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Sindrei [870]
3 years ago
9

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

that extends from the axis of the disk to the rim. At t = 0 this line lies along the x-axis and the disk is rotating with positive angular velocity ω0z. The disk has constant positive angular acceleration αz. At what time after t = 0 has the line on the disk rotated through an angle θ?
Express your answer in terms of the variables ω0z, αz, and θ.
Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

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A. What is the basic building block of matter?
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4 0
3 years ago
Read 2 more answers
A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​
iragen [17]

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

M=\frac{y'}{y}=-\frac{q}{p}

where

M is the magnification

y' is the size of the image

y is the size of the object

q is the distance of the image  from the mirror

p is the distance of the object from the mirror

In this problem, we notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

p-q=40 cm (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

y'=\frac{1}{3}y

Solving the equation for p, we find the distance between the object and the mirror:

\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm

So, the mirror is 30 cm from the object.

#LearnwithBrainly

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3 years ago
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
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Answer

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(a).

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c)

Work done

W= m g h + \dfrac{1}{2}m(v_f-v_i)^2

Where V = final speed

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now,

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