A form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.
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 )
Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things
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Answer:
P = 7196 [kPa]
Explanation:
We can solve this problem using the expression that defines the pressure depending on the height of water column.
P = dens*g*h
where:
dens = 1028 [kg/m^3]
g = 10 [m/s^2]
h = 700 [m]
Therefore:
P = 1028*10*700
P = 7196000 [Pa]
P = 7196 [kPa]
During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .
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