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adell [148]
2 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]2 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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The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo
Anna [14]

Answer:

the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.

w is angular frequency

formula:2π/T

now k is spring constant

F/R-->mw²

putting values:70*(2π/0.2)²

=4.9x10⁶

so we can say that SHM is not affected by the amplitude of the bounce.

6 0
3 years ago
A 900-N lawn roller is to be pulled over a 5-cm (high) curb. Radius of roller is 25 cm. What minimum pulling force is needed if
Feliz [49]
I believe b 30 degrees 
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Alja [10]
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7 0
3 years ago
Which image illustrates refraction
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4 0
3 years ago
Read 2 more answers
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
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