Question

СРОЧНОООО....РЕШИТЕ ЗАДАЧИ (с дано и с решением)ПРОШУУУ!!!!! 1-Ракета начинает двигатьсяускоренно с ускорением 35м/с^2,определите путь котoрый она пройдет за 10сек и ее скорость. 2-Тело опускают с высоты 30м без начальной скорости, определить время падения и скорость в конце падения

Answer 1

1.
S=at²/2
S=35*100/2=1750
2.
g=10м/с²
S=at²/2
t=√2S/g
t=√6
S=v/2g*t
v=2gS/t
v=600/√6

Answer 2

Problem 1: Ракета начинает двигаться ускоренно с ускорением 35 м/с^2. Определите путь, который она пройдет за 10 секунд, и ее скорость.

To solve this problem, we can use the equations of motion. The equations we will use are:

1. s = ut + (1/2)at^2 (Equation 1) 2. v = u + at (Equation 2)

Where: - s is the distance traveled by the rocket - u is the initial velocity of the rocket (which is 0 in this case) - a is the acceleration of the rocket (given as 35 m/s^2) - t is the time taken (given as 10 seconds) - v is the final velocity of the rocket

Let's solve the problem step by step:

Step 1: Calculate the distance traveled by the rocket (s): Using Equation 1, we can calculate the distance traveled by the rocket: s = ut + (1/2)at^2

Since the initial velocity (u) is 0, the equation simplifies to: s = (1/2)at^2

Substituting the given values: s = (1/2)(35)(10)^2

Simplifying the equation: s = 1750 meters

Therefore, the rocket will travel a distance of 1750 meters.

Step 2: Calculate the final velocity of the rocket (v): Using Equation 2, we can calculate the final velocity of the rocket: v = u + at

Substituting the given values: v = 0 + (35)(10)

Simplifying the equation: v = 350 m/s

Therefore, the final velocity of the rocket will be 350 m/s.

So, the rocket will travel a distance of 1750 meters and will have a final velocity of 350 m/s.

Problem 2: Тело опускают с высоты 30 м без начальной скорости. Определить время падения и скорость в конце падения.

To solve this problem, we can use the equations of motion again. Since the body is falling freely under the influence of gravity, we can use the following equations:

1. s = ut + (1/2)gt^2 (Equation 3) 2. v = u + gt (Equation 4)

Where: - s is the distance fallen by the body - u is the initial velocity of the body (which is 0 in this case) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - t is the time taken - v is the final velocity of the body

Let's solve the problem step by step:

Step 1: Calculate the time taken for the body to fall (t): Using Equation 3, we can calculate the time taken for the body to fall: s = ut + (1/2)gt^2

Since the initial velocity (u) is 0, the equation simplifies to: s = (1/2)gt^2

Substituting the given values: 30 = (1/2)(9.8)t^2

Simplifying the equation: t^2 = 30 / (1/2)(9.8) t^2 = 6.12 t ≈ 2.47 seconds

Therefore, the time taken for the body to fall is approximately 2.47 seconds.

Step 2: Calculate the final velocity of the body (v): Using Equation 4, we can calculate the final velocity of the body: v = u + gt

Substituting the given values: v = 0 + (9.8)(2.47)

Simplifying the equation: v ≈ 24.17 m/s

Therefore, the final velocity of the body at the end of the fall is approximately 24.17 m/s.

So, the time taken for the body to fall is approximately 2.47 seconds, and the final velocity of the body at the end of the fall is approximately 24.17 m/s.