The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0° north of the equator.

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = meters per second squared

Step-by-step explanation:

The formula is:

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

Centripetal Acceleration at Equator = meters per second squared

b)

At 30.0° north of the equator:

Now,

Speed =

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

Centripetal Acceleration at 30 degrees north of equator = meters per second squared

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.

8 2/3 is the answer

step-by-step explanation: just use a calculator