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
d: all of these because more coil will increase the current,voltage,and the electrical field hope this helped
The Indian Ocean.
The rest of your options are not oceans, they are bodies of water. (e.g: lakes,rivers,gulfs)
B. the reason we must wear seat belts
Answer:
An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.
Explanation:
