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QuestionEl radio del Sol es 6.96 x 108 m? A una distancia de la Tierra, 1ot1 , Intensidad de Ia radiacion solar medida por salelites sobre Ia atmostera e5 3,87 :102...

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QuestionEl radio del Sol es 6.96 x 108 m? A una distancia de la Tierra, 1ot1 , Intensidad de Ia radiacion solar medida por salelites sobre Ia atmostera e5 3,87 :1026 Jis , Cual es lemperatura en Ia superlicie del Sol? Asuma que el Sol tiene una emisividad de 5790 K 4570 K 1.12 X 1015 8000 KQuestionLa masa de una molecula hello es 6 64 x 10-27 kg: Silas moleculas del helio se encuentran una temperatura de 333 Ken Un conlenedor ccual es Ia velocidad rmS de Ias moleculas de helio? 1441 m/s 1599 m/

Question El radio del Sol es 6.96 x 108 m? A una distancia de la Tierra, 1ot1 , Intensidad de Ia radiacion solar medida por salelites sobre Ia atmostera e5 3,87 :1026 Jis , Cual es lemperatura en Ia superlicie del Sol? Asuma que el Sol tiene una emisividad de 5790 K 4570 K 1.12 X 1015 8000 K Question La masa de una molecula hello es 6 64 x 10-27 kg: Silas moleculas del helio se encuentran una temperatura de 333 Ken Un conlenedor ccual es Ia velocidad rmS de Ias moleculas de helio? 1441 m/s 1599 m/s 2000 mts 2881 ns



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The solar corona is a hot ( 2 MK) extended atmosphere surrounding the Sun's cooler visible surface. The coronal gas pressure is about $0.03 \mathrm{Pa} .$ What's the coronal density in particles per cubic meter? Compare with Earth's atmosphere.

It's exercise. We're gonna talk about the energy of a photon. So remember that E the energy is equal to age blanks, constant times the frequency f off the photo, where age is 6.63 times 10 to the minus 34 jewels. Second, mhm. Also remember that the power delivered by an electromagnetic sources equal to the energy divided by the time over which the power was delivered overreach. The the energy was delivered, actually, and in our case, specifically we have to consider an antenna that works with the power of 50,000 watts. So whenever that one watch is equal to one Jew, your second and also the frequency of the electromagnetic radiation emitted by the antenna is equal to 870 kg hertz. And given this information, our goal is to find how many photons per second I'm going to call it an are emitted by this antenna. So I noticed that the energy it made it since, uh, it's the power is 50,000 Jews per second. The energy in one second is 50,000 Jew. Okay, so I'm going to write it as five times into the fourth, feels this is a total energy. Now, what is the energy of a single Fulton? I'm gonna call it a gamma. That's H f. So that's 6.63 times 10 to the minus 34 jewels. Second F is 870 1 kg. Hurt is 10 to the minus three hurts. Okay, This means that the gamma is equal to five point 57. I'm sorry. 77 times. Stand to the miners. 28 Jews and the number of photons will be the total energy emitted divided by the energy of a single Fulton. Okay, so this is five times 10 to the fourth visuals, divided by 5.77 times, 10 to the minus 28 shoe. And this is equal to 8.66 times 10 to the 31 Fulton's. This concludes our problems.

The incident. Power is given by I A cause I intento equals 10 to the minus 26. What for? Mater Square times. The area is by our square. So far, e the radius is three or five over to major square of that time school sign of zero degree solving. This comes out to be certain foreign three times 10 to the negative. 22. What?

You know what is the intensity off? A radio wave that has powered off 5000 once and it is being received at 4 to 40 light years away. So the intensity goes inverse off the Rangers square. In this case, we have intensity is equal to power, divided by the area, the area off the sphere around the source. So this be divided by four by our square. And this gives us 50,000 divided by four by and the radius is for to or three. I'm miss 9.46. Time stands the 15 I swear. And this gives us the intensity off. True 0.47. Time stands the managed Terry. What? Ramires. Where?

Hello, everyone and this problem. We're told that there is a broadcasting station for television show that is a meeting radio signals over a hemisphere, and you're asked to find various quantities about these waves, such as the radiation pressure it exerts on a completely reflecting home five kilometers away from the station. So we have the distance to the home is five kilometers, where five times 10 33 m speed of flight is three times and 3 m per second. As usual, the electric primitive ITI is 8.85 times in three minus 12 for aspirin meters. The power that the station emits debate with is 316 times 23 watts were 316 kilobytes. And because it's free because it's radiating signal over hemisphere, not the entire sphere, you have that the area over which this power distribute is distributed is a half times four by R squared. So the first thing we are as to find is the pressure to radiation pressure provided that the home is a completely reflecting service. Okay, so for completely respectful service service, we have the radiation. Pressure is two times the intensity divided by the speed of flight. Okay, but we're not giving the intensity. So how do we find the intensity? Well, the intensity is the power over the area. So it's the It's the energy transmitted per unit, time per unit area. And so we're giving the area as well because we're no the distance. So we could just work out I two b p over a half times four pi r squared. I'm not going to give a value for this. We can just leave it in terms of symbols and just work with that and simplify everything and put into values at the very end. Okay, so given that this just the intensity and this is the radiation pressure that we have to calculate, we can combine the two to find that the radiation pressure is two times p over two pi r squared, oversee So here to choose, cancel and then we can multiply this one oversee into the pi r squared. Let me have that. The radiation pressure is the power divided by pi r squared times the speed of flight and putting in the values from your bob Here we find that the radiation pressure is 1.34 times and threw minus 11. Paschal's okay, so that's the answer to part A. On the next board were asked to find the electric and magnetic field amplitude. So we know that there is a relationship between these amplitude is that says that the maximum electric field amplitude is equal to see times or the speed of light times deep magnetic field amplitude. So given this relationship, it's actually enough to work out just what IMAX is. And in terms of IMAX, we know that the intensity is given by ah half times Epsilon zero has the speed of flight times imax squared Serie arranging this equation over here for IMAX find that imax the square root of two times I divided by absolute zero times. See? And then again, we use our trusty formula from up here for the intensity. So I just use this I over here again Choose cancel. We're gonna have that Thea Electric field amplitude is squaring off the power divided by pi r squared times. Absolutely. You can see putting it all the values we find that this is equal to 1.23 ball square meters, okay, and then to find a magnetic field amplitude. We just simply divide this value over here by sea. And we get that The magnetic field amplitude is 4.10 times 10 to the minus nine. Tesla's Okay, easy enough. And then, in part, Sceviour has to find what the average energy density toe wave carries is And so that is given by, you know, the energy density. You is absolute zero times e uh, he squared, but the M X squared. But because we're doing an averaging, you're averaging the oscillations over time. And so we pick up a factor of half. So you bar the average energy density is a half times episode zero times um, X squared and that works have to be equal to 6.70 times 10 to the minus 12 jewels for meters cube. Okay. And then in the last party were asked to find, Whereas what percentage off the of this energy density is carried by the electric and the magnetic fields, and that is just 50 50. So half of the energy density is carried by the, uh half. The energy density is carried by the electric grave or the electric field and the other half is carried by them in that field. And the reason for this is that we're doing it everything. So obviously, if you're calculating the energy density, strictly speaking, you're going to be finding that the energy density also oscillates, just as do the electric and magnetic fields. But it's always gonna be true that they carry half off the total energy debt.


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