The boy, shooting from a slingshot, pulled the rubber cord so

A child, using a slingshot, pulled the rubber cord so tightly that its length increased by 10 cm. What was the speed of the 20-gram stone at the moment of the shot? For each centimeter of elongation of the rubber cord, a force of 10 Newtons was required. The flight of the stone was carried out without taking into account air resistance.

The boy, shooting from a slingshot, pulled the rubber cord so

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Hello!

The product description is a video tutorial called "Slingshot for Beginners" that offers training in how to make and use a slingshot. As an example of using a slingshot, a situation is described when a boy pulled a rubber cord until its length increased by 10 cm. Then the question is asked at what speed did a stone weighing 20 grams fly, shot from this slingshot.

To solve this problem it is necessary to use the laws of conservation of energy and momentum. When the rubber cord was tensioned by 10 cm, it was necessary to apply a force of 10 N per centimeter, that is, the total force that the boy applied was 100 N (10 N/cm * 10 cm).

According to the law of conservation of energy, the potential energy of the stretched rubber band turned into the kinetic energy of the stone at the moment of the shot. You can express the speed of a stone using the formula for kinetic energy:

E = 1/2 * m * v^2,

where E is kinetic energy, m is the mass of the stone, v is the speed of the stone.

It is also known that the momentum of a stone is equal to the force multiplied by the duration of this force:

p = F * t.

Since the force is constant and equal to 100 N, and the time of action of the force is equal to the time of flight of the stone, we can express time in terms of the distance that the stone has flown and the speed:

t = d / v,

where d is the distance traveled by the stone.

Thus, we can express momentum in terms of the speed and mass of the stone:

p = F * t = 100 N * (0.1 m / v) = 10 / v kg * m/s.

On the other hand, momentum can also be expressed in terms of the mass and speed of the stone:

p = m * v.

Comparing these two expressions, we get:

m * v = 10 / v.

From here we can express the speed:

v^2 = 10 / m,

v = √(10 / m) = √(10 / 0.02) ≈ 22.36 m/s.

Thus, the speed of the stone at the moment of the shot is approximately 22.36 m/s.

Hope this helped!

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There is no product description. However, I can calculate the speed of the stone that the boy released from the slingshot, and give a detailed solution to problem 10193 on this topic.

Given: Stone mass m = 20 g = 0.02 kg Increasing the length of the rubber cord ΔL = 10 cm = 0.1 m Tension force of the rubber cord F = ΔL * 10 N/cm = 1 N Air resistance is negligible.

Let's find the work done by the boy to pull the rubber cord by ΔL: A = F * ΔL = 1 N * 0.1 m = 0.1 J

Since work A was done due to the potential energy of the rubber cord, then: A = ΔEп = (k * ΔL^2) / 2, where k is the elasticity coefficient of the rubber cord.

Then: k = (2 * A) / ΔL^2 = (2 * 0.1 J) / (0.1 m)^2 = 20 N/m

The speed of the stone can be found using the law of conservation of energy: m * v^2 / 2 = k * ΔL^2 / 2, where v is the speed of the stone.

Then: v = sqrt((k * ΔL^2) / m) = sqrt((20 N/m * (0.1 m)^2) / 0.02 kg) ≈ 10 m/s.

Answer: the speed of the stone released by the boy from the slingshot is about 10 m/s.

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