Answer: 0j
Explanation:
At that point potential energy is zero and kinetic energy is maximum.. P. E=mgh=0
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:
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Hey there!
When light changes speed, it REFRACTS.
Your answer is going to be option C.
Hope this helps you.
Have a great day!
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