when a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s. (a) what is the magnitude of the angular acceleration of the fan?

Answers

Answer 1

The angular acceleration of the fan is 0.740 rad/s^2,

Angular acceleration which represents the rate at which the angular velocity changes over time. The unit used to measure angular acceleration is radians per square second (rad/s2), according to the International System of Units. The Greek alphabet symbol alpha (α) is used to denote angular acceleration.

To calculate the angular acceleration of the fan, the formula α = Δω/Δt is used. Here, α represents angular acceleration, Δω represents the change in angular speed, and Δt represents the change in time.

In this scenario, Δω is equal to 10.0 - 6.30 = 3.70 rad/s, and Δt is equal to 5.00 s. By substituting these values into the formula, we obtain α = 3.70/5.00 = 0.740 rad/s^2.

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Related Questions

the surface of the sun appears sharp in visible light because

Answers

"The surface of the sun appears sharp in visible light because the photosphere is thin compared to the other layers in the sun."

Most of the electromagnetic energy that reaches the earth begins in the photosphere, the area of the sun that is visible to us. The photosphere is referred to as the sun's surface, despite the fact that it is a gaseous entity.

The gas in the photosphere appears to have a sharp surface, but in reality, it is heavier lower in the Sun and less dense higher up. It is more transparent the less thick it is. The area of the gas that is visible to us is where it has largely become translucent. About 300 km of this layer are deep.

The photosphere is the line separating the core of the Sun from its atmosphere. It is the part of the Sun's surface that is visible to us. The photosphere is not like a planet's surface; even if you could stand in the sun, you couldn't do so on the photosphere.

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in one cycle a heat engine absorbs 480 j from a high-temperature reservoir and expels 320 j to a low-temperature reservoir. if the efficiency of this engine is 56% of the efficiency of a carnot engine, what is the ratio of the low temperature to the high temperature in the carnot engine?

Answers

The ratio of the low temperature to high temperature of the Carnot engine is 2.38.

What is the efficiency of Carnot engine?

The efficiency of the Carnot engine can be defined as the ratio of network done per cycle by the engine to the heat energy absorbed by the engine per cycle by the working substance from the source.

Efficiency = 1 - (Tlow/Thigh)

Heat absorbed by engine = 480J

Heat expelled by engine = 320J

Efficiency of the engine = 56% of efficiency of Carnot engine

The ratio of low temperature to high temperature in the Carnot engine.

Let's assume the efficiency of the Carnot engine is 'ηc' = 1 - T₂/T₁

Where, T₂ = Low temperature and T₁ = High temperature

To calculate the efficiency of the engine given, η = (Q1 - Q2)/Q1

η = (480 - 320)/480

η = 160/480

η = 1/3

η = 33.33%

Now, η = 56% × ηc

0.56ηc = 1/3ηc = (1/3)/0.56 = 0.58

As we already know, ηc = 1 - T₂/T₁

T₂/T₁ = 1 - ηc

T₂/T₁ = 1 - 0.58

T₂/T₁ = 0.42

T₁/T₂ = 1/0.42

T₁/T₂ = 2.38

Therefore, the ratio of low temperature to high temperature in the given Carnot engine with an efficiency of 56% will be about 2.38.

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