Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
Answer
The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.
Explanation
Hope that this helps you and have a great day:)
Answer:
Wheel A.
Explanation:
The lesser the moment of inertia, the greater the angular acceleration. Then, the moments of inertia of each wheel are described below:
Wheel A
Wheel B
The wheel A accelerates faster in response to the torque.
Answer:
Any other language I don't know this language
