Question

Пожалуйста помогите срочно!! Определите, с какой силой действует на ремень безопасности водитель массой m = 100 кг, если после нажатия на педаль тормоза его автомобиль сбрасывает скорость от 108 км/ч до нуля за 6 с на прямолинейном горизонтальном участке дороги. Считайте, что в горизонтальном направлении на водителя действует только ремень безопасности. Помогите пожалуйста. Очень нужна помощь.

Answer 1

Ответ:

500H

Объяснение:

дано:

Vo=108 км/час=30 м/с

V=0

t=6 c

m=100

F-?

F=ma

a=V-Vo/t

a=0-30/6= -5

F=|ma|

F=|100*(-5)|=500H

Answer 2

I can help you with your question.

According to Google Translate, your question is:

Answer in detail. Please help urgently !! Determine the force with which the driver with a mass of m = 100 kg acts on the seat belt, if after pressing the brake pedal his car drops the speed from 108 km / h to zero in 6 s on a straight horizontal section of the road. Assume that only the seat belt acts on the driver in the horizontal direction. Please help. I really need help.

To solve this problem, we need to use the formula for the average force:

$$F_{avg} = \frac{\Delta p}{\Delta t}$$

where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval.

The initial momentum of the driver is:

$$p_i = mv_i$$

where $m$ is the mass of the driver and $v_i$ is the initial speed of the car.

The final momentum of the driver is zero, since the car stops:

$$p_f = 0$$

The change in momentum is:

$$\Delta p = p_f - p_i = -mv_i$$

The time interval is given as 6 s, so we can plug in the values and find the average force:

$$F_{avg} = \frac{-mv_i}{6}$$

We need to convert the speed from km/h to m/s, so we multiply by 1000/3600:

$$v_i = 108 \frac{km}{h} \times \frac{1000}{3600} \frac{m}{km} = 30 \frac{m}{s}$$

The mass of the driver is given as 100 kg, so we substitute these values and get:

$$F_{avg} = \frac{-100 \times 30}{6} = -500 N$$

The negative sign means that the force is opposite to the direction of motion, which makes sense since the seat belt is pulling the driver back to stop him from moving forward.

Therefore, the force with which the driver acts on the seat belt is 500 N. I hope this helps.