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2 years ago

A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.

Physics

1 answer:

Serggg [28]2 years ago

7 0

Answer:

(a) α = 35.20 rad/s^2

(b) θ = 802°

(d) a = 156.64 cm/s^2

Explanation:

(a) To find the angular acceleration of the disc you use the following formula:

$\alpha=\frac{\omega-\omega_o}{t}$ (1)

w: angular speed of the disc = 31.4 rad/s

wo: initial angular speed = 0 rad/s

t: time = 0.892s

You replace the values of the parameters in the equation (1):

$\alpha=\frac{31.4rad/s-0rad/s}{0.892s}=35.20\frac{rad}{s^2}$

The angular acceleration of the disc, for the given time, is 35.20rad/s^2

(b) To calculate the angle describe by the disc in such a time you use:

$\theta=\frac{1}{2}\alpha t^2$ (2)

$\theta=\frac{1}{2}(35.20rad/s^2)(0.892s)^2=14.00rad$

In degrees you have:

$\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°$

The angle described by the disc is 802°

(c) To calculate the tangential speed of the microbe for t=0.892s, you use the following formula:

$v=\omega r$ (3)

w: angular speed for t = 0.892s = 31.4rad/s

r: radius of the disc = 4.45cm

$v=(31.4rad/s)(4.45cm)=139.73\frac{cm}{s}$

The tangential speed is 139.73 cm/s

(d) The tangential acceleration is calculated by using the following formula:

$a=\alpha r$

α: angular acceleration for t=0.892s

$a=(35.20rad/s^2)(4.45cm)=156.64\frac{cm}{s^2}$

The tangential acceleration is 156.64cm/s^2

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Explanation:

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Read 2 more answers

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2 years ago

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netineya [11]

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Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

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