Answer:
v=20m/S
p=-37.5kPa
Explanation:
Hello! This exercise should be resolved in the next two steps
1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed
Q=VA
for he exitt
Q=flow=5m^3/s
A=area=0.25m^2
V=Speed
solving for V

velocity at the exit=20m/s
for entry

2.
To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

where
P=presure
α=9.810KN/m^3 specific weight for water
V=speed
g=gravity
solving for P1

the pressure at exit is -37.5kPa
Air pollution level is higher in Location B, because poisonous fumes are produced when coal burns.
Explanation:
It is given that,
Mass of lithium, 
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :


q is the charge on an electron

v = 78608.58 m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :



r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.
882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91