Question

Самолет выходит из пикирования,двигаясь в вертикальной плоскости по дуге окружности радиусом 1 км. Какова скорость самолета в нижней точке траектории, если летчик испытывал 5-тикратную перегрузку? Помогите,пожалуйста с формулой.

Answer 1

Перегрузка есть N = P/mg
N = 5
Вес, таким образом, есть
P = Nmg.
С другой стороны,
Р = m(g + v²/R), где v  - скорость самолёта в нижней точке траектории
v²/R - центростремительное ускорение
R - радиус дуги траектории.
Т.о.
Ng = g + v²/R
откуда
v = √(gR(N - 1) = √10*1000(5 - 1) = 200 м в сек

Answer 2

To find the speed of the airplane at the bottom point of its trajectory, we can use the formula for centripetal acceleration:

a = v^2 / r

where: - a is the centripetal acceleration - v is the velocity of the airplane - r is the radius of the circular path

In this case, the radius of the circular path is given as 1 km. We need to find the velocity of the airplane at the bottom point of the trajectory.

To find the velocity, we can use the formula for centripetal acceleration in terms of gravitational acceleration and the load factor:

a = g * n

where: - g is the acceleration due to gravity (approximately 9.8 m/s^2) - n is the load factor (in this case, 5)

We can equate the two formulas for centripetal acceleration:

v^2 / r = g * n

Rearranging the equation to solve for velocity v:

v = sqrt(g * n * r)

Now we can substitute the given values into the equation:

v = sqrt(9.8 m/s^2 * 5 * 1000 m)

Calculating the result:

v = sqrt(49000 m^2/s^2) ≈ 221.36 m/s

Therefore, the speed of the airplane at the bottom point of the trajectory, given a 5G load factor and a radius of 1 km, is approximately 221.36 m/s.