The acceleration of the object after 3 seconds of fall is -9.8 m/s².
The given parameters;
- initial velocity of the object, u = 0
- time of motion of the object, t = 3 seconds
Acceleration is the change in velocity per change in time of motion.
The acceleration of the object after 3 seconds of fall is calculated as follows;
- Since the object is in free fall, the object experiences only acceleration due to gravity.
- the magnitude of this acceleration due to gravity is 9.8 m/s²
- the direction of this acceleration is downwards
Thus, the acceleration of the object after 3 seconds of fall is -9.8 m/s².
Learn more here: brainly.com/question/13197713
Answer:
See explanation
Explanation:
We have a mass 
revolving around an axis with an angular speed 
, the distance from the axis is 
. We are given:
![\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]](https://tex.z-dn.net/?f=%5Comega%20%3D%2010%20%5Brad%2Fs%5D%5C%5Cr%3D0.5%20%5Bm%5D%5C%5Cm%3D13%5BKg%5D)
and also the formula which states that the kinetic rotational energy of a body is:
.
Now we use the kinetic energy formula
where 
is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:
After replacing in the previous equation we get:
now we have the following:
therefore:
then the moment of inertia will be:
![I = 13*(0.5)^2=3.25 [Kg*m^2]](https://tex.z-dn.net/?f=I%20%3D%2013%2A%280.5%29%5E2%3D3.25%20%5BKg%2Am%5E2%5D)
