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 is Physical Fitness.
To solve this you must set up what is called a proportion. A proportion is a way of comparing two comparing values where one of the four values is missing. In your problem the missing value is the height of the smallest tree in the model.
To set up a proportion, you need all of your values. The easiest way to do this is to list them:
Highest tree in real life: 40ft
Highest tree in model: 10ft
Smallest tree in real life: 4ft
Smallest tree in model: x
So know you can set your proportion like this:
40/4 = 4/x
(When setting up a proportion, you always want to have the values belong to each other. For example don't put the height of the small tree in the model underneath the value of the highest tree in real life.)
So know to find what the x values equals, we need to cross multiply. And then all that's left after that is to solve for x.
40 times x = 4 times 4
40x = 16
x = 2.5
The smallest tree in the model should equal 2.5 feet.
Hope this helps! :)
Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 
Therefore, factor
Thus, the energy will change half times as the result of the collision.
