To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

Here



Mass inside the orbit in terms of Volume and Density is

Where,
V = Volume
Density
Now considering the volume of the star as a Sphere we have

Replacing at the previous equation we have,

Now replacing the mass at the gravitational acceleration formula we have that


For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

At the same time the general expression for the centripetal acceleration is

Where
is the orbital velocity
Using this expression in the left hand side of the equation we have that



Considering the constant values we have that


As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.
So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
I believe it is A bit not completely sure