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LenKa [72]
4 years ago
7

An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angu

lar velocity of 2000 rad/s. The engine’s rotation slows with an angular acceleration of magnitude 80.0 rad/s2. (a)Determine the angular velocity after 10.0 s. (b) How long does it take the rotor to come to rest?
Physics
1 answer:
kondor19780726 [428]4 years ago
8 0

Answer:

(a) ωf = 1200 rad/s

(b) t  = 25 s

Explanation:

Kinematic Equation to the rotor

ωf= ω₀ + α*t  Formula (1)

Where:

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf : final angular velocity  ( rad/s)

(a) Angular velocity after 10.0 s

Data

α = -80 rad/s²

t = 10 s

ω₀ = 2000 rad/s

We apply the formula (1):

ωf = ω₀ - α*t

ωf = 2000 -  80*(10)

ωf = 2000 - 800

ωf = 1200 rad/s

(b) Time the rotor stops

Data

α = -80 rad/s²

ω₀ = 2000 rad/s

ωf = 0

We apply the formula (1):

ωf = ω₀ - α*t

0 = 2000 -  80*(t)

80*(t) = 2000

t = 2000 / 80

t  = 25 s

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gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

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The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

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So, the total time is 14.303+3.5355s=17.838s

7 0
3 years ago
The power output of a tuba is 0.35 W. At what distance is the sound
STALIN [3.7K]

The distance of the sound from the tuba is 4.82 m.

<h3>Area of the tube</h3>

The area of the tuba is calculated as follows;

I = P/A

where;

  • I is intensity of sound
  • P is power
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A = P/I

A = 0.35 / (1.2 x 10⁻³)

A = 291.67 m²

<h3>Distance of the sound</h3>

Area = 4πr²

r = \sqrt{\frac{A}{4\pi} } \\\\r = \sqrt{\frac{291.67}{4\pi} } \\\\r = 4.82 \ m

Thus, the distance of the sound from the tuba is 4.82 m.

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Answer:

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Δx = v₀ t + ½ at²

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