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I need help with it too
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I need help with it too
since u found the reflection coefficient in part which is Ir / I = (Z1-Z2)^2 / (Z1 + Z2 )^2
Here the c.r.o. is actually showing us the change in voltage (on y-axis) with respect to change in time. As stated, Y gain is 2V/cm.. So the peak voltage would be 3.5*2v=7V . In no way, can we deduce the amplitude of the wave because the c.r.o. has been set to show the variation of voltage on y axis.See one wave pattern occupies four 1 cm boxes along x axis. As time base setting is 0.5ms/cm , this means that the period of wave is (0.5ms*4)=2ms=2*10-3 s. Frequency is 1/period. So, 1/2*10-3 = 0.5*10^3 Hz =0.5kHz.
Nosince u found the reflection coefficient in part which is Ir / I = (Z1-Z2)^2 / (Z1 + Z2 )^2
idk whether my final ans is correct i solved it in a hurry
so the reflection coefficient for fat would b equal to
Ir / I = e^ - μx where μ = 48 given in que and x = thickness needed to be calculated
so( ans in part 1) Ir / I = e^- 48 x
gives u the ans
Thank youView attachment 60295
According to flemings left hand rule, the force will be downwards. So in magnetic field region, it will be a curve and out it will be straight.
I will try In sha Allah.
Thank you!I will try In sha Allah.
Don't thank me now.I haven't done anything yet.XDThank you!
Isn't the uplink the signal transmitted from earth? Which frequency is higher and why?the uplink frequency is a tiny fraction of signal transmitted from the earth...the satellite regenerates and amplifies it an transmits it to earth at a different frequency..
This allows it to effectively receive and re-transmit information without risk of overlap or mixed reception with the received signal
c)i) Ir/I= 4/225
yea ...i said the same....the satellite regenertes it back to earth as downlink after amplification ..Isn't the uplink the signal transmitted from earth? Which frequency is higher and why?
the formula is same u just have to take the reflection into account...
u knw Debrogleis wavelength ...every moving particle has a wavelength associated with it ...photons incident on the surface have a wavelength of 590nm.....View attachment 60306 Can someone explain me the b part
http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-119.htmlSomeone please explain this part.
http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-218.htmlIn part ii, don't we have to convert power into energy first??? And why are we subtracting the masses? We don't consider the overall total mass? And please explain 2 as well. SOMEONE View attachment 60323
Thanks a millionc)i) Ir/I= 4/225
then
c)ii) Intensity = I*e(-48*2*x).As the ultrasound will be attenuated going from the surface to the boundary and from the boundary to surface, so it moves twice the distance. Thats the intensity that comes at the barrier, 4/225 of that is reflected back. so
(4/225)*I*e(-48*2*x)=0.012*I
cancel I with I
ln (0.012/0.018) = -48*2*x
x=4.22x10^-3 m
=0.42cm
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