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Katyanochek1 [597]
2 years ago
12

The rotating blade of a blender turns with constant angular acceleration of 1.51 rad/s^2. How much time does it take to reach an

angular velocity of 36.8 rad/s, starting from rest?
Physics
1 answer:
yawa3891 [41]2 years ago
6 0

time to reach an angular velocity of 36.8 is 24.370 s.

<h3></h3><h3>What is angular acceleration?</h3>

The temporal rate at which angular velocity changes is referred to as angular acceleration. Naturally, there are two forms of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, just as there are two types of angular velocity, namely spin angular velocity and orbital angular velocity. As opposed to orbital angular acceleration, which is the angular acceleration of a point particle around a fixed origin, spin angular acceleration describes the angular acceleration of a rigid body about its centre of rotation.

w(t) = w(0) + α*t

also w(0) =0

=> time = 36.8/1.51= 24.370 s

to learn more about angular acceleration go to - brainly.com/question/21278452

#SPJ4

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  v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

         Em₀ = U = m g h

Final point. Lower

        Em_{f} = K = ½ m v²

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Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

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And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

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y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

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t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

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