We found the centripetal acceleration of the earth as it revolves around the sun. compute the centripetal acceleration of a point on the surface of the earth at the equator caused by the rotation of the earth about its axis. (enter the magnitude. the radius of the earth is 6,371 km.)

This question involves the concepts of centripetal acceleration and linear speed.

The centripetal acceleration of a point on the surface of the Earth at the equator caused by rotation of the Earth about its axis is "0.033 m/s²".

The centripetal acceleration is given by the following formula:

where,

a = centripetal acceleration = ?

r = radius = 6371 km = 6371000 m

T = time for a single rotation = (24 hrs)(3600 s/1 hr) = 86400 s

s = distance travelled in single rotation = 2πr = 2π(6371000 m)

s = 40,030,173.6 m

v = linear speed =

Therefore,

a = 0.033 m/s²

Learn more about centripetal acceleration here:

link

The centripetal acceleration by the rotation of the Earth is 3.31 10⁻² m/s²

Explanation:

The equation for centripetal acceleration is

      = v² / r

Where magnitude v is the distance it travels in a complete rotation; distance is the distance of a circle

       V = d / t = 2π r /T

With R the radius of the earth and T the period (1 day)

       T = 1 day (24h / 1day) 3600s / 1h)

       T = 86400 s

      V = 2 π 6.371 10⁶/86400

      V = 4.6 10² m / s

        = (4.6 10²)² / 6.371 10⁶

        = 3.31 10⁻² m/s²

The centripetal acceleration by the rotation of the Earth is 3.31 10⁻² m/s²

explanation:

areas w steep slopes, west coast, rocky mountains, and appalachian mountains

cell differentiation  is how generic embryonic  cells  become specialized  cells. this occurs  through a process called gene expression. gene expression is the specific combination of genes that are turned on or off (expressed or repressed), and this is what dictates how a  cell  functions.

explanation: