Solution to problem 14.5.18 from the collection of Kepe O.E.

Problem 14.5.18 from the collection of Kepe O.?. (which, as far as I know, is about mathematics) is formulated as follows:

"In triangle ABC, a bisector AL is drawn that intersects side BC at point L. The circumcircle of triangle ABC intersects side BC at point K, different from L. Prove that BL = LK."

Thus, we are talking about a geometric problem related to a triangle and its circumcircle. To solve it, you need to apply the appropriate properties and formulas that are described in geometry textbooks.

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Problem 14.5.18 from the collection of Kepe O.?. is formulated as follows:

"There is a charge q on the surface of a ball of radius R. Determine the electric field strength at a point located at a distance r from the center of the ball (r < R)."

To solve this problem, you can use the formula for the electric field strength of a point charge:

E = k * q / r^2

where E is the electric field strength, k is the Coulomb constant (k = 1 / (4πε0), where ε0 is the electric constant), q is the magnitude of the charge, r is the distance from the point to the charge.

To find the electric field strength on the surface of the ball, it is necessary to take the limit of this formula as r tends to R. We obtain:

E = k * q / R^2

Thus, the electric field strength on the surface of the ball is equal to k * q / R^2.

To find the electric field strength at a point at a distance r from the center of the ball (r < R), it is necessary to use field superposition. At a point located at a distance r from the center of the ball, two charges can be considered: charge q, located on the surface of the ball, and charge -q, located at the center of the ball. Then the electric field strength at this point will be equal to:

E = k * q / r^2 - k * q / R^2

where the first term corresponds to the field strength from the charge on the surface of the ball, and the second - from the charge in the center of the ball.

Thus, the solution to problem 14.5.18 from the collection of Kepe O.?. consists in using the formula for the electric field strength of a point charge and the superposition of fields to find the strength at a point located at a distance r from the center of the ball.

The product is the solution to problem 14.5.18 from the collection of Kepe O.?.

In this problem there is a drum 2 on which threads are wound, to which weights 1 and 3 are attached with a mass of m1 = 2m3 = 2 kg. The moment of inertia of the drum relative to the axis of rotation is equal to I = 0.05 kg • m2.

It is necessary to determine the angular momentum of a system of bodies relative to the axis of rotation if the angular velocity of the system is equal to ? = 8 rad/s, and radii R = 2r =20 cm.

Solving this problem allows us to obtain an answer equal to 1.12.

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