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2 answers:
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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.
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:
Aphelion: 6404 W/m2
Perihelion: 14978 W/m2
Explanation:
The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.
The distances we need are the aphelion and perihelion of Mercury.
Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:
Where
e: eccentricity
c: linear eccentricity
a: semi major axis
The linear eccentricity is equal to the distance of the focus of the center of the ellipse.
a = 0.39 AU = 5.83e10 m
In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:
Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m
Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m
And the solar energy fluxes will be: