Respuesta :
Answer
given,
diameter of the cylinder = 1600 m
radius = 800 m
acceleration due to gravity = 9.8 m/s²
a = r ω²
[tex]\omega = \sqrt{\dfrac{g}{r}}[/tex]
[tex]\omega = \sqrt{\dfrac{9.8}{800}}[/tex]
[tex]\omega = 0.11\ rad/s[/tex]
we know time period
[tex]T = \dfrac{2\pi }{\omega }[/tex]
[tex]T = \dfrac{2\pi }{0.11}[/tex]
T = 57.12 s
The rotation period that will provide normal gravity will be 57.12 seconds.
From the information given, the diameter of the cylinder is 1600m. Therefore, the radius will be:
= Diameter / 2.
= 1600 / 2
= 800m
Acceleration due to gravity = 9.8m/s²
Since a = rw², we'll use the subject of the formula to find w and this will be:
w = √g/✓r
w = ✓9.8 / √800
w = 0.11 rad/s
Therefore, the rotation period will be:
T = 2π/w
T = 2π/0.11
T = (2 × 3.142) / 0.11
T = 57.12 seconds
In conclusion, the correct option is 57.12 seconds.
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