The earth's radius is 6.37×106m; it rotates once every 24 hours.
What is the speed of a point on the earth's surface located at 1/4 of the length of the arc between the equator and the pole, measured from the equator? (Hint: what is the radius of the circle in which the point moves?)

Speed of the point = 1.541 × 10⁶ m/h = 428 m/s

Explanation:

Modelling the earth as a sphere,

The equator and the pole have a length of an arc that's fully 90° of the whole 360° circular circumference.

A point located 1/4 of the length of the arc between the equator and the pole has an angle (1/4) × 90° arc = 22.5°.

On the attached image, θ = 22.5°, so Φ = 90° - 22.5° = 67.5°

To obtain the radius of the point on the globe

From the image again, it shows that trigonometric relations can be used to obtain this radius

Sin Φ = (radius of the point)/(radius of the earth)

Sin 67.5 = radius of the point/(6.37 × 10⁶)

Radius of the point = 6.37 × 10⁶ × 0.9239 = 5.885 × 10⁶ m

Speed of the point on the globe = (circumference of the circular path mapped out by the point during rotation)/(24 hours)

circumference of the circular path mapped out by the point during rotation = 2πr = 2π × 5.885 × 10⁶ = 3.6977 × 10⁷ m

Speed of the point = 3.6977 × 10⁷/24 = 1.541 × 10⁶ m/h = 428 m/s

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

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