Physics: Post your doubts here!

[My Name]

No

View attachment 60305

Someone please explain
My Name Rizwan Javed qwertypoiu nehaoscar

I will try In sha Allah.

 

[Akshajistari]

I will try In sha Allah.

Thank you!

 

[My Name]

Thank you!

Don't thank me now.I haven't done anything yet.XD

 

[ahmedish]

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?

 

[ahmedish]

View attachment 60299View attachment 60300
HELP WITH C ii please!

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

 

[ashcull14]

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

 

[ashcull14]

No

View attachment 60305

Someone please explain
My Name Rizwan Javed qwertypoiu nehaoscar

the formula is same u just have to take the reflection into account...

 

[ashcull14]

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

 

[funky brat]

Someone please explain this part.

 

[ashcull14]

Someone please explain this part.

http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-119.html
solution 603

 

[taimurahmed]

Someone help me in determining the relation of force with speed. Will we use F=mv^2/r or F=qvB

 

[Mahnoorfatima]

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

 

[ashcull14]

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 View attachment 60323

http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-218.html
Solution 1043

 

[Akshajistari]

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

 

[ahmedish]

Someone help me in determining the relation of force with speed. Will we use F=mv^2/r or F=qvBView attachment 60321 View attachment 60321

speed with force of magnetic field... You will use F=BeV
B and e are constant so its a straight line

 

[ahmedish]

how is no of photons = power / energy of photon?

Each photon contributes to the total power, so power =n * energy of photon
it also depends on rate so the question here matters. bc its not a rule u have to know

Is the power produced in a cell equal to the the total power in the circuit?

Its equal to current through main circuit or cell * Potential difference across cell.

 

[Saad the Paki]

Each photon contributes to the total power, so power =n * energy of photon
it also depends on rate so the question here matters. bc its not a rule u have to know

Its equal to current through main circuit or cell * Potential difference across cell.

By p.d across cell you mean the terminal p.d right?

 

[Midnight dream]

M/J/09/4
Question 9b)
Please anyone.

 

[ahmedish]

By p.d across cell you mean the terminal p.d right?

No. Power lost through the cell is still power generated. The question will specify what it needs, really, so itll be obvious which to include/

 

[funky brat]

http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-119.html
solution 603

Thanks but I dont get this, somehow. :/