Question

З видовження пружини на 0.012 м виникла сила пружності 0,004 кН. Яка виникне сила пружності, якщо пружину видовжили на 24 мм 100 балів. ​

Answer 1

Ответ: 8 Н

Объяснение: Див. додаток

Answer 2

З видовження пружини на 0.012 м виникла сила пружності 0.004 кН. Яка виникне сила пружності, якщо пружину видовжили на 24 мм?

Calculation

To calculate the force of elasticity, 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. The formula for Hooke's Law is:

F = k * x

Where: - F is the force exerted by the spring, - k is the spring constant, and - x is the displacement or elongation of the spring.

In this case, we are given that the elongation of the spring is 0.012 m and the force of elasticity is 0.004 kN. We need to find the force of elasticity when the spring is elongated by 24 mm.

First, let's convert the elongation from millimeters to meters: 24 mm = 24/1000 m = 0.024 m

Now, we can rearrange the formula to solve for the force of elasticity: F = k * x 0.004 kN = k * 0.012 m

To find the spring constant (k), we divide both sides of the equation by 0.012 m: k = 0.004 kN / 0.012 m = 0.333 kN/m

Now, we can substitute the new elongation value into the formula to find the force of elasticity: F = k * x F = 0.333 kN/m * 0.024 m

Calculating the result: F = 0.007992 kN

Therefore, the force of elasticity when the spring is elongated by 24 mm is approximately 0.007992 kN.