Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
12 m
Explanation:
The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

where
s is the distance
u is the initial velocity
t is the time
a is the acceleration
For this problem,

and
u = 0, since we are considering the first second of motion
So, substituting t = 1 s, we find

<span>In the labeled portion of the curve ,you use the heat of vaporization to calculate the heat absorbed in the 4th portion. It is indicated in the picture that it is the region where vaporization occurs, that is why you need to consider this portion to calculate.</span>
The twice as heavy weight will hit the ground with more force, or impact.