Question

Помогите пжлста дам 50 баллов Поршень гидравлического пресса площадью 200 см действует на сдавливаемое тело с силой 18 кн. Площадь малого поршня 5 см. Рассчитай, с какой силой действует меньший поршень на масло в прессе. Н. Ответ (округли до десятых): меньший поршень действует с силой _ Н​

Answer 1

Дано:

S1=200 см2=0,02 м2

F1=18 кН=18000 Н

S2=5 см2=0,0005 м2

F2-?

Решение:

F1/S1=F2/S2

F2=F1×S2/S1=18000×0,0005/0,02=450 Н

Ответ: Меньший поршень действует с силой 450 Н.

Answer 2

Hydraulic Press Piston Force Calculation

To calculate the force exerted by the smaller piston on the oil in the press, we can use the principle of Pascal's law, which states that the pressure applied to a fluid in a closed container is transmitted equally in all directions.

Given: - Large piston area (A1) = 200 cm^2 - Force on the large piston (F1) = 18 kN - Small piston area (A2) = 5 cm^2 - Force on the small piston (F2) = ?

Calculation

We can use the formula for pressure:

Pressure (P) = Force (F) / Area (A)

Using Pascal's law, the pressure is the same throughout the fluid. Therefore, we can equate the pressures on the two pistons:

Pressure on large piston (P1) = Pressure on small piston (P2)

We can express the force in terms of pressure and area:

F = P * A

For the large piston: F1 = P1 * A1

For the small piston: F2 = P2 * A2

Solution

We can rearrange the equation to solve for F2:

F2 = P2 * A2

Since the pressure is the same throughout the fluid, we can write:

P1 = P2

Substituting the values, we get:

F1 / A1 = F2 / A2

Solving for F2:

F2 = (F1 * A2) / A1

Now, let's calculate the force exerted by the smaller piston:

F2 = (18 kN * 5 cm^2) / 200 cm^2

F2 ≈ 0.45 kN

So, the force exerted by the smaller piston on the oil in the press is approximately 0.45 kN.