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nirvana33 [79]
2 years ago
14

A Blu-ray disc is approximately 11 centimeters in diameter. The drive motor of a Blu-ray player is able to rotate up to 10,000 r

evolutions per minute. (a) Find the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates
Physics
1 answer:
stepladder [879]2 years ago
8 0

Answer:

the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates is 57.6 m/s

Explanation:

Given information:

diameter of the disc, d  = 11 cm, r = 5.5 cm = 0.055 m

angular speed ω = 10000 rev/min = (10000 rev/min)(2π rad/rev)(1/60 min/s)

                             = 1000π/3 rad/s

to calculate the maximum angular speed we can use the following formula

ω = v/r

v = ωr

  = (1000π/3)(0.055)

  = 57.6 m/s

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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r
Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

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Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

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\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

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3 0
3 years ago
The amount of work done by two boys who apply 200 N of force in an
Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

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           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

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A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
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The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

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so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

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<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

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