Answer:
The sphere's volume charge density is 2.58 μC/m³.
Explanation:
Given that,
Radius of sphere R= 8.40 cm
Electric field 
Distance r= 16.8 cm
We need to calculate the sphere's volume charge density
Using Gauss's law
Put the value into the formula
Hence, The sphere's volume charge density is 2.58 μC/m³.
It is the intervention of citizens in decision-making regarding the management of resources and actions that have an impact on the development of their communities.
Let's assume that the gas is behaving ideally. Hence, we can use the ideal gas equation: PV = nRT. We only know the value of the volume. Since it was not mentioned, let's assume that the temperature at those altitude do not have a significant difference. So, we can assume temperature, as well as the number of moles is constant, because it is a closed system. The other parameter, pressure, can be calculated in terms of height:
P = ρgh, where ρ is the air density equal to 0.0765 lb/ft³, g is the acceleration due to gravity equal to 32.2 ft/s², and h is the height in feet. Let's solve for the pressure, P₁.
P₁ = ( 0.0765 lb/ft³)(32.2 ft/s²)(99 feet) = 243.8667 lb/ft-s²
For consistency, let's convert V₁ to ft³ using the conversion 12 inches=1 foot.
V₁ = (6 in³)(1 foot/12 inches)³ = 0.003472 ft³
Then, let's use this to find nRT.
P₁V₁ = nRT
nRT = (243.8667 lb/ft-s²)(0.003472 ft³)
nRT = 0.8467 lb-ft²/s²
Now, let's do the same procedure for P₂:
P₂ = ( 0.0765 lb/ft³)(32.2 ft/s²)(33 feet) = 81.2889 lb/ft-s²
Then, we use the ideal gas equation knowing that nRT=0.8467
P₂V₂ = nRT = 0.8467 lb-ft²/s²
81.2889 lb/ft-s²(V₂) = 0.8467 lb-ft²/s²
V₂ = 0.010416 ft³
Finally, let's convert this to in³:
V₂ = 0.010416 ft³ (12 in/1ft)³
V₂ = 18 ft³
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