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OpenStudy (somy):
here are the question, i just need to understand them
OpenStudy (somy):
@thushananth01
OpenStudy (anonymous):
1. The gradient of the graph would give velocty, u see in any of the other graph, u dont find a constant gradient as it saying it reaches terminal velocty means gradient shouldbe a straight line, so there u go
OpenStudy (somy):
owhhhh okaaay q 1 cleared :D
OpenStudy (anonymous):
2. Apply law of conservation of momentum
total initial momentum = total final momentum
let velocty of X be v1 and velocty of y be v2
0 = mv1 -2mv2
mv1 = 2mv2
v1 = 2v2-----2
Kinetic energy of X = 0.5 m (2v2)^2 note that i hve substitited for v1 from eq 2
kinetic energy of y = 0.5* 2m ( v2)^2
divide x and y u get 2/1
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OpenStudy (anonymous):
E = TL/AX
make x the subject
T=EAX/L-----1
Use equation 1 for Tension in P
Tp = EAX/L---3
use eq 1 for tension Q note that A and L are different
\[T_{q} =\frac{ E \times \frac{1}{2}A \times X}{2l}\]
\[T_{q} =\frac{ E \times A \times X}{4l}------2\]
divide 3 by 2 u get the answer
OpenStudy (anonymous):
4. For this u have to find the area under the graph which means u hve to count the number of boxes approximately tht would give the energy
OpenStudy (anonymous):
5. let lambda = b
x = bD/a
3 * 10^-3 = 600*10^-9 *D/a----1
Make D/a the subject in equation 1
x'=bD/a
x' = 350*10^-9 * (D/a)
OpenStudy (anonymous):
@Somy
OpenStudy (somy):
im still doing 2
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OpenStudy (anonymous):
ohhhh ok
OpenStudy (somy):
T_T
OpenStudy (anonymous):
whts up
OpenStudy (somy):
im still not getting it
lemme try again
OpenStudy (anonymous):
sure :)
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OpenStudy (somy):
yaaaaaaay got it!!!
OpenStudy (anonymous):
hehe good :)
OpenStudy (somy):
q 3
T L A X what are they?
OpenStudy (anonymous):
E = FL/AX youngmodulus equation
OpenStudy (somy):
ohhh its FL/Ax
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OpenStudy (somy):
oops lol ok let me try
OpenStudy (somy):
tell me one thing Tension is force right?
OpenStudy (anonymous):
ofcourse :)
OpenStudy (somy):
okay cool wait then
OpenStudy (somy):
im not really getting what u r doing but i'll try like this
first step is making F as subject right?
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OpenStudy (somy):
if length is 2 L then extension will be 2 x right?
OpenStudy (anonymous):
no they hve said extension is same
OpenStudy (somy):
ohw wait then
OpenStudy (somy):
got ittt!!!
OpenStudy (anonymous):
yayyyXd
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OpenStudy (somy):
4 is that a joke? i really have to count those tiny boxes?
OpenStudy (anonymous):
exctly i din bother to
OpenStudy (somy):
then wht do i do?
OpenStudy (anonymous):
u gotta count
OpenStudy (somy):
okay i think i just ruined my KE question im confused again
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OpenStudy (anonymous):
wts wrng
OpenStudy (somy):
why the hell i wrote 1/4 intead of 4/1
OpenStudy (somy):
instead*
OpenStudy (anonymous):
do it again
OpenStudy (somy):
here
my q is what about mv parts? shouldn't they be cancelled?
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OpenStudy (anonymous):
yes thts right
OpenStudy (somy):
oh wait i did it wrong
OpenStudy (somy):
oh no wait lol i did it right
my brain is exploding
OpenStudy (anonymous):
haha ski[ p
OpenStudy (somy):
3 done
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OpenStudy (somy):
i mean 4 done
OpenStudy (somy):
where did u bring 600*10^-9 from?
OpenStudy (anonymous):
wavelength given
OpenStudy (somy):
its not 600nm its 700nm
OpenStudy (anonymous):
lol sorry my bad
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OpenStudy (somy):
oh thnx God it was mistake lol i thought it'd be something new
OpenStudy (somy):
im getting ans as 1.5
OpenStudy (anonymous):
is it corrct?
OpenStudy (anonymous):
Nope
OpenStudy (somy):
oh wait after i get my answer i should divide it by 2 because they say slit separation is doubled?
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OpenStudy (anonymous):
My working
OpenStudy (somy):
waaaaaaaaaaait if Slit separation is doubled then Fringe separation will be doubled too??
OpenStudy (anonymous):
noooo u hve to find it
OpenStudy (somy):
here my working
OpenStudy (somy):
my final answer is like double of original answer
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OpenStudy (anonymous):
i am getting
d/a = 3*10^-3/700*10^-9
x = 350*10^-9 * d/2a
OpenStudy (somy):
oh yeha lol i forgot 2a part
OpenStudy (somy):
then that means the ans i got in D/a will be multiplied by 1/2 because a is now doubled right?
OpenStudy (anonymous):
yup
OpenStudy (somy):
gotttt the answer!!!!!!!!!! thank u a looooooot like really a lot!! @thushananth01
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