Solution to problem 7.9.10 from the collection of Kepe O.E.

Problem 7.9.10 from the collection of Kepe O.?. is related to the topic of analytical geometry and is formulated as follows:

Given points A(-2, 1), B(3, 5) and C(5, -3). Find the coordinates of point D such that AD is the median of triangle ABC.

To solve the problem, you can use the properties of the medians of a triangle. In particular, the median divides the side of the triangle into two equal parts, and also intersects the opposite side at a point that divides it into two segments proportional to the lengths of the adjacent sides.

Using these properties, you can find the coordinates of point D as follows:

1. Let's find the coordinates of the middle of side BC connecting points B and C. To do this, you can use the formulas for finding the coordinates of a point lying on a segment defined by two points: x_D = (x_B + x_C) / 2 = (3 + 5) / 2 = 4 y_D = (y_B + y_C) / 2 = (5 - 3) / 2 = 1

Thus, the coordinates of point D are (4, 1).

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Problem 7.9.10 from the collection of Kepe O.?. is formulated as follows:

“Given two equal weight and inextensible ropes, each of length $L$. At a distance $l$ from one of the ends of each rope, a weight of mass $m$ is suspended from them. The ropes are thrown over a block that can move without friction along a horizontal surface. What minimum mass $M$ should the block have so that after releasing the weight system the ropes do not cut the block?

To solve this problem, it is necessary to calculate the tension forces in the ropes and the direction of movement of the block when releasing the loads. Using the equilibrium conditions for the block, the required mass of the block can then be determined.

When hanging weights on ropes, the tension force in each rope becomes equal to $T = \frac{mg}{2\cos\theta}$, where $m$ is the mass of the load, $g$ is the acceleration of gravity, $l$ - the distance from the end of the rope to the load, and $\theta$ is the angle formed by the rope with the horizon.

After releasing the loads, the tension force in the ropes will be equal to $T = \frac{Mg}{2}$, where $M$ is the mass of the block.

The minimum block mass that will not cause the ropes to be cut is $M_{min} = \frac{2m}{\cos\theta}$. With this mass of the block, the tension force in the ropes will be equal to the tension force with suspended loads.

Thus, the solution to problem 7.9.10 from the collection of Kepe O.?. consists in calculating the tension force in the ropes with suspended loads, determining the tension force in the ropes after releasing the loads, and calculating the minimum mass of the block that will not lead to cutting the ropes.

Solution to problem 7.9.10 from the collection of Kepe O.?. consists in determining the polar angle of a point at the time when its radius is 1 m. For this, the polar radius equation r = sin?t is given, and it is also known that d?/dt = 0.4 rad/s at t0 = 0 and ?0 = 0.

To solve the problem, it is necessary to find the moment of time t when r = 1m. To do this, we solve the equation sin ?t = 1, from which we obtain ?t = π/2 + 2πn, where n is an integer. Since we are interested in the value of the polar angle at time t0 = 0, we can take n = 0.

This means that at time t0 = 0 we have ?0 = π/2. Next, to find the value of the polar angle at time t, it is necessary to integrate the equation d?/dt = 0.4 rad/s from t0 = 0 to t:

? - ?0 = 0.4(t - t0) ? - π/2 = 0.4t ? = 0.4t + π/2

Substituting the found value of the polar angle ?t = π/2 into the expression for ?(t), we obtain:

π/2 = 0.4t + π/2 0.4t = 0 t = 0

Thus, the value of the polar angle is ? at the moment of time when r = 1m, it is equal to 0.2 rad.

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