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2 years ago

What is the most important part of a successful Election Day?

Engineering

2 answers:

wlad13 [49]2 years ago

8 0

Answer: voting of course

Explanation:

Ymorist [56]2 years ago

7 0

The most successful part of election day is casting your vote by mail or in person.

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Give five general principles involved in the process of sewage filtration?

aleksandr82 [10.1K]

Answer:

Some general principles are given below in the explanation segment.

Explanation:

Sewage treatment seems to be a method to extract pollutants from untreated sewage, consisting primarily of domestic sewage including some solid wastes.

<u>The principles are given below:</u>

Unless the components throughout the flow stream become greater than the ports or even the gaps throughout the filter layer, those holes would be filled as either a result of economic detection.
The much more common element of filtration would be the use of gravity to extract a combination.
Broadcast interception or interference.
Inertial influence.
Sieving seems to be an excellent method to distinguish particulates.

8 0

3 years ago

How many watts are consumed in a circuit having a power factor of 0. 2 if the input is 100 vac at 4 amperes?.

Anna71 [15]

The watts that are consumed is 80 watts.

<h3>What power factor?</h3>

The term power factor has to do with the measure of the efficiency of the use of energy. Recall that power is defined as the rate of doing work. The magnitude of the power factor shows the extent to which the power is used.

Now, to obtain the watts are consumed in a circuit having a power factor of 0. 2 if the input is 100 vac at 4 amperes we have; V × I × PF = 100V × 4A × 0.2 = 80 watts.

Learn more about power factor:brainly.com/question/10634193

#SPJ4

6 0

1 year ago

A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo

Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, $e_{min}$ = 0.51

Void ratio in the loosest state, $e_{max}$ = 0.87

Now,

Dry density, $\gamma_d=\frac{\gamma_t}{1+w}$

$=\frac{18}{1+0.05}$

= 17.14 kN/m³

Also,

$\gamma_d=\frac{G\gamma_w}{1+e}$

here, G = Specific gravity = 2.7 for sand

$17.14=\frac{2.7\times9.81}{1+e}$

e = 0.545

Relative density = $\frac{e_{max}-e}{e_{max}-e_{min}}$

= $\frac{0.87-0.545}{0.87-0.51}$

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

S = 0.2477

S = 0.2477 × 100% = 24.77%

7 0

3 years ago

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

$A=\frac{\pi d^{2}}{4}$