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bulgar [2K]
3 years ago
12

the earth has a radius of 6.38×10^16 meter and turns around once on its axis in 24 hour.what is the radial acceleration of perso

n at the earth equator​
Physics
1 answer:
Scrat [10]3 years ago
5 0

337493603.8m/s²

Explanation:

Radius of the earth = 6.38 x 10¹⁶m

time = 24hr (86400s)

Unknown:

Centripetal acceleration = ?

Solution:

The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.

     centripetal acceleration  = \frac{v^{2} }{r}

  where v is the velocity and r is the radius

   also;

            v  = wr

  where w is the angular velocity

substituting in the equation for centripetal acceleration gives;

                 

          a = w²r

 also w = \frac{2 x pi}{T}

      therefore;

                 a = \frac{4 \pi  ^{2} r }{T^{2} }

 a = \frac{ 4 x 3.142^{2}  x 6.38 x 10^{16} }{86400^{2} }

 a = 337493603.8m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

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3 years ago
Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di
vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

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Explanation:

If we go by the hint; The area of the disk can be expressed as:

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Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

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The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

d = \pi r^2 e

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replacing d = \frac{\pi (\frac{D}{2})^2 }{10, 000} in the equation above; we have:

r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }

r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}

r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}

r_e = 500 ly

The distance (s) between each civilization = 2(r_e)

= 2 (500 ly)

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