Respuesta :
If an object is at a pole such that the apparent weight is mg and the object is in large circular path as the earth rotates. The following blank spaces are filled up with the words, is, n-mg = may =0, equator, n-mg = may>0, smaller, pole.
If an object is at a pole it [is] accelerating, At pole accelaration of gravity is maximum. effect of centrifugal force is less.
We Cam Such that write truly n-mg = may =0 such that the apparent weight is exactly mg, because normal force is equal to effective or apparent weight.
At a the equator, the object is moving in very large circular path as the earth rotates. Because, the axis of earth is line joining two pole so at equator distance from the axis to surface is largest.
This means that technically n-mg = may>0. Because at equator normal force is greater than mg due to the presence of centepetal force.
That is the weight force towards the center of the circular path must be be slightly smaller, if Com provide a Contepetal force.
as, weight force = mg-centepetal force.
weight force <mg
Thus n<mg at the pole, At pole Centepetal force is zero so, weight force and mg are equal.
To know more about equator here
https://brainly.com/question/25235769
#SPJ4
