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krek1111 [17]
2 years ago
14

Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α. express your answer in terms of ω1

and α.
Physics
2 answers:
Ivahew [28]2 years ago
6 0

the time t1 it takes to accelerate:

\large{\boxed{\bold{t_1=\frac{\omega_1 }{\alpha}}}}

<h3>Further explanation </h3>

The speed of the rotating object is called angular velocity. This value is part of circular motion.

The angular velocity is denoted by w including the vector quantity. The unit used is usually rpm (rotation per minute) or radians / sec

Average angular velocity is the ratio of angular displacement to the time interval

\rm \omega=\dfrac{\Delta \theta}{\Delta t}

Circular motion with constant acceleration (angular acceleration denoted by α) can be formulated:

\rm \omega=\omega o+\alpha t

ωo = initial angular velocity

ω = angular velocity after t

 the flywheel accelerate to ω1, so:

wo = 0 because The flywheel is assumed to be at rest at time t = 0

so the equation becomes:

ω1 = ωo + αt

ω1 = 0 + αt

ω1 = αt

\large{\boxed{\bold{t=\frac{ \omega_1}{\alpha }}}

Learn more

the average velocity

brainly.com/question/5248528

resultant velocity

brainly.com/question/4945130

velocity position

brainly.com/question/2005478

VMariaS [17]2 years ago
3 0
T1 = ω1/α.............
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A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
Stella is driving down a steep hill. She should keep her car __________ to help _________. a. light, it speed up in a higher b.
frez [133]

Answer:

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1 year ago
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Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl
timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

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Arrow 3 mass = 90 g   speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

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According to newton's second law

        F = m a

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Then Force is also equal to zero.

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1 year ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
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