Explanation:
The given data is as follows.
Fluid is water so, density 
Weight flow rate = 500 lbf/s = 2224.11 N/sec
Cross-sectional area (A) = 
= 0.05184 
Hence, weight flow rate will be given as follows.
w = 
2224.11 N/sec = 
V =
m/s
= 4.373 m/s
Thus, we can conclude that average velocity in the given case is 4.373 m/s.
Answer: 45 joules of energy
Explanation:
c adding research resources during an investigation
Current will be

now just pluf in the values and Voila..