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Agata [3.3K]
4 years ago
8

A technician is checking refrigerant system pressures. Both high- and low-side service ports are located on the A/C compressor.

When the compressor is engaged, the high-side pressure instantly goes over 375 psig. What is the most likely cause of this condition
Physics
1 answer:
konstantin123 [22]4 years ago
5 0

Answer:

Two major causes are outline bellow

1. The presence of air in the system

2. Clogged condenser

Explanation:

1. The presence of air in the system

One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.

2. Clogged condenser

Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve

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Calculate the average speed in metres per second from Glasgow to Edinburgh
mariarad [96]
This is the same question as the one previously but with more details, so I will just use my previous answer.

1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.

So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
5 0
3 years ago
A fireman standing on a 15 m high ladder
ludmilkaskok [199]

Answer:

31.6 m/s

Explanation:

Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

Writing v₁ in terms of v₂:

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

Plugging in values:

579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²

v₂ = 31.6 m/s

8 0
3 years ago
What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

3 0
3 years ago
I need help on this pls
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5 0
3 years ago
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An inductor is connected to an AC generator. As the generator's frequency is increased, the current in the inductor A. Increases
Luda [366]

Answer:

B) Decrease

Explanation:

Resistance offered due to inductor in any AC circuit is given as

X_L = \omega L

here we know that

\omega = angular frequency

\omega = 2\pi f

now by ohm's law we know that

i = \frac{V}{X}

here we have

i = \frac{V}{\omega L}

so current is inversely depends on the frequency of AC

so if frequency is increased then the current will decrease

6 0
3 years ago
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