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Mila [183]
2 years ago
10

A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

is the centripetal acceleration of the tip of the blades at t=0~\text{s}t=0 s?
Physics
1 answer:
iogann1982 [59]2 years ago
5 0

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

a=\omega^2 r

At the instant t = 0,

\omega=3.93 rad/s

So the centripetal acceleration of the tip of the blades is

a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2

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shutvik [7]

Period = 6 seconds and frequency = 0.167Hz .

<u>Explanation:</u>

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of  \frac{1}{t} . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency =   \frac{1}{t}

⇒ frequency = \frac{1}{time}

⇒ frequency = \frac{1}{6}

⇒ frequency = 0.167Hz

Therefore , Period = 6 seconds and frequency = 0.167Hz .

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3 years ago
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inysia [295]

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land.  Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

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2 years ago
The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the
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The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

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Learn more about elastic potential energy: brainly.com/question/156316

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