thank You nid... +rep

but is  "phase diff.=2pi/lambda *path diff "  a given formula or u derived it on ur own???

[sweetie]

cn u xplain Q 7bii

[zxcvbnm]

november 05, question 8, last part...can someone uplaod the curve drawn? please asap

[sizbeauty]

can sm1 plz help me out wid june08:
Q4(part c)
Q6(part b)don't we just add up the powers???

[zxcvbnm]

june 09, variant 1, question 5,b

[sweetie]

june 09, variant 1, question 5,b

https://studentforums.biz/index.php/topic,8636.60.html

[zxcvbnm]

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

thankyou sweetie and nid!
but where did you get the formula from? for phase difference?

[zxcvbnm]

oh i realised that we can also use (2n+1)(lambda)/2...but i've forgotten what the n stands for...please clarify this asap!