The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) what is the radial acceleration of an object at the earth’s equator? give your answer in m/s2 and as a fraction of g. (b) if arad at the equator is greater than g, objects will fly off the earth’s surface and into space. what would the period of the earth’s rotation have to be for this to occur?

a) 0.03374 m/s²

b) 0.00343g

c) 5067 seconds or lower.

Explanation:

R = Radius of Earth = 6380000 m

T = Time period

= Radial acceleration

From Kepler's third law

g = Acceleration due to gravity = 9.81 m/s²

Radial acceleration is 0.03374 m/s²

1 g = 9.81 m/s²

Radial acceleration is 0.00343g

b) Here, we have considered

Hence, the time period of the earth’s rotation would be 5067 seconds.

Increasing the radial acceleration would decrease the time period further.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..