Answer:
The net force is zero
Explanation:
When an object is falling, there are two forces acting on it:
- The force of gravity, which is equal to the weight of the object, which pushes the object downwards
- The air resistance, which acts against the motion of the object, so it pulls upward
While the magnitude of the force of gravity is constant, the magnitude of the air resistance increases as the velocity of the falling object increases: at some point of the motion, the air resistance becomes equal in magnitude to the force of gravity. At this point, the net force on the object becomes zero, and according to Newton's second law, the acceleration of the object becomes also zero:
But zero acceleration means that the velocity of the object is now constant: this is known as terminal velocity.
The rod's mass moment of inertia is 5kgm².
<h3>Moment of Inertia:</h3>
The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.
The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.
If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;
or = M / L = dm / dl
dm = (M / L) dl
Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.
![I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BM%7D3L%7D%5B%28%5Cfrac%7BL%5E3%7D%7B2%5E3%7D%20%20%20-%20%5Cfrac%7B-L%5E3%7D%7B2%5E3%7D%20%29%5D%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B1%7D%7B12%7DML%5E2)
Mass of the rod = 15 kg
Length of the rod = 2.0 m
Moment of Inertia, I = 
= 5 kgm²
Therefore, the moment of inertia is 5kgm².
Learn more about moment of inertia here:
brainly.com/question/14119750
#SPJ4
Answer:
Please find detailed explanation of second class levers below
Explanation:
Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).
Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).
I believe the answer you’re looking for is A. Spiraling around the source
