Question

Задача 2. Найдите жесткость пружины, которая под действием силы 50 H удлинилась на 10 см.

Answer 1

Ответ:

500н/м

Объяснение:

F=kx( сила упругости)

k=F/x

k=50Н/0,1м=500н/м

Answer 2

Finding the Stiffness of a Spring

To find the stiffness or spring constant of a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or elongation of the spring. Mathematically, this can be expressed as:

F = k * x

Where: - F is the force applied to the spring - k is the spring constant or stiffness - x is the displacement or elongation of the spring

In this case, we are given that a force of 50 H (H = Newton) causes the spring to elongate by 10 cm (0.1 m). We can use this information to find the spring constant.

F = 50 H x = 0.1 m

Using Hooke's Law, we can rearrange the equation to solve for the spring constant:

k = F / x

Substituting the given values:

k = 50 H / 0.1 m

Calculating the value:

k = 500 H/m

Therefore, the stiffness or spring constant of the spring is 500 H/m.

Please note that the search results did not provide specific information related to the given problem. However, the solution is derived from the well-known Hooke's Law equation, which is widely used to calculate the stiffness of a spring.