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Zinaida [17]
3 years ago
6

A Blu-ray disc is approximately 8 centimeters in diameter. The drive motor of the Blu-ray player is able to rotate up to 10,000

revolutions per minute, depending on what track is being read. (a) Find the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates.

Physics
2 answers:
Ray Of Light [21]3 years ago
8 0

Answer:

The question is incomplete. Here is the complete question:

A Blu-ray disc is approximately 8 centimeters in diameter. The drive motor of the Blu-ray player is able to rotate up to 10,000 revolutions per minute, depending on what track is being read. (a) Find the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates. (b) Find the maximum linear speed (in meters per second) of a point on the outermost track as the disc rotates.

Answer:

(a) 1047.2 rad/sec

(b) 41.9 m/s

Explanation: Please see the attachments below

Elodia [21]3 years ago
7 0

Answer:

1047.193radians/s.

Explanation:

10,000revs/min = 166.6666revs/s

1 revolution = 360 degrees = 2\pi radians

so 10,000 revolutions= 2*\pi *166.666 = 1047.193radians.

so maximum angular speed is.

1047.193radians/second.

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Rzqust [24]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

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    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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You may look at what group they are in
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