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sineoko [7]
3 years ago
10

This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa

dcasts (or aliens find us from ours). The closest stars are 4 light years away from us. How far away must you be from a 781 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

2.77287\times 10^{15}\ m

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

\nu = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

The distance is 2.77287\times 10^{15}\ m

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olganol [36]

Answer:

The force is     F_c  =  789.03 \  N    

Explanation:

From the question we are told that

   The tangential  resistive force is F_t  =   115 \ N

   The mass of the wheel is  m  = 1.80 kg

  The diameter of the wheel is  d =  50.0 cm  = 0.5 \ m

   The diameter of the sprocket is  d_c  =  8.50 \ cm =0.085 \ m

  The angular acceleration considered is  \alpha  =  4.30\ rad/s^2

Generally the radius of the wheel is

       r = \frac{d}{2}

=>     r = \frac{0.5}{2}

=>     r = 0.25 \ m

Generally the radius of the sprocket is

       r_c = \frac{d_c}{2}

=>     r_c = \frac{0.085}{2}

=>     r_c = 0.0425 \ m

Generally the moment of inertia of the wheel is mathematically represented as

      I  =  m  *  r^2

=>    I  =  1.80  *  0.25^2

=>    I  = 1.1125 \ kg \cdot m^2

Generally the torque experienced by the wheel due to the forces acting on it  is mathematically represented as

      \tau =  F_c *  r_c  -  F_t  * r

Here  F_c is the force acting on the sprocket

So  

      \tau =  F_c *  0.0425 - 115  * 0.25

       \tau = 0.0425F_c  -  28.75

Generally the torques that will cause the wheel to move with \alpha  =  4.30\ rad/s^2 is mathematically represented as

       \tau  =  I  * \alpha

So

        0.0425F_c  -  28.75  =   I  * \alpha

        0.0425F_c  -  28.75  =   1.1125  *4.30    

       0.0425F_c  -  28.75  =   1.1125  *4.30    

        F_c  =  789.03 \  N    

5 0
3 years ago
The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A
love history [14]

Answer:

Part a)

EMF = 14 \times 10^{-3} V

Part b)

EMF = 15.67 \times 10^{-3} V

Explanation:

As we know that magnetic flux through the loop is given as

\phi = B.A

now we have

\phi = B\pi r^2

now rate of change in flux is given as

\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}

now we know that

A = \pi r^2

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Now plug in all data

EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)

EMF = 14 \times 10^{-3} V

Part b)

Now the radius of the loop after t = 1 s

r_1 = r_0 + \frac{dr}{dt}

r_1 = 0.30 + 0.037

r_1 = 0.337 m

Now plug in data in above equation

EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)

EMF = 15.67 \times 10^{-3} V

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