Let's denote the orbital period of planet X as T and its average distance from the sun as A. For planet Y, let its orbital period be T_1, implying that if planet Y's mean distance from the sun is twice that of planet X:
This indicates that the orbital period for planet Y increases by a factor of
It is stated that a straight rod has one endpoint at the origin (0,0) and the opposite endpoint at (L,0), with a linear density defined by , where a is a constant and x is the x coordinate.
Thus, the infinitesimal mass is expressed as:
The total mass can be calculated by integrating the above expression as follows:
Consequently,
Now, we can calculate the center of mass, of the rod as:
Now, it follows that
x_{cm}=\frac{1}{\frac{aL^3}{3}}\int_{0}^{L}ax^3dx=\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}
Therefore, the center of mass, is located at:
Answer:
x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction
Step-by-step explanation:
we understand that
The quadratic equation solution formula for the structure is described as
for this particular equation we recognize
thus
insert values into the equation
therefore
x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction