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liubo4ka [24]
3 years ago
12

The diesel fuel cooler is being discussed. Technician A says that high-pressure fuel systems generate a lot of heat. Technician

B says that temperature affects the density of the diesel fuel. Which technician is correct
Engineering
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

Both the technicians are correct.

Explanation:

As the pressure increases the temperature will also increase in accordance with the Boyle's law hence greater amount of heat is formed.

Pressure\propto Temperature

When the temperature increases the intermolecular spaces increase as the molecules of the fuel gain energy and their kinetic energy increases. This is limited to temperatures below dissociation/combustion temperature of diesel .

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Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg
babunello [35]

Answer:

0.

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

\sigma_A = -4.10^{-5}C/m^2

\sigma_B= -7.10^{-5}C/m^2

\sigma_C = -3.1^{-5}C/m^2

We find the electric field for each one, this formula is given by,

E= \frac{\sigma_i}{2\epsilon_0}

Substituting each value from the three charged sheets, we have

E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})

E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

The electric field is

E_{NET}= E_A+E_B+E_C

E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_{NET} = 0

Force on each sheet is,

F=E_{NET}\sigma ds

F=0

The total force is 0

5 0
3 years ago
How do you remove a manual transmission?
Citrus2011 [14]

Answer:

Step 1

Elevate the front of the vehicle using the floor jack and support the vehicle with two jack stands. Make sure the vehicle is stable.

Step 2

Disengage all electrical components connected to the transmission. Indicate by marking the position of the drive shaft for its reinstallation. From the output shaft, remove the rear U joint. Jam the cloth to keep the liquid from dripping out of the extension housing.

Step 3

Loosen the shift linkages and the speedometer cable from the transmission manually. Place the transmission jack under the transmission, and then take a socket wrench and remove the support nut, the cross-member, and the rear support insulator from the rear engine. Support the engine with a jack stand and use the transmission jack to withdraw the transmission toward the rear of the vehicle.

Explanation:

4 0
4 years ago
A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?
Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

#SPJ1

5 0
2 years ago
Plz answer all of these questions!
statuscvo [17]

Answer: all you need to to is go to help me with this question.cm

Explanation:

5 0
3 years ago
Identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
Gre4nikov [31]

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.

8 0
3 years ago
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