A is the answer of course
I know that the relationship between altitude and atmospheric density is that the higher the altitude, the lesser the density, and the lower the altitude the higher the density. Lower density float to the top, and higher density is 'heavy' so it comes down
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Answer:
v = 306.76 Km/h
Explanation:
given,
height of the aircraft = 3000 m
differential pressure reading = 3300 N/m²
density of air = 0.909 Kg/m³
speed of aircraft = ?
Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.
using Bernoulli's equation
where ρ is the density of the air at 3000 m
v = 85.21 m/s
v = 306.76 Km/h
