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solniwko [45]
3 years ago
13

A plant might be emitting some dangerous pollutants that are environmentally harmful, but completely eliminating them would be s

o expensive that the plant would have to close, throwing many local inhabitants out of work. Assuming there is an obligation both to preserve jobs and to protect the environment. What is the best technique that should be used to resolve this problem?
Select one:
a. The convergence and divergence techniques
b. The Utilitarian approach
c. The creative middle way
d. The line drawing technique
Engineering
1 answer:
valentinak56 [21]3 years ago
5 0

Answer: c. The creative middle way

Explanation:

As there is both an obligation to preserve jobs and to protect the environment, a creative middle way which involves compromise would be most effective.

The company involved should process and remove the worst pollutants alone while leaving others so that the process will not be so expensive that they have to close down.

They will do this till a better and more environmentally beneficial solution can be found at which point they can then clean up the previous pollutants with the hope that they have not irrecoverably damaged the environment.

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A single- shear pin connection.

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5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h
frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

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maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

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The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

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