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Alekssandra [29.7K]
3 years ago
8

Explain the importance of water quality in aquaculture business.

Engineering
2 answers:
mars1129 [50]3 years ago
5 0

Answer:

Mantain the image of the business

Explanation:

Good water quality gives good images to others and furthers business

Scorpion4ik [409]3 years ago
4 0

Answer:

this is very important the agricultural because there can be a risk of hight salt concentrations that can limit the amount of water plants can take

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How do Solar Engineers Help Humans?<br> (2 or more sentences please)
Gennadij [26K]
Solar engineers work in any number of areas of engineering products that help harness energy from the sun. They may research, design, and develop new products, or they may work in testing, production, or maintenance. They may collect and manage data to help design solar systems.
8 0
3 years ago
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%

Where:

V_{gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{fe} - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%

Where:

m_{fe}, m_{gr} - Masses of the ferrite and graphite phases, measured in grams.

\rho_{fe}, \rho_{gr} - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%

\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%

If \rho_{gr} = 2.3\,\frac{g}{cm^{3}}, \rho_{fe} = 7.9\,\frac{g}{cm^{3}}, m_{gr} = 3.2\,g and m_{fe} = 96.8\,g, the volume percentage of graphite is:

\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%

\%V_{gr} = 10.197\,\%V

The volume percentage of graphite is 10.197 per cent.

5 0
3 years ago
all of the following are steps in the problem solving process except a. try, b. reflect, c. debug, d. define
IceJOKER [234]

Answer:

a

Explanation:

5 0
3 years ago
*6–24. The beam is used to support a dead load of 400 lb&gt;ft, a live load of 2 k&gt;ft, and a concentrated live load of 8 k. D
lisabon 2012 [21]

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load  Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

A_{ymax} = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

3 0
3 years ago
A piston–cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine t
drek231 [11]

Answer:

See explanation

Explanation:

Given:

Initial pressure,

p

1

=

15

psia

Initial temperature,

T

1

=

80

∘

F

Final temperature,

T

2

=

200

∘

F

Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

R

=

0.04513

Btu/lbm.R

C

v

=

0.158

Btu/lbm.R

Find the work done during the isobaric process.

w

1

−

2

=

p

(

v

2

−

v

1

)

=

R

(

T

2

−

T

1

)

=

0.04513

(

200

−

80

)

w

1

−

2

=

5.4156

Btu/lbm

Find the change in internal energy during process.

Δ

u

1

−

2

=

C

v

(

T

2

−

T

1

)

=

0.158

(

200

−

80

)

=

18.96

Btu/lbm

Find the heat transfer during the process using the first law of thermodynamics.

q

1

−

2

=

w

1

−

2

+

Δ

u

1

−

2

=

5.4156

+

18.96

q

1

−

2

=

24.38

Btu/lbm

7 0
3 years ago
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