Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
Therefore,
Volume of sphere=
Using the formula
Hence, the density of the compressed sphere=
Option a is correct.
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:
W = 18.88 J
Explanation:
Given that,
Constant force, F = 11.8 N (in +x direction)
Mass of an object, m = 4.7 kg
The object moves from the origin to the point (1.6i – 4.6j) m
We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :
So, the work done by the given force is 18.88 J.
Work= Force in the direction of displacement*displacement.
You know the force in the direction of displacement (horizontally) and the displacement. So,
W=130*11=1430
Therefore, the work done is 1,430 Joules
