Answer:
Part 1 
Part 2 
Part 3 
Explanation:
Given
Number of protons 
Radius of nucleus 
Distance of the electrons 
Part 1
Electric field produced by just outside its surface

Part 2
Electric field produced by just outside its surface

Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1 
Part 2 
Part 3 
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
Answer:
Organic matter decomposition serves two functions for the microorganisms, providing energy for growth and suppling carbon for the formation of new cells. ... Dead plant residues and plant nutrients become food for the microbes in the soil
Explanation: