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adell [148]
3 years ago
10

1. How far away must you be from a 675 kHz radio station with power 50.0 kW for there to be only one photon per second per squar

e meter? Assume no reflections or absorption, as if you were in deep outer space. 2. Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.
Physics
1 answer:
goldenfox [79]3 years ago
6 0

a) 0.321 ly

b) 0.321 light years is not far in astronomical terms. Alien life would need to transmit at tremendous power in order for their radio transmissions to be detectable. Their radio signal also needs to be stronger than background noise in order to be distinguishable. Therefore it is unlikely that radio transmissions from alien life will ever be detected.

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Pls help me this is being timed.
GrogVix [38]

Answer:

Increase air resistance

Explanation:

Gravity forces the parachute down but air resistance pushing up on the flat surface of the parachute causes it to fall slower to the ground.

5 0
3 years ago
Only living things have energy T or F?
Angelina_Jolie [31]

Answer:

. living things only have energy : true.

8 0
2 years ago
Read 2 more answers
True or False: If a point charge has electric field lines that point into it, the charge must be positive.
Licemer1 [7]

Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges

8 0
3 years ago
P wave for 100km time= distance/speed , T=100km/6.1km/s
shutvik [7]
I dont know what you are asking sorry can u elaborate
8 0
3 years ago
58.5 million excess electrons are inside a closed surface. What is the net electric flux through the surface?
Anni [7]
The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.

Now let's look at the rest of the problem:

The formula you used is correct:

           Net flux through the surface = (net charge inside) / ε₀

and          ε₀ = 8.85 x 10⁻¹² farad/meter.

What's the net charge inside the surface in this problem ?

It's    (5.85 x 10⁷ electrons) x (the charge on each electron)

     =  (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)

     =      -9.36 x 10⁻¹² coulomb .   

Finally,      (net charge inside) / ε₀

             =  (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)

             =        -1.058  newton-m²/coulomb .

The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing.  The only thing left is
the order of magnitude.  You most likely took one of the negative
exponents and made it positive.  You got an answer that's 10²² too
small.  Big deal.  You could claim "that's close", and see whether you
can convince a teacher.
8 0
3 years ago
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