what will be the centipetal acceleration on a man whose mass is 80 kg,when resting on a ground at the equator if the radius of the earth'R' is 6.4x10 power 6 meter.
I will be happy to check your work.
To calculate the centripetal acceleration, we need to use the formula:
Acceleration (a) = (Velocity (v))² / Radius (r)
To find the velocity, we need to consider that the equator completes one full revolution in 24 hours or 86400 seconds. The circumference of a circle is given by the formula:
Circumference = 2πr
Where r is the radius of the Earth. Since the Earth's radius is given as 6.4x10^6 meters, the circumference is:
C = 2π(6.4x10^6) = 40.32x10^6 meters
Now, to find the velocity:
Velocity (v) = Circumference / Time
v = (40.32x10^6m) / (86400s)
v ≈ 465.7 m/s
Now that we have the velocity, we can substitute it into the formula for acceleration along with the radius of the Earth:
Acceleration (a) = (Velocity (v))² / Radius (r)
a = (465.7 m/s)² / (6.4x10^6 m)
a ≈ 0.034 m/s²
Therefore, the centripetal acceleration of a person resting on the ground at the equator, with a mass of 80 kg, is approximately 0.034 m/s².