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Eva8 [605]
4 years ago
5

Which process applies to cleaning refrigerant for immediate reuse by oil separation and single or multiple passes through device

s, like replaceable core filter driers, which reduce moisture and acidity
Engineering
1 answer:
Dovator [93]4 years ago
7 0

Answer:

Recycling.

Explanation:

Recycling can be defined as the process of gathering, processing and conversion of waste materials such as thrash or scraps into new materials or products. Thus, recycling facilitates the reuse of materials that could have been disposed off as waste or thrash.

Hence, recycling is a process which applies to cleaning refrigerant for immediate reuse by oil separation and single or multiple passes through devices, like replaceable core filter driers, which reduce moisture and acidity. This simply means that, refrigerants can be cleaned for reuse by using an oil separation technique to remove the residual refrigerant oil and then a filter-drier to reduce moisture and acidity.

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Jordan's dad made a new recipe for dinner. Jordan looked at the food and saw that it was white, yellow, and purple in color. She
zhenek [66]

Answer:

Jordan used her eyes to see the food, her touch to feel the food, and her nose to smell the food, and lastly, but most importantly, she used her mouth to taste the food.

7 0
2 years ago
A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?
Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0
4 years ago
How many 10" diameter circles can be cut from a semicircular shape that has a 20"
N76 [4]

9514 1404 393

Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

3 0
3 years ago
Read 2 more answers
Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc
wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is  22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know  distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14  + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so  velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = \frac{v^2}{R}

ar = \frac{38.05^2}{63.66}

ar = 22.74 m/s²

so magnitude of total acceleration is

A = \sqrt{a^2 + ar^2}

A = \sqrt{3.37^2 + 22.74^2}

A = 22.98 m/s²

so magnitude of acceleration is  22.98 m/s²

8 0
3 years ago
In order to defend against side channel power analysis, we should: ______________
wariber [46]

Answer:

Some examples of predator and prey are lion and zebra, bear and fish, and fox and rabbit. ... The words "predator" and "prey" are almost always used to mean only animals that eat animals, but the same concept also applies to plants: Bear and berry, rabbit and lettuce, grasshopper and leaf

Explanation:

8 0
3 years ago
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