Question

Ротор мощной паровой турбины делает 1 оборот за 0,02 с. Какова частота его вращения? (полное решение: дано, найти, решение, ответ)

Answer 1

Ответ:

Объяснение:

Дано:

n =  1 оборот

t= 0,02 с

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ν = ?

решение

T = t/n = 0.02 c - это период обращения

ν = 1/Т = 1 / 0,02 = 50 Гц

или 50 оборотов в секунду или 50*60 = 3000 оборотов в минуту

Answer 2

Problem Analysis

We are given that the rotor of a powerful steam turbine makes 1 revolution in 0.02 seconds. We need to find the frequency of its rotation.

Solution

The frequency of rotation is defined as the number of revolutions per unit time. In this case, we need to find the number of revolutions per second.

To find the frequency, we can use the formula:

Frequency = 1 / Time period

The time period is the time taken for one complete revolution. In this case, the time period is given as 0.02 seconds.

Substituting the values into the formula, we get:

Frequency = 1 / 0.02 = 50 Hz

Therefore, the frequency of rotation of the rotor is 50 Hz.

Answer

The frequency of rotation of the rotor is 50 Hz.

Explanation

The frequency of rotation is calculated by dividing 1 by the time period. In this case, the time period is given as 0.02 seconds. Therefore, the frequency is 1 divided by 0.02, which equals 50 Hz.