The linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation is 392.91 m/s
We find the linear speed of a point on the 32nd parallel from v = rω where r = radius of 32nd parallel north = Rcos32° (since it is the radius of the small circle at the 32nd parallel) where R = radius of earth = 6,371,000 m and ω = angular speed of the earth = 2π/T where T = period of earth = 24h = 24 × 60 × 60 s = 86400 s.
So, v = rω
v = Rcos32° × 2π/T
v = 2πRcos32°/T
substituting the values of the variables into the equation, we have
v = 2πRcos32°/T
v = 2π × 6,371,000 m × cos32°/86400 s.
v = 2π × 6,371,000 m × 0.8480/86400 s.
v = 33947512.5035 m/86400 s
v = 392.91 m/s
So, the linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation is 392.91 m/s
Learn more about linear speed here:
https://brainly.com/question/6969329
