Answer:
the last one, It moves away from a mid-ocean ridge.
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
Answer:
W=1705.2 J
Explanation:
Given that
mass ,m= 60 kg
Acceleration due to gravity ,g= 9.8 m/s²
Height ,h= 2.9 m
As we know that work done by a force given as
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 60 x 9.8 x 2.9 J
W=1705.2 J
Therefore the answer will be 1705.2 J.
<span>25,000 miles per hour
hope that this helps</span>