A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

Where

initial moment of inertia and angular velocity

is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

Where is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

The final moment of inertia of the satellite about the centre of mass

Where is the antenna's mass and

I is the length of the antenna

So, the Final rotation rate of the satellite is:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

solve it with the formula 1−0.4=0.6, 0.6Vρg=Vρbg where ρb 

Explanation:

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.Thus, the fraction of immersed volume remains the same.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.When fluid is at rest: f=⇒ρb=0.6ρ.In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.Thus, the fraction of immersed volume remains the same.Body will float with 40% of the volume above water surface.