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Rashid [163]
2 years ago
8

A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever

sers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Physics
1 answer:
Amanda [17]2 years ago
4 0

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

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Explanation:

The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.

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To Polish

Jeśli się uwzględni,

Odległość między dwoma kolejnymi grzebieniami wynosi 20 m

Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami

Zatem jego długość fali wynosi λ = 20 m

Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)

Następnie,

Okres T = 20 sekund

Z równania falowego

v = fλ

Gdzie

v to prędkość

f oznacza częstotliwość, a

λ jest długością fali

Częstotliwość jest związana z okresem

f = 1 / T

Następnie,

v = λ / T

Zatem v = 20/20

v = 1 m / s

Prędkość propagacji fali wynosi 1m/s

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