Answer:
V = 4.63 10² m / s
ac = 3.37 10⁻² m/s²
Explanation:
This is a uniform circular motion, so we can calculate centripetal laceration of the object
ac = v² / r
Where v is the speed of the object and r the radius of the Earth, 6.37 106 m.
a) The speed of the object can be calculated from the distance traveled in the time called period (T)
V = d / t
T = 24 h (3600 c / 1 h) = 86400 s
Circle length d = 2π r
V = 2π 6.37 10⁶/86400
V = 4.63 10² m / s
With this data we calculate the acceleration of the body
ac = (4.63 10²)² / 6.37 10⁶
ac = 3.37 10⁻² m/s²
To express the acceleration as a percentage of the acceleration of gravity we divide the two magnitudes
Fraction = ac / g
Fraction = 3.37 10⁻⁻² / 9.8
Fraction = 3.4 10⁻³
ac = 3.4 10⁻³ g
b) The acceleration vector is directed to the center of the circle which, in this case, being in Ecuador coincides with the center of the Earth
c) In this case the person continues to perform a circular movement, in a constant time of 24 h, what has changed is that since we are on a sphere the circle radius for 35ºN is different from the radius of the Tiera.
Let's use some trigonometry
sin 35 = r / R
Where r is the radius of the circle and R is the radius of the Earth
r = R sin 35
r = 6.37 10⁶ sin 35
r = 3.65 10⁶ m
We calculate the speed and acceleration at this point
v = 2π 3.65 10⁶/86400
v = 2.65 10² m / s
ac = (2.65 10²)² / 3.65 10⁶
ac = 1.9 10⁻² m/s²
d) The centripetal acceleration of the person at 35º is directed to the center of the circle and the acceleration of the mass is directed to the center of the earth, in this has one of 35 below the horizontal.
