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user100 [1]
3 years ago
11

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is

spun at 440.9 rad/s 440.9 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds 0.569 seconds , what is the magnitude of the average angular acceleration of the disk?
Physics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

-775 rad/s^2

Explanation:

Knowing the initial and final angular speed is \omega_0 = 440.9 rad/s and \omega = 0 rad/s, respectively. We can use the following formula for equation of motion to calculate the average angular acceleration in t = 0.569 seconds

\Delta \omega = \alpha t

\omega - \omega_0 = \alpha t

0 - 440.9 = \alpha 0.569

\alpha = \frac{-440.9}{0.569} = -775 rad/s^2

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spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o
satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = - \frac{m_{1} m_{2} }{r^2}

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 2 \frac{M}{r^2} = \frac{m}{(d-r)^2}

          (d-r) ² = \frac{m}{2M} r²

           d² - 2rd + r² = \frac{m}{2M} r²

           r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - \frac{7.36 10^{22} }{2 \  5398 10^{24}} = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  \frac  {2d \frac{+}{-}   \sqrt{4d^2 - 4 a d^2}} {2a}

           r = [2d ± 2d \sqrt{1-a}] / 2a

           r = \frac{d}{a}   (1 ± √ (1.65 10⁻³)) =  \frac{d}{a} (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = \frac{3.84 10^8}{0.99385} 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

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