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vodka [1.7K]
4 years ago
5

To produce cooling the refrigerant much

Engineering
1 answer:
lawyer [7]4 years ago
7 0
Yes that is correct good job ❤️
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A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistance
yuradex [85]

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

\frac{Vout}{V2 - V1 }  = \frac{R2}{R1}

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

4 0
3 years ago
A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half it
horsena [70]

Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

P_{1}\cdot V_{1} = P_{2} \cdot V_{2} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in litres.

If we know that \frac{V_{1}}{V_{2}} = 2 and P_{1} = 150\,kPa, then the resulting pressure is:

P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}

P_{2} = 300\,kPa

The resulting pressure is 300 kilopascals.

6 0
3 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

F = \tau A

where

\tau = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

where

\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are appli
lakkis [162]

Answer:

Complete question for your reference is below.

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar

Explanation:

Please find attached file for complete answer solution and explanation.

3 0
4 years ago
Explain what is meant by the "ductile-brittle transition" of steel and how this relates to the cracking problems with the Libert
makkiz [27]

 

Explanation:

A material can convert from ductile into brittle material by following reasons

1.At low temperature

2.Due to notches present in materials

When temperature is very low then a ductile material starts to behave like brittle material and it leads to failure of the material.

When material consists large number of notches then it losses it ductility and behave like brittle materials.

During the world war second ships are failed because temperature of sea became to low and due to this steel (ductile material) converted into brittle material and failed without giving any warning.

8 0
3 years ago
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