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Yanka [14]
3 years ago
5

Ventajas motor avion

Engineering
1 answer:
Sauron [17]3 years ago
4 0

Answer:

Explanation:

Usar motores eléctricos en aviones ofrece numerosas ventajas reales. A diferencia de los motores de combustión interna los motores eléctricos no necesitan aire para funcionar, lo que significa que pueden mantener toda su capacidad y potencia incluso a altitudes elevadas donde el aire es más tenue.

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A mass of 0.3 kg is suspended from a spring of stiffness 0.4 N/mm. The damping is 3.286335345 kg/s. What is the undamped natural
mina [271]

Answer:

 f=5.81 Hz

Explanation:

Given that

m= 0.3 kg

K= 0.4 N/mm

K= 400 N/m

C= 3.28

We know that undamped natural frequency given as

\omega= \sqrt{\dfrac{K}{m}}

Now by putting the values

\omega= \sqrt{\dfrac{K}{m}}

\omega= \sqrt{\dfrac{400}{0.3}}

  ω = 36.51 rad/s

We know that

ω = 2 π f

36.51 = 2 x π x f

f=5.81 Hz

So the undamped natural frequency is 5.81 Hz.

6 0
3 years ago
Say that a variable A in CFG G is necessary if it appears in every derivation of some string w ∈ G. Let NECESSARY CFG = {hG, Ai|
ale4655 [162]

Answer:

Explanation:

solution

8 0
3 years ago
Read 2 more answers
Transcript
posledela

Answer:

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Explanation:

8 0
3 years ago
A converging-diverging nozzle has an area ratio of 5.9. (1) Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd
nata0808 [166]

Answer:

Check the explanation

Explanation:

The Total pressure is the overall of fixed or static pressure p, the dynamic pressure q, as well as gravitational head. Total pressure can also be referred to as the measure of the overall energy of the airstream, and is the same to static pressure plus velocity pressure.

kindly check the step by step solution in the attached image below to Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd critical points.

5 0
3 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0
3 years ago
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