a level physics wave questions
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ok, QN 1.
A satellite passing the planet Neptune communicates with its controller on earth using a microwave transmitter with output power 22.0W and wavelength 79600 (miu meters). Neptune is 4.35*10^12 m from earth at the time when communication took place.
assuming that the power transmitted by the satellite is radiated uniformly in all directions, calculate the power received on the earth by a dish aerial with effective area 260m^2.
actual power received at dish aerial is 1.2*10^-15 W.
1.)suggest why the actual power received is greated than calculated above.
2.) Calculate the actual rate at which photons of microwaves arrive at the dish aerial.
QN2.)
intensity of a wave depends on the amplitude. the intensity is also proportional to the square of the frequency.
a wave has frequency 3.0HZ amplitude 1.5cm and intensity I.
what is the intensity of a similar wave of freq. 6.0HZ and amplitude 0.5cm???
thanks. my physics abit noob
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I only know F=ma
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dot dot dot that one sec one students also knowOriginally posted by airgrinder:I only know F=ma -
I'm not so sure for the first part... But this is what I would do.Originally posted by unclebutcher:ok, QN 1.
A satellite passing the planet Neptune communicates with its controller on earth using a microwave transmitter with output power 22.0W and wavelength 79600 (miu meters). Neptune is 4.35*10^12 m from earth at the time when communication took place.
assuming that the power transmitted by the satellite is radiated uniformly in all directions, calculate the power received on the earth by a dish aerial with effective area 260m^2.
actual power received at dish aerial is 1.2*10^-15 W.
1.)suggest why the actual power received is greated than calculated above.
2.) Calculate the actual rate at which photons of microwaves arrive at the dish aerial.
Power received = 260 / (4 * pi * Rneptune^2) * 22 = 2.41 * 10^-23 W
1) Why is it greater? Not sure about this...
2) E=hf=hc/lambda for photons mah...
1 second = 1.2*10^-15 J
Hence, got 1.2*10^-15/hf = 480M photons/s -
hey thanks eagle
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This is maths...Originally posted by unclebutcher:QN2.)
intensity of a wave depends on the amplitude. the intensity is also proportional to the square of the frequency.
a wave has frequency 3.0HZ amplitude 1.5cm and intensity I.
what is the intensity of a similar wave of freq. 6.0HZ and amplitude 0.5cm???
intensity directly proportional to amplitude => 1/3 I
intensity proportional to square of freq => 1/3 I * (6/3)^2 = 4/3 I -
teach me if u know 1)Originally posted by unclebutcher:hey thanks eagle -
why is P= area/ what is 4pi Rneptune^2??? i dont get the formula...Originally posted by eagle:I'm not so sure for the first part... But this is what I would do.
Power received = 260 / (4 * pi * Rneptune^2) * 22 = 2.41 * 10^-23 W
2) E=hf=hc/lambda
what is hf and hc/lambda? -
wrong leh...Originally posted by airgrinder:I only know F=ma
F = d (mv)/dt
When mass is constant, it reduces to F=ma -
If Mr Eagle haven't answer by tonite, then I answer tomorrow. Damn sleepy now.
Will give a look tmr! -
for the 1st qn, is it because of the presence of cosmic rays that slightly affects the readings?Originally posted by eagle:I'm not so sure for the first part... But this is what I would do.
Power received = 260 / (4 * pi * Rneptune^2) * 22 = 2.41 * 10^-23 W
1) Why is it greater? Not sure about this...
2) E=hf=hc/lambda for photons mah...
1 second = 1.2*10^-15 J
Hence, got 1.2*10^-15/hf = 480M photons/s -
weird the answer is 4/9IOriginally posted by eagle:This is maths...
intensity directly proportional to amplitude => 1/3 I
intensity proportional to square of freq => 1/3 I * (6/3)^2 = 4/3 I -
I not sure also... I imagined a giant spherical area from the satellite till it reaches Earth. Then I took the fraction of the aerial disc's surface area over the the spherical area, then multiply by 22WOriginally posted by unclebutcher:why is P= area/ what is 4pi Rneptune^2??? i dont get the formula...
Energy of a photon is hf, which is planck's constant * frequencyOriginally posted by unclebutcher:what is hf and hc/lambda?
since c = frequency * lambda (from v = f*lambda),
blah blah blah... u shud understand
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i sec 4 i dunnoOriginally posted by unclebutcher:dot dot dot that one sec one students also know
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oops... Intensity proportional to square of amplitude... You should be able to solve it now...Originally posted by unclebutcher:weird the answer is 4/9I
A bit sleepy... haha -
your answer is right.....
I not sure also... I imagined a giant spherical area from the satellite till it reaches Earth. Then I took the fraction of the aerial disc's surface area over the the spherical area, then multiply by 22W -
OH! you are almost there... I would say rays by other sources, including the suns, the stars, etc etc... They all may contain microwaves, which will increase the total power arriving at the disc...Originally posted by tanjun:for the 1st qn, is it because of the presence of cosmic rays that slightly affects the readings?
Good job! -
I not sure also... I imagined a giant spherical area from the satellite till it reaches Earth. Then I took the fraction of the aerial disc's surface area over the the spherical area, then multiply by 22W
hmm can explain further? i get the multiply by 22w part but i dun get how you arrive at the fraction of (260)/ 4 PI R(neptune)^2 -
Imagine a giant sphere, with the centre at the satellite, the surface at Earth. The rays from the satellite is directed at all parts.Originally posted by unclebutcher:hmm can explain further? i get the multiply by 22w part but i dun get how you arrive at the fraction of (260)/ 4 PI R(neptune)^2
The aerial disc will just be a small part of this giant sphere... So we take by proportion... the amount of energy that reached that particular area... -
ok i get it now, but how come got 4piR^2? shouldnt it be simply piR^2? how come got extra 4? sorry my physics damn noobOriginally posted by eagle:Imagine a giant sphere, with the centre at the satellite, the surface at Earth. The rays from the satellite is directed at all parts.
The aerial disc will just be a small part of this giant sphere... So we take by proportion... the amount of energy that reached that particular area... -
haha... this is maths... surface area of a sphere...Originally posted by unclebutcher:ok i get it now, but how come got 4piR^2? shouldnt it be simply piR^2? how come got extra 4? sorry my physics damn noob -
oOriginally posted by eagle:haha... this is maths... surface area of a sphere...
differentiate 4/3piR^3
=
4piR^2
THANKS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
for ur help!!!!!!!!!!!!!!!!!!
next wk common test!!!!!!!!!!
grr!!!!!!!!!!!!!!!!! -
ok me retire for the day from sgf le
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why greater? Maybe because cosmic rays also fall on the dish?Originally posted by eagle:I'm not so sure for the first part... But this is what I would do.
Power received = 260 / (4 * pi * Rneptune^2) * 22 = 2.41 * 10^-23 W
1) Why is it greater? Not sure about this...
2) E=hf=hc/lambda for photons mah...
1 second = 1.2*10^-15 J
Hence, got 1.2*10^-15/hf = 480M photons/s -
answered above already...Originally posted by menokki:why greater? Maybe because cosmic rays also fall on the dish?