A uniform thin rod of mass =3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is =0.929 kg·m2 ?

The length of the rod for the condition on the question to be met is

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is

           The mass of each small bodies is  

           The moment of inertia of the three-body system with respect to the described axis is  

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

Where   is the moment of inertia of the rod about the describe axis which is mathematically represented as  

And   the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

Thus  

Hence

=>  

substituting vales  we have  

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards