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wlad13 [49]
2 years ago
8

The A/C compressor will not engage when the A/C is turned on. The static refrigerant pressure is 75 psi and the outside temperat

ure is 72 degrees F. Technician A says that a poor connection at the pressure cycling switch could be the cause. Technician B says that a faulty A/C clutch coil could be the cause. Who is correct
Engineering
1 answer:
VikaD [51]2 years ago
3 0

In the case above,  poor connection at the pressure cycling switch  and also a faulty A/C clutch coil could be the cause.

<h3>What is likely the reason when an A/C compressor will not engage if A/C is turned on?</h3>

The cause that hinders the A/C Compressor from engaging are:

  • Due to low pressure lockout.
  • Due to a poor ground
  • Due to bad clutch coil.
  • Dur to an opening in the wire that links to the clutch coil.
  • Due to a blown fuse.

Note that the pressure switches is known to be one that control the on/off function of any kind of AC compressor and as such, if there is switch failure, it can hinder the AC compressor from functioning at all.

Therefore, technician A and B are correct.

Learn more about refrigerant pressure from

brainly.com/question/10054719

#SPJ1

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A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung wit
Anna71 [15]

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.            

<h3>What is clockwise torque?</h3>

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

                                             \begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g                                                                         ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g                                                                                                      ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass                                                                               ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force                                                                             ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

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7 0
1 year ago
Does anyone know what this is​
sammy [17]

Answer:

Looks like mold that got frosted over

Explanation:

4 0
2 years ago
Read 2 more answers
A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

7 0
3 years ago
A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?
Charra [1.4K]

Answer:

Angle Iron

Explanation:

Mark brainliest please.

3 0
2 years ago
Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–
Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

C_p = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]

Substitute all parameters in the equation

Δs = 5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]

Δs = 5 kg × 0.14629 kJ/kg.K

    = 0.73145 kJ/K

b)

Δs = m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

           T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = 5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]

    = 5 kg (0.13795 kJ/kg.K)

    = 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0
3 years ago
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