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Physics: Post your doubts here!

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since 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
 
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View attachment 60287
why can't the ans be A..... why it is B
helppp plzzz
thanks in advance:)
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.
Hope it helps you :)
 
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since 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
No :cry:

Screen Shot 2016-04-13 at 6.28.08 PM.png

Someone please explain
My Name Rizwan Javed qwertypoiu nehaoscar
 
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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
Isn't the uplink the signal transmitted from earth? Which frequency is higher and why?
 
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c)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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Isn't the uplink the signal transmitted from earth? Which frequency is higher and why?
yea ...i said the same....the satellite regenertes it back to earth as downlink after amplification ..
The satellite gets power from solar cell. So, the transmitter (satelite) doesnt consist of a higher power. On the other hand, the ground station (Earth) can have much higher power. As we want less attenuation and better signalto-noise ratio, lower frequency is more suitable for downlink and higher frequency is commonly used for uplink
sorry for being ambigous
 
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View attachment 60306 Can someone explain me the b part
u knw Debrogleis wavelength ...every moving particle has a wavelength associated with it ...photons incident on the surface have a wavelength of 590nm.....
and this wavelength is inversely proportional to momentum of photons
thus λ = h/ p where h is plancks constant = 6.63*10^-34
therfore momentum of one one photon would b = 6.63*10^-34 / 590nm = 1.12*10^-27 kgm/s
since we need the total momentum of photons ...we multiply it with the no. of photons
(9.5*10^15) * (1.12*10^-27) = 1.06*10^-11 kgm/s
 
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In 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 image.jpg
 
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c)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
Thanks a million
 
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