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Tech A says that it is best to use a knife or other type of sharp tool to cut away the insulation when

Marina86 [1]

Tech A because it is best to use a knife

3 0

3 years ago

Optimum engine oil pressure at operating temperature and moderate engine load should be __________ ps

kobusy [5.1K]

Answer:

What Is Normal Oil Pressure Temperature? Oil temperatures typically range from 200 degrees to 200 degrees Fahrenheit, depending on the weather. The oil pressure ranges from 100psi to 10-15psi depending on the weather at start up.

Explanation:

3 0

2 years ago

Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volu

Lunna [17]

Answer:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

Explanation:

For this case we know the following info :

$V_i = 0.3 m^3$ represent the initial volume

$V_f = 0.1 m^3$ represent the final volume

We know that the pressure and volume are related with the following expression:

$P = aV^{-2}= \frac{a}{V^2}$

Where a is a constant given $a = 6.5 Kpa m^6$

And we need to calculate the work associated to this process.

We have a compression, and by definition the work is defined with the following general expression:

$W = \int_{V_i}^{V_f} P dV$

If we replace the expression for P we got:

$W = \int_{V_i}^{V_f} a V^{-2} dV$

If we integrate we got:

$W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}$

Using the fundamental theorem of calculus we have:

$W = -a (\frac{1}{V_f} -\frac{1}{V_i})$

And replacing the values we got:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

5 0

4 years ago

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

DiKsa [7]

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

$n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm$

2) The speed of the rotor is the motor speed. The slip is given by:

$Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm$