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Degger [83]
2 years ago
11

The refrigerant is being recovered from an A/C system. Five minutes after the recovery process is complete, the low-side pressur

e loses the vacuum and the pressure rises above zero. This condition indicates:
Physics
1 answer:
SashulF [63]2 years ago
8 0

Answer:

The low side pressure of an A/C system losing vacuum and the pressure rising above zero indicates that there is too much refrigerant in the system.

Explanation:

Considering an A/C system, the condenser fan might be malfunctioning if the low side pressure of the air conditioner is excessive. On the other hand, it's also conceivable that the system has been overcharged with refrigerant.

Stated the scenario that the refrigerant of the system was being recovered, it is an indication that the system is merely overcharged. Even with the engine off, you will notice high pressures.

Either too much oil is present, or there is too much refrigerant in the air conditioning system. In either case, until you let some of that pressure out—ideally, a mechanic should do this—the issue won't go away on its own.

To know more about the pressure scenarios related to AC systems, refer to:

brainly.com/question/17072827

#SPJ4

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On the way home from school, Taylor's car runs out of gas. He has to walk 25m north and 10m west in order to reach the nearest g
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Answer:

<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>

Explanation:

<u>Displacement  and Distance</u>

These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is calculated independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:

d = 25 m + 10 m = 35 m

To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.

The magnitude of the displacement is the distance from (0,0) to (-10,25):

D=\sqrt{(25-0)^2+(-10-0)^2}

D=\sqrt{625+100}=\sqrt{725}

D = 26.93 m

The distance is 35 m and the magnitude of the displacement is 26.93 m

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