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:
An acute injury is sudden and severe such as a broken bone. A chronic injury develops and worsens over an extended period of time like shin splints
Explanation:
Conservation tillage practices help reduce soil erosion and maintain soil nutrient levels.
<u>Explanation:</u>
The approach that helps in the reduction of doing tillage practices and also reducing its frequency. this is done for obtaining certain benefits for both environment and economic. This mainly focuses on providing sustainability by leaving some plants remaining in the soil.
It aims in decreasing the emission of gases of greenhouse effects like carbon dioxide. Using these practices helps in reducing the erosion and runoffs. This will promote health of the soil because the nutrients are not take off form the soil due to soil erosion and runoffs.
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
where
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.
