Question

Из реактивной установки массой 0,5 Т, находящейся первоначально в покое, в горизонтальном направлении выбрасывается последовательно две порции вещества со скоростью 100 м/с относительно установки. Масса каждой порции 25 кг. Какой станет скорость установки после выброса второй порции?

Answer 1

М=0,5тонн=500кг

М=500-25=475 кг

Из закона сохранения импульса mv=MV (v - скорость вещества; m - масса вещества; V - скорость тела) получаем:
V=mv/M=100*25/475= м/с.

дальше сам

Answer 2

a=V  

   T     a=15м/с 

               10 с

a=1,5м/с в кв.

S=Vt+at  в кв

             2         S= 15м/с+1,5м/с в кв*100с в кв=240м

Answer 3

Calculation of the Final Velocity of the Reactor

To calculate the final velocity of the reactor after the ejection of the second portion of substance, we can use the principle of conservation of momentum. According to this principle, the total momentum before the ejection is equal to the total momentum after the ejection.

The momentum of an object is defined as the product of its mass and velocity. Therefore, we can write the equation:

Initial momentum = Final momentum

The initial momentum is the momentum of the reactor before the ejection, and the final momentum is the momentum of the reactor after the ejection.

Let's denote the mass of the reactor as M, the velocity of the reactor before the ejection as V, the mass of each portion of substance as m, and the velocity of each portion of substance relative to the reactor as v.

The initial momentum of the reactor is given by:

Initial momentum = M * V

After the ejection of the first portion of substance, the momentum of the reactor changes. The momentum of the first portion of substance is given by:

Momentum of the first portion = m * v

Therefore, the final momentum of the reactor after the ejection of the first portion is:

Final momentum after the first ejection = (M + m) * (V - v)

Similarly, after the ejection of the second portion of substance, the momentum of the reactor changes again. The momentum of the second portion of substance is given by:

Momentum of the second portion = m * v

Therefore, the final momentum of the reactor after the ejection of the second portion is:

Final momentum after the second ejection = (M + 2m) * (V - 2v)

According to the principle of conservation of momentum, the initial momentum of the reactor is equal to the final momentum of the reactor:

M * V = (M + 2m) * (V - 2v)

We can rearrange this equation to solve for the final velocity of the reactor (V):

V = (2m * v) / (M + 2m) + 2v

Now, let's substitute the given values into the equation and calculate the final velocity of the reactor.

Given: - Mass of the reactor (M) = 0.5 T = 500 kg - Velocity of each portion of substance relative to the reactor (v) = 100 m/s - Mass of each portion of substance (m) = 25 kg

Substituting these values into the equation, we get:

V = (2 * 25 * 100) / (500 + 2 * 25) + 2 * 100

Simplifying the equation, we find:

V = 5000 / 550 + 200

Calculating the values, we get:

V ≈ 9.09 m/s

Therefore, the final velocity of the reactor after the ejection of the second portion of substance is approximately 9.09 m/s.