Solution to problem 15.5.10 from the collection of Kepe O.E.

15.5.10 The system consists of two rods connected by a sleeve. Rod 1 with mass m1 = 4 kg moves with a speed v1 = 1 m/s in horizontal guides and is bent at point A at an angle of 60°. then the movement sets in motion rod 2 with mass m2 = 2 kg. It is necessary to determine the kinetic energy of the rod system. Solution: Since rod 1 moves along horizontal guides, its kinetic energy is equal to: Ek1 = (m1 * v1^2) / 2 = (4 * 1^2) / 2 = 2 J Rod 2 moves under the influence of the force transmitted from the rod 1 through the bushing. Since the rods are connected by a sleeve, the speed of rod 2 is equal to the speed of rod 1: v2 = v1 = 1 m/s Then the kinetic energy of rod 2 is equal to: Ek2 = (m2 * v2^2) / 2 = (2 * 1^2) / 2 = 1 J Thus, the kinetic energy of the rod system is: Ek = Ek1 + Ek2 = 2 J + 1 J = 3 J Answer: 3 J.

Solution to problem 15.5.10 from the collection of Kepe O..

The solution to problem 15.5.10 from the collection of Kepe O.. is a digital product intended for students and teachers who study mechanics. This file contains a detailed solution to problem 15.5.10 from the collection of Kepe O.., which describes the movement of two rods connected by a sleeve under the action of a force transmitted from one rod to another.

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Solution to problem 15.5.10 from the collection of Kepe O.?. describes the movement of two rods connected by a sleeve under the influence of a force transmitted from one rod to the other. The file contains a detailed solution to the problem, which is presented in the form of text, diagrams and formulas, which allows you to better understand and remember the material.

In this problem, rod 1 with mass m1 = 4 kg moves with a speed v1 = 1 m/s in horizontal guides and is bent at point A at an angle of 60°. The movement sets in motion rod 2 with mass m2 = 2 kg. The rods are connected to each other by a sleeve. It is necessary to determine the kinetic energy of the rod system.

According to the solution to the problem, the kinetic energy of rod 1 is equal to Ek1 = (m1 * v1^2) / 2 = (4 * 1^2) / 2 = 2 J. Also, since the rods are connected by a sleeve, the speed of rod 2 is equal to the speed of rod 1: v2 = v1 = 1 m/s. The kinetic energy of rod 2 is equal to Ek2 = (m2 * v2^2) / 2 = (2 * 1^2) / 2 = 1 J. Then the kinetic energy of the rod system is equal to: Ek = Ek1 + Ek2 = 2 J + 1 J = 3 J.

The answer to the problem is 3 J, but it does not correspond to the value specified in the product description (2.75). This may be due to inaccuracy in calculations or to the use of a different technique for solving the problem.

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The product in this case is the solution to problem 15.5.10 from the collection of problems in physics, authored by Kepe O.?.

The problem considers the motion of a system consisting of two rods connected to each other by a bushing. The first rod has a mass m1 = 4 kg and moves at a speed v1 = 1 m/s in horizontal guides. It is bent at point A at an angle of 60°. The second rod has a mass m2 = 2 kg and is at rest at the initial moment.

When the first rod moves, the bushing transmits movement to the second rod, which begins to move at speed v2. It is necessary to determine the kinetic energy of the system of rods at the moment when the second rod reaches its maximum speed.

To solve the problem it is necessary to use the laws of conservation of energy and momentum. During the solution process, it should be taken into account that the system of rods moves in a horizontal plane and is not subject to external forces.

The result of the calculations is the answer 2.75.

Problem 15.5.10 from the collection of Kepe O.?. refers to the topic of probability theory and sounds like this:

"There are 10 white and 6 black balls in a basket. 4 balls are drawn at random. What is the probability that among them there will be 2 white and 2 black?"

To solve this problem, you need to use a combinatorics formula that calculates the number of combinations of n elements by k:

C(n, k) = n! / (k! * (n-k)!)

where n is the total number of elements, k is the number of elements in the combination.

Thus, to solve the problem, you need to calculate the number of combinations that can be used to select 2 white and 2 black balls from 10 white and 6 black. Then you need to divide this number by the total number of combinations of 16 balls that can be used to extract 4.

So, the number of combinations of 10 white balls, 2 each, and 6 black balls, 2 each:

C(10, 2) * C(6, 2) = (10! / (2! * 8!)) * (6! / (2! * 4!)) = 45 * 15 = 675

Total number of combinations of 16 balls of 4:

C(16, 4) = 16! / (4! * 12!) = 1820

Thus, the probability that the drawn balls will consist of 2 white and 2 black is equal to:

P = C(10, 2) * C(6, 2) / C(16, 4) = 675 / 1820 = 0.37 (rounded to two decimal places).

Answer: The probability that among the 4 balls drawn there will be 2 white and 2 black is 0.37.

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