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denis23 [38]
2 years ago
13

You jump off a truck and accelerate toward the surface of the Earth. Does the Earth accelerate toward you?

Physics
1 answer:
Triss [41]2 years ago
7 0

When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
  • Gravitational attraction force is GMm/r² between the earth and a person.
  • M= mass of the earth

m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>
  • According to Newton's second law, Force = M×acceleration
  • Acceleration= Force / M
  • Here, Force = GMm/r²,

so acceleration of earth= Gm/r²

  • As this acceleration is very small, so we can't notice it.

Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

brainly.com/question/72250

#SPJ4

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In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en
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Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω =  22.1 rev/minutes = 22.1 \frac{2\pi }{60} = 2.31 rad/s

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the speed of a person "stuck" to the wall

v = ω R

  = 2.31 x 6.4

  = 14.8 m/s

the normal force of the wall on a rider

F = m a

a  = ω^2 R

   =  \frac{v^{2} }{R^{2} } R

   = \frac{v^{2} }{R}

F = \frac{mv^{2} }{R}

  = \frac{(54)(14.8)^{2} }{6.4}

  = 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W =  μ N

m g =  μ \frac{mv^{2} }{R}

g = μ \frac{v^{2} }{R}

μ = \frac{gR}{v^{2} }

  = \frac{(9.8) (6.4)}{14.8^{2} }

  = 0.29

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