Question

Определите коеефициент трения между змеей и землей,если змея массой 120 г движится равномерно со скоростью 1 м/с ,при этом сила трения равна 0,15 Н

Answer 1

По второму закону ньютона ma=F+F...
Проектируем
ОХ:  ma=-Fтр
ОУ: 0=N-mg
Fтр=MN (M коэффициент трения)
Fтр=Mmg   M=Fтр/mg= 0.15/10*0.12=0.15/1.2=0.125, только я не понимаю, зачем нужно ускорение

Answer 2

Coefficient of Friction Between Snake and Ground

To determine the coefficient of friction between the snake and the ground, we can use the formula:

Coefficient of Friction (μ) = Frictional Force (F) / Normal Force (N)

Given: - Mass of the snake (m) = 120 g = 0.12 kg - Velocity of the snake (v) = 1 m/s - Frictional force (F) = 0.15 N

First, we need to calculate the normal force (N) acting on the snake. The normal force can be calculated using the formula:

Normal Force (N) = Mass (m) x Acceleration due to gravity (g)

The acceleration due to gravity (g) is approximately 9.81 m/s^2.

Let's calculate the normal force (N): Normal Force (N) = 0.12 kg x 9.81 m/s^2

Now, we can calculate the coefficient of friction (μ) using the formula mentioned above.

Calculation

Normal Force (N) = 0.12 kg x 9.81 m/s^2 = 1.1772 N

Coefficient of Friction (μ) = Frictional Force (F) / Normal Force (N) Coefficient of Friction (μ) = 0.15 N / 1.1772 N

Result

Coefficient of Friction (μ) ≈ 0.1275

Therefore, the coefficient of friction between the snake and the ground is approximately 0.1275.