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:
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Answer:
Transformer
Explanation:
Transformers can increase (step up), or it can decrease (step down) voltage.
Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ = 
let's calculate
x₂ = 
x₂ = 1.33 m
(a)
The velocity of the meteorite just before hitting the ground is:
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
(b) 1 megaton of tnt is equal to
To find to how many megatons the meteorite energy loss
corresponds, we can set the following proportion
And so we find
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of
. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
And so we find
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
