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valentina_108 [34]
3 years ago
15

Determine an expression in standard form for the voltage gain VoVs. Hv(jω)=Vo(jω)Vi(jω)=R2R111+jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R2R111

+jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R2R111−jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R1R211+jωCR2 HintsReferenceseBook & Resources Hint #1

Engineering
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

Applying Kirchhoff's current law at inverting terminal

Explanation:

Detailed explanation is given in the attached document.

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
3 years ago
It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)
ycow [4]

Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

for example ; bronze and brass

so from above discussion it is clear that option (d) will be the correct option

8 0
3 years ago
Read 2 more answers
What is hardness and how is it generally tested?
drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:  

  1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;  

  2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

  3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

   Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

    Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.  

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

    Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a   136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

3 0
3 years ago
Military glorification through mosaics, relief carvings and triumphant arches are often associated with?
emmasim [6.3K]

Military glorification through mosaics, relief carvings and triumphant arches are often associated with roman architectural monument and is the correct choice.

<h3>What is a Monument?</h3>

This can be defined as a structure which is usually large and is used to commemorate the history of an influential person. This helps to serve as an example and also a reminder as to why the performance of good deeds are important

The use of military glorification through mosaics, relief carvings and triumphant arches were also often used by the Romans in other to commemorate war victories and the succession of a new ruler which is known as the emperor.

Read more about Roman monuments here brainly.com/question/15786258

#SPJ1

8 0
2 years ago
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