Solution to problem 3.2.23 from the collection of Kepe O.E.

Of course I can answer in Russian!

Problem 3.2.23 from the collection of Kepe O.?. is formulated as follows:

"Determine at what values ​​of the parameter $a$ the system of equations

$\begin{cases} x^2 + y^2 = a \ xy + x + y = 0 \end{cases}$

has no solutions, has a single solution and has two solutions."

To solve this problem, you can use the substitution method or the method of eliminating unknowns. Substituting $y$ from the second equation into the first, we obtain a quadratic equation for $x$. Solving it, we find the values ​​of $x$, and then, substituting them into the second equation, we find the corresponding values ​​of $y$.

In order for the system to have no solutions, the discriminant of the resulting quadratic equation must be negative. In order for a system to have a single solution, the discriminant must be zero, and in order for the system to have two solutions, the discriminant must be positive.

So, for $a < \frac{9}{4}$ the system has no solutions, for $a = \frac{9}{4}$ the system has a unique solution, and for $a > \frac{9}{4 }$ system has two solutions.

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Problem 3.2.23 from the collection of Kepe O.?. is formulated as follows:

“A board weighing 2 kg lies on a smooth horizontal table. A load weighing 4 kg is placed on the board. The coefficient of friction between the board and the table is 0.2. Determine the friction force between the board and the table if the load is at rest.”

To solve the problem, it is necessary to use the equilibrium conditions of the body, i.e. the sum of all forces acting on the body must be equal to zero.

The force of gravity acting on the load is equal to the mass of the load multiplied by the acceleration of gravity g = 9.8 m/s². The reaction force of the support acting on the board is equal to the combined force of gravity of the board and the load. The friction force between the board and the table is equal to the product of the friction coefficient and the support reaction force.

Thus, the sum of all forces acting on the body is zero:

Фтр - Фтяж - Фр = 0

where Ftr is the friction force between the board and the table, Ft is the gravitational force of the load, Fр is the support reaction force.

Since the load is at rest, the force of gravity and the reaction force of the support are equal in magnitude and directed in opposite directions:

Ftyaz = Fr

Hence,

Ftr = Ftry * friction coefficient = Fр * friction coefficient

Substituting the values, we get:

Ftr = 2kg * 9.8m/s² * 0.2 = 3.92N

Answer: the friction force between the board and the table is 3.92 N.

A square copper frame has a side length of 0.1 m and a resistance of half 5 ohms. The frame is pushed into the region of a magnetic field with an induction of 1.6 Tesla, while the lines of magnetic induction are perpendicular to the plane of the frame.

The frame performs harmonic oscillations in its plane with a frequency of 50 Hz and an amplitude of 0.05 m. It is necessary to determine the maximum value of the current induced in the frame.

To solve the problem, we use Faraday’s law of electromagnetic induction:

?MDS = -dF / dt

where ?MDS is electromotive force, F is magnetic flux, t is time.

The magnetic flux through the frame area can be expressed as follows:

Ф = B * S * cos(a)

where B is the magnetic field induction, S is the area of ​​the frame, α is the angle between the plane of the frame and the direction of the magnetic field.

Since the lines of magnetic induction are perpendicular to the plane of the frame, then α = 90°, and cos(α) = 0. Therefore, the magnetic flux through the frame is zero.

Consequently, ΔMDS induced in the frame is also zero. Consequently, the maximum value of the current induced in the frame will also be zero.

Answer: the maximum value of the current induced in the frame is zero.

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