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

the correct answer is

explanation

the given function is

the excluded values are obtained by solving:

we factor to obtain:

the excluded values are

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x