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2 years ago

What is the importance of knowing the parts and function of the ignition system in servicing your own vehicle

Engineering

1 answer:

lisov135 [29]2 years ago

4 0

Answer:

El conocer la función del sistema nervioso permite comprender la estructura del cuerpo. Puesto que, toda acción que se se realice en ella dependerá del sistema nervioso para que sea eficaz

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What input is required from ADC to allow the INS to calculate W/V?

Kazeer [188]

TAS is an important input which is required from ADC to allow the INS calculate W/V.

INS is abbreviation for Inertial Navigation System and it can be defined as a navigation device that makes use of motion sensors, a computer, and rotation sensors, so as to continuously calculate by dead reckoning the velocity, position, and orientation of a moving object.

In the Aviation and Engineering filed, True Airspeed (TAS) is an important input which is required from ADC to allow the Inertial Navigation System (INS) calculate W/V.

Read more on Inertial Navigation System here: brainly.com/question/26052911

#SPJ12

4 0

2 years ago

Why is an integrated circuit (IC) referred to as a special device?"<br> points more than 10 plz help

sladkih [1.3K]

Answer:

Because IC microcomputers are smaller and more versatile than previous control mechanisms, they allow the equipment to respond to a wider range of input and produce a wider range of output. They can also be reprogrammed without having to redesign the control circuitry

Explanation:

5 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent

monitta

Answer:

$F_r = 200N$

Explanation:

Given

Let the two forces be

$F_1 = 130N$

$F_2 = 110N$

and

$\tan(\theta) = \frac{12}{5}$

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

This means that we need to first calculate $\cos(\theta)$

Given that:

$\tan(\theta) = \frac{12}{5}$

In trigonometry:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

By comparing the above formula to $\tan(\theta) = \frac{12}{5}$

$Opposite = 12$

$Adjacent = 5$

The hypotenuse is calculated as thus:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 12^2 + 5^2$

$Hypotenuse^2 = 144 + 25$

$Hypotenuse^2 = 169$

$Hypotenuse = \sqrt{169$

$Hypotenuse = 13$

$\cos(\theta)$ is then calculated using:

$\cos(\theta)= \frac{Adjacent}{Hypotenuse}$

$\cos(\theta)= \frac{5}{13}$

Substitute values for $F_1$ , $F_2$ and $cos(\theta)$ in

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

$F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}$

$F_r^2 = 16900 + 12100 + 11000$

$F_r^2 = 40000$

Take square roots of both sides

$F_r = \sqrt{40000$

$F_r = 200N$

<em>Hence, the resultant force is 200N</em>

4 0

3 years ago

A student wants to restate some ideas she found in a journal article by a prominent expert in economics. She combines her own wo

kipiarov [429]

Explanation:

Althought she referenced the article at the end, is imposible to know which part of the article is hers and which part is the expert's so that would be plagiarism.

If she used quotation marks in the words of the expert it would be clear and no plagiarism could be accused.

4 0

3 years ago