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Reika [66]
2 years ago
4

Artificial gravity is produced in a space station by rotating it, so it is a noninertial reference frame. The rotation means tha

t there must be a centripetal force exerted on the occupants; this centripetal force is exerted by the walls of the station. The space station in Arthur C. Clarke's 2001: A Space Odyssey is in the shape of a four-spoked wheel with a diameter of 150 m. If the space station rotates at a rate of 2.45 revolutions per minute, what is the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station?
Physics
1 answer:
icang [17]2 years ago
7 0

Answer:

Explanation:

centripetal acceleration = ω² R

ω is angular speed of rotation and R is radius of circular path

ω = 2π n

n is no of rotation per sec

= 2.45 / 60

ω = 2π x  2.45 / 60

= .2564 rad / s

centripetal acceleration = ω² R

= .2564² x 150/2

= 4.93 m /s²

gravitational acceleration provided = 4.93 m /s²

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Answer:

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4 0
2 years ago
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<span>The distance covered by the tectonic plate, in meters, is
</span>d=9000km=9\cdot 10^6 m<span>
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8 0
2 years ago
A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
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In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

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x(t)=at

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jeka94

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