3. To determine the speed of the current of a wide and turbulent river, Gavrila and Glafira conducted the following experiment. At a certain point on the shore 100 m from the bridge, they set up a siren emitting sound signals at equal intervals. Glafira took another identical siren and positioned herself at the beginning of the bridge on the same bank. Gavrila got into a motorboat located at the shore, halfway between the first siren and the beginning of the bridge. The experimenters started simultaneously, with the speed of the motorboat relative to the water, as well as the speed of the bicycle, equal to 20 km/h and directed perpendicular to the shore. It turned out that sound signals from both sirens reached Gavrila simultaneously. Determine the speed of the current of the river at a distance of 62 m from the shore. The riverbank is straight, and the speed of the current at each point is directed along the shore.
Answer: 12.4 km/h = 3.44 m/s. Solution. Introduce a coordinate system, the x-axis of which is directed along the shore, the origin of coordinates at the starting point of Gavrila. The siren on the shore has coordinates (L, 0), Glafira is moving along the line x = -L (in our case L = 50 m). Since the experimenters are at the same distance from the shore, the equality of the times needed for the sound signal to travel gives a condition on the coordinates (x, y) of Gavrila: x + L = sqrt((x - L)^2 + y^2).

From this, we obtain 4xL = y^2, which describes a parabola. Differentiation gives a relationship for the speeds: 4UL = 2yV, where V and U are the speed of the boat and the speed of the current at that point, respectively. At a distance h from the shore, we have U = hV / 2L.

Answer to variant 192: 12.24 km/h = 3.4 m/s.

Answer to variant 193: 12.4 km/h = 3.44 m/s.

Answer to variant 194: 12.24 km/h = 3.4 m/s.

## Problem 4

Variant 191 A monatomic ideal gas is the working substance of a heat engine. First, there is an isobaric compression from state 1 to state 2, then an isochoric heating to state 3, the cycle is closed by the process 3-1, in which the pressure is a linear function of the volume. The temperatures in states 1 and 3 are the same, the pressure in the cycle changes by a factor of 2. Determine the efficiency of the heat engine.

Solution. The state parameters are: (p0, nv0), (p0, v0), (np0, v0), n = 2. The work done by the gas over the cycle A = 1/2(n-1)^2 p0 v0.

On the segment 1-2, heat is removed from the system, on 2-3, heat is added (3/2)(n-1) p0 v0. It is necessary to find the point B on the segment 3-1, where the heat supply stops. The equation of the process 3-1: p = np0 - p0(v - v0) / v0 = (n+1)p0 - v p0 / v0, T = p v / R = ((n+1)p0 v - p0 v^2 / v0) / R. Let's find the amount of heat received by the gas during the expansion from state 3 to volume v:

Q(v) = 3/2((n+1)p0 v - p0 v^2 / v0 - np0 v0) + 1/2((n+1)p0 - p0 v / v0 + np0)(v - v0) = 5/2(n+1)p0 v - 2 p0 v^2 / v0 - (5n+1)/2 p0 v0.

Q(v) is a quadratic function, having a maximum at v = 5(n+1)v0 / 8. For n ≥ 5/3, it lies on the segment [v0, nv0]. Thus, in the considered case, the heat supplied to the system is calculated by the formula:

Q+ = [3/2(n-1) + (5n-3)^2 / 32] p0 v0

The final answer:

η = 16(n-1)^2 / [48(n-1) + (5n-3)^2] = 16 / 97

## Answer 16 / 97

Variant 192 A monatomic ideal gas is the working substance of a heat engine. First, there is an isobaric expansion from state 1 to state 2, then an isochoric cooling to state 3, the cycle is closed by the process 3-1, in which the pressure is a linear function of the volume. The temperatures in states 1 and 3 are the same, the pressure in the cycle changes by a factor of 2. Determine the efficiency of the heat engine.

Solution. The state parameters are: (np0, v0), (np0, nv0), (p0, nv0), n = 2. The work done by the gas over the cycle A = 1/2(n-1)^2 p0 v0.

On the segment 1-2, heat is added to the system in the amount 5/2(n-1)np0 v0, on 2-3, heat is removed. It is necessary to find the point B on the segment 3-1, where the heat supply begins. The equation of the process 3-1: p = np0 - p0(v - v0) / v0 = (n+1)p0 - v p0 / v0, T = p v / R = ((n+1)p0 v - p0 v^2 / v0) / R. Let's find the amount of heat received by the gas during the compression from state 3 to volume v:

Q(v) = 3/2((n+1)p0 v - p0 v^2 / v0 - np0 v0) + 1/2((n+1)p0 - p0 v / v0 + p0)(v - nv0) = 5/2(n+1)p0 v - 2 p0 v^2 / v0 - n(n+5)/2 p0 v0.

Q(v) is a quadratic function, having a maximum at v1 = 5(n+1)v0 / 8. For n ≥ 5/3, it lies on the segment [v0, nv0], since Q(nv0) = 0, heat is supplied to the system for v1 ≤ v ≤ nv0. Thus, in the considered case, the heat supplied to the system is calculated by the formula:

Q+ = [5/2(n-1) + (3n-5)^2 / 32] p0 v0

The final answer:

η = 16(n-1)^2 / [80n(n-1) + (3n-5)^2] = 16 / 161

Answer 16 / 161

Variant 3 Analogous to 191 Answer 4 / 15

Variant 4 Analogous to 192 Answer 4 / 31