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astraxan [27]
3 years ago
6

A(n)______ is a device used to ensure positive position of a valve or damper actuator A. calibrator B. positioner C. actuator D.

characteristic cam
Engineering
1 answer:
zlopas [31]3 years ago
4 0

Answer: C) actuator

Explanation:

Actuator is the device that used to provides the power and manipulate the motion of the moving parts of the valve and damper is used to control the flow of the fluid. Actuator is the device or the mechanism which are used to control valve automatically and valve is a device which is used to control and regulate the fluid by rotating the flow.

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Why do we need technical sketching?
Scrat [10]

Technical Drawings give a better understanding of what is needed and required in the project.

Explanation:

8 0
3 years ago
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A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
3 years ago
Type the correct answer in the box. Spell all words correctly.
Otrada [13]

Answer:

solar engineering field

5 0
3 years ago
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I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original
Liono4ka [1.6K]

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

Aug=[A eye(size(A))]

Above command is tested in MATLAB command window and is attached in figure below

8 0
3 years ago
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