Question

. Жидкость, налитая в конический сосуд высотой 0,18 м и диаметром основания 0,24 м, переливается в цилиндрический сосуд, диаметр основания которого 0,1 м. Как высоко будет стоять уровень жидкости в сосуде?

Answer 1

Объем цилиндрического сосуда равен V = πr²*h, Конического сосуда - в три раза меньше. Значит, если перелить из конического сосуда в цилиндрический, то высота получится в 3 раза меньше.
радиус конического сосуда равен 0,24 : 2 = 0,12(м)
V=3,14*0,12² *0,18=3,14*0,0144*0,18=0,0081(м³) объем конического сосуда
при диаметре 0,1 м, радиус=0,05м объем конического сосуда будет равен:
V=3,14*0,05²*h=0,0081
h=0,0081 : 0,00785 ≈ 1.03(м) в коническом сосуде при диаметре
1,03 : 3 ≈ 0,343 (м) высота в цилиндрическом сосуде (уровень жидкости)

Answer 2

Calculation of the Liquid Level in the Cylindrical Vessel

To calculate the height at which the liquid level will stand in the cylindrical vessel, we can use the principle of conservation of volume. Since the volume of the liquid remains constant, we can equate the volume of the liquid in the conical vessel to the volume of the liquid in the cylindrical vessel.

Let's denote the height of the liquid level in the cylindrical vessel as h.

The volume of a cone can be calculated using the formula:

V_cone = (1/3) * π * r^2 * h_cone,

where r is the radius of the base of the cone and h_cone is the height of the cone.

The volume of a cylinder can be calculated using the formula:

V_cylinder = π * r^2 * h_cylinder,

where r is the radius of the base of the cylinder and h_cylinder is the height of the cylinder.

Since the volume of the liquid remains constant, we can equate the volumes of the cone and the cylinder:

(1/3) * π * r_cone^2 * h_cone = π * r_cylinder^2 * h_cylinder.

Given the dimensions of the vessels, we can substitute the values into the equation and solve for h.

The conical vessel has a height of 0.18 m and a base diameter of 0.24 m. Therefore, the radius of the base of the cone (r_cone) is half of the diameter, which is 0.12 m.

The cylindrical vessel has a base diameter of 0.1 m. Therefore, the radius of the base of the cylinder (r_cylinder) is half of the diameter, which is 0.05 m.

Substituting these values into the equation, we can solve for h:

(1/3) * π * (0.12)^2 * 0.18 = π * (0.05)^2 * h.

Simplifying the equation:

(1/3) * 0.0144 * 0.18 = 0.0025 * h.

0.002432 = 0.0025 * h.

Solving for h:

h ≈ 0.9736 m.

Therefore, the liquid level in the cylindrical vessel will stand at approximately 0.9736 meters.

Please note that the above calculation assumes that the vessels are completely filled with liquid and that there are no other factors affecting the liquid level.