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MariettaO [177]
3 years ago
14

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 8.0 s, it rotates 35 rad.

Physics
1 answer:
Harman [31]3 years ago
8 0

Answer:

(a) 1.093 rad/s^2

(b) 4.375 rad/s

(c) 8.744 rad/s

(d)  67.845 rad

Explanation:

initial angular velocity, ωo = 0

time, t = 8s

angular displacement, θ = 35 rad

(a) Let α be the angular acceleration.

Use second equation of motion for rotational motion

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

35 = 0 + 0.5 x α x 8 x 8

α = 1.093 rad/s^2

(b)  The average angular velocity is defined as the ratio of total angular displacement to the total time taken .

Average angular velocity = 35 / 8 = 4.375 rad/s

(c) Let ω be the instantaneous angular velocity at t = 8 s

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 1.093 x 8 = 8.744 rad/s

(d) Let in next 5 seconds the angular displacement is θ.

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5

θ = 67.845 rad

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Nesterboy [21]

Answer:

you havent given the full question

but im guessing momentum

momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity or the impetus gained by a moving object.

Explanation:

as the child is pushed, it gathers momentum as its weight allows it be pushed forward, and the velocity is the speed driven by the amount of force the parent pushes on the child whilst they are swinging. The momentum is the result of this action

the equation that links these factors together are

p = mv

p = momentum

m = mass

v = velocity

hope i got it right ._.

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What is the main reason why people use machines
Butoxors [25]

Answer:

Machines are faster and more efficent

Explanation:

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A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal
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Answer:

below

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Net accelerating force becomes  12-8 = 4 N

F = ma

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1 year ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

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