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2 years ago

How many buttons in an airbus a380 cockpit

Engineering

2 answers:

AlekseyPX2 years ago

5 0

Answer:

Explanation:

A330 has over 200

Ket [755]2 years ago

3 0

Explanation:

It depends on the cockpit. A glider for instance wouldn't have many buttons at all, whereas something like an Airbus A330 has over 200 buttons on the overhead panel, not including the circuit breakers. Generally the larger the aircraft, the more buttons it has.

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Pilot testing:

andrew11 [14]

Answer:

c. allows planners to work out any problems before the program is launched

Explanation:

Pilot testing is simply aimed at getting it right before the launch of a program, it is also called pilot run, pilot project, feasibility run, etc. Pilot testing is the rehearsal or practice done for an idea, program, research study or invention with few participants prior to lunching out the main program. The main purpose of pilot testing is to determine how feasible a project is, it can also help to evaluate the cost of an idea, invention, research study, etc.

6 0

2 years ago

Which of the following terms is defined as small bumps and slashes within a fluid power system?

-BARSIC- [3]

Answer:

friction

Explanation:

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7 0

2 years ago

Read 2 more answers

An engine cylinder has a stroke of 320mm and bore 280MM. Calculate the mass of air contained in the cylinder if it is filled wit

belka [17]

Answer:

The mass of the air is 0.0243 kg.

Explanation:

Step1

Given:

Stroke of the cylinder is 320 mm.

Bore of the cylinder is 280 mm.

Pressure of the air is 101.3 kpa.

Temperature of the air is 13°C.

Step2

Calculation:

Stroke volume of the cylinder is calculated as follows:

$V=\frac{\pi }{4}d^{2}L$

$V=\frac{\pi}{4}\times(\frac{280}{1000})^{2}\times(\frac{320}{1000})$

V = 0.0197 m³.

Step3

Assume air an ideal gas with gas constant 287 j/kgK. Then apply ideal gas equation for mass of the air as follows:

PV=mRT

$m=\frac{PV}{RT}$

$m=\frac{101.3\times 1000\times 0.0197}{287\times (13+273)}$

m= 0.0243 kg.

Thus, the mass of the air is 0.0243 kg.

4 0

2 years ago

Air flows through a device in which heat and work is exchanged. There is a single inlet and outlet, and the flow at each boundar

Pepsi [2]

Answer:

11548KJ/kg

10641KJ/kg

Explanation:

Stagnation enthalpy:

$h_{T} = c_{p}*T + \frac{V^2}{2}$

given:

cp = 1.0 KJ/kg-K

T1 = 25 C +273 = 298 K

V1 = 150 m/s

$h_{1} = (1.0 KJ/kg-K) * (298K) + \frac{150^2}{2} \\\\h_{1} = 11548 KJ / kg$

Answer: 11548 KJ/kg

Using Heat balance for steady-state system:

$Flow(m) *(h_{1} - h_{2} + \frac{V^2_{1} - V^2_{2} }{2} ) = Q_{in} + W_{out}\\$

Qin = 42 MW

W = -100 KW

V2 = 400 m/s

Using the above equation

$50 *( 11548- h_{2} + \frac{150^2 - 400^2 }{2} ) = 42,000 - 100\\\\h_{2} = 10641KJ/kg$

Answer: 10641 KJ/kg

c) We use cp because the work is done per constant pressure on the system.

7 0

2 years ago

Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific vol

ivanzaharov [21]

Answer:

(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg

Explanation:

Solution

T₁ = 35°C = 308

the first step to take is to Use the Table A-17: Ideal gas properties for air:

Now,

At T₁ = 308 K

V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)

So,

V₁ =219.818 kJ/kg

Thus,

Vr₁ = 596 + [308-305/310-305] (572.3 - 596)

= 581.78

so,

Vr₂/Vr₁ = 1/10

Vr₂ =58.178

Applying Table A-17, at Vr₂ = 58.178

Then,

(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)

T₂ = 747.5 and K = 474.5 °C

V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)

so,

V₂ =549.996 kJ/kg

Hence,

(b) q - w = v₂ -v₁

= 0 -w = 549.996- 219.818

w = 330.178 kJ/kg

4 0

2 years ago