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erastova [34]
3 years ago
5

Technician A says that vacuum-operated fuel pressure regulators maintain an equal pressure drop across the fuel injectors all th

e time. Technician B says that intake manifold pressure is high at idle. Which technician is correct?
A) Technician A only
B) Technician B onlyC) Both technicians A and B
D) Neither technician A nor B
Physics
1 answer:
yuradex [85]3 years ago
5 0

Answer:

C) Both technicians A and B

Explanation:

Fuel pressure regulators are a vacuum operated spring-loaded diaphragm that enables a vehicle's fuel delivery system to maintain a constant pressure. When the vehicle is at idle the regulator is open allowing fuel to bypass the delivery system and go back into the tank.

In pounds per square inch, the atmosphere exerts 14.7 PSI at sea level on average. The vacuum inside an engine's intake manifold, by comparison, can range from zero up to 22 inches Hg or more depending on operating conditions. Vacuum at idle is always high and typically ranges from 16 to 20 inches Hg in most vehicles.

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Given the diagram showing gas molecules in different containers, a reasonable inference could be made that there is a relationsh
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Answer:

The correct answer should be A

Explanation:

4 0
3 years ago
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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
A measure of randomness or disorder
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It is simply called Entropy.

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3 years ago
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On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai
xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,

the horizontal component of this initial velocity is given by

v_x = v\cos \theta

v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}

The vertical component is given by

v_y = v\sin\theta

v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}

3 0
3 years ago
A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





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3 years ago
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