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Rzqust [24]
3 years ago
5

Which of the following is the class of ingredient which is the base for most baked items?

Engineering
1 answer:
Ierofanga [76]3 years ago
3 0

Answer:flour

Explanation:

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Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______
Alina [70]

Answer:

Multiplying impulse response by t  ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

5 0
3 years ago
Low-level coding means that...
svet-max [94.6K]

Answer:

a.) a component item is coded at the lowest level at which it appears in the BOM structure is the correct answer.

Explanation:

  • Low-level coding is a kind of programming language used in BOM structures and it carries basic commands that are identified by a computer.
  • The two types of low-level coding are
  • Assembly language.
  • machine language.
  • The advantages of using low-level coding are programs develop by using low-level code are very memory effective and quick and there no need to use interpreters for the conversion of the source to machine code.
7 0
3 years ago
Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances
lidiya [134]

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

4 0
2 years ago
Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

4 0
3 years ago
Read 2 more answers
Using the degree day method calculate the annual kwh use in springfiled il with a heat loss of 12kwh an inddor twmperature of 7
vladimir2022 [97]

Answer:

The correct solution is "21024 KWh/degree day".

Explanation:

The given query is incomplete. Below is the attachment of complete query is provided.

The given values are:

Indoor design temperature:

= 70°F

Now,

According to the question,

The heat loss annually will be:

= 12\times 24\times 365

= 105120 \ KW

Degree days will be:

= 75-65

= 5

Hence,

Annual KWh use will be:

= \frac{Heat \ loss \ annually}{degree \ days}

On substituting the values, we get

= \frac{105120}{5}

= 21024 \ KWh/degree \ day

8 0
3 years ago
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