The center of gravity is a SIKKE bro you really thought I’d give the answer of something so simple
Answer:

Explanation:
The planet can be thought as a solid sphere rotating around its axis. The moment of inertia of a solid sphere rotating arount the axis is

where
M is the mass
R is the radius
For the planet in the problem, we have


Solving the equation for R, we find the radius of the planet:

3.11 is the answer I think
<h2>Hey there! </h2>
<h2>Natural Resource:</h2>
- The resources, for example, sunshine, air, water, soil, plants, animals, minerals, rivers etc. which after natural formation remain distributed on the earth, are known as natural resource.
- The natural resources are found in solid, liquid or gaseous states, and in metallic or non metallic form.
<h2>
Hope it help you </h2>
Answer:
E. two times the original diameter
Explanation:
Resistance of a wire is:
R = ρ L/A
where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
For a round wire with diameter d:
R = ρ L / (¼ π d²)
The two wires must have the same resistance, so:
ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)
The wires are made of the same material, so ρ₁ = ρ₂:
L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)
The new length is four times the old, so 4 L₁ = L₂:
L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)
1 / (¼ π d₁²) = 4 / (¼ π d₂²)
Solving:
1 / (d₁²) = 4 / (d₂²)
(d₂²) / (d₁²) = 4
(d₂ / d₁)² = 4
d₂ / d₁ = 2
So the new wire must have a diameter twice as large as the old wire.