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Annette [7]
3 years ago
11

Briefly explain how a jet moves

Engineering
1 answer:
sergejj [24]3 years ago
8 0

Answer:

Jet engines move the airplane forward with a great force that is produced by a tremendous thrust and causes the plane to fly very fast. All jet engines, which are also called gas turbines, work on the same principle. The engine sucks air in at the front with a fan. ... Spinning the turbine causes the compressor to spin.

Explanation:

yw

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Type the correct answer in the box. Spell all words correctly.
Semmy [17]

Answer:

Anne is a mechanical engineer.

Explanation:

There are many different types of engineers <em>(mechanical, industrial, electrical, chemical and civil) </em>around the world.<em> </em>When a job involves <em>designing machines</em>, it falls under <u>mechanical engineering</u>. Such job requires a person to be as creative as possible, so he/she can come up with an innovative design that will be reasonable when used.

Designing machines includes<em> spacecraft</em>s. Therefore, Anne is a mechanical engineer working for NASA.

3 0
4 years ago
Which two factors does the power of a machine depend on? A. work and distance B. force and distance C. work and time D. time and
pshichka [43]
Work and time because the power depends on how much it can do in so much time
8 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 30 . It has been determ
UNO [17]

Answer:

fracture will occur since ( 31.8 Mpa√m ) is greater than the K_{Ic of the material ( 30 Mpa√m )

Explanation:

Given the data in the question;

To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.

On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.

so we first of all solve for the parameter Y in the condition where fracture occurred.

K_{Ic = 30 Mpa√m

σ = 237 MPa

2α = 2.78 mm = 2.78 × 10⁻³ m  

so

Y = K_{Ic / σ√πα

we substitute

Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )

Y =  (30 Mpa) / (237)( 0.06608187 )

Y = 30 / 15.6614

Y = 1.9155

Next we solve for Yσ√πα for the second case;

σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m

so

Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )

= 1.9155 × 355 × 0.0467269

= 31.8 Mpa√m

so

( 31.8 Mpa√m ) > K_{Ic ( 30 Mpa√m )

Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the K_{Ic of the material ( 30 Mpa√m )

4 0
3 years ago
Which of these is not applicable to tuning for some Overshoot on Start-up? Select one: 1)-No overshoot during normal modulating
alekssr [168]

Answer: 5) KC-0.78 KU

Explanation:

 KC-0.78 KU is not defined for tuning of overshoot on the Start-up as, this method is only applicable for there is some turning constants and also there is no overshoot during the normally modulating control. But some overshoot at start up are applicable. For the continuous cycling method, some overshoot is same as for closed loop tuning.    

4 0
4 years ago
A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2.
Viefleur [7K]

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

3 0
3 years ago
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