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nevsk [136]
1 year ago
9

3. (5%) you would like to physically separate different materials in a scrap recycling plant. describe at least one method that

might be used to separate polymers, aluminum alloys, and steels from one another.
Engineering
1 answer:
Likurg_2 [28]1 year ago
8 0

One of the methods that are used to separate polymers, aluminium alloys, and steels from one another is the Gravitation Separation method.

One straightforward technique is to run the mixture through a magnet, which will keep the steel particles on the magnet and separate them from the polymer.

What is the Gravitation Separation method?

When it is practicable to separate two components using gravity, i.e., when the combination's constituent parts have different specific weights, gravity separation is a technique used in industry. The components can be in suspension or in a dry granular mixture.

Polymers, Steel and Aluminium alloys can be readily split apart. The technique depends on how the two components are combined. The approach used is gravitational density. Due to the significant difference in relative specific mass values between steel and polymers (which range from 1.0 to 1.5), it is possible to separate them using flotation in a liquid that is safe and has the right density.

Therefore, the Gravitation Separation method is used to separate polymers, aluminium alloys and steels.

To learn more about the Polymer from the given link

brainly.com/question/2494725

#SPJ4

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A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is V
Radda [10]

Answer:

a) 0.0663 m³/s

b) 3.312 N/(m/s)²

c) 16.665 m/s

d) 0.1105 m³/s

Explanation:

See attached pictures.

3 0
3 years ago
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0
3 years ago
What do Engineers aim to do
Basile [38]

Answer:

Engineers design, evaluate, develop, test, modify, install, inspect and maintain a wide variety of products and systems.

Explanation:

6 0
2 years ago
Read 2 more answers
A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

4 0
2 years ago
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