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shutvik [7]
3 years ago
8

A compact disc stores music in a coded pattern of tinypits 10^(-7) m deep. the pits are arranged in a track thenspirals outward

toward the rim of the disc; the inner and theouter radii of this spiral are 25.0 mm and 58.0 mmrespectively. as the disc spins inside a CD player, the trackis scanned at a constant linear speed of 1.25 m/s
a)What is the angular speed of the CD when scanning theinnermost part of the track? the outtermost part of thetrack?
b)the maximum playing time of a CD is 74.0 minutes. whatwould be the length of the track on such a maximum-duration CD ifit were stretched out in a straight line?
c) what is the average angular acceleration of amaximim-duration CD during its 74.0 minutes playing time? take the direction of rotation of the disc to be positive.
Physics
1 answer:
il63 [147K]3 years ago
5 0

Answer:

Explanation:

v = ω R

v is linear speed and ω is angular speed

ω = v / R

a ) Inner radius = 25 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (25 x 10⁻³ )

= .05 x 10⁻³

= 5 x 10⁻⁵ rad / s

outer  radius = 58 x 10⁻³ m

speed v = 1.25 m/s

ω = 1.25 / (58 x 10⁻³ )

= .0215 x 10⁻³

= 2.15 x 10⁻⁵ rad / s

b )

linear constant speed v = 1.25 m /s

time = 74 min = 74 x 60 s

distance tracked = speed x time

= 1.25 x 74 x 60

= 5550 m

c ) time given

= 74 min = 74 x 60 s

angular acceleration

= (  2.15 - 5 ) x 10⁻⁵ /  (74 x 60 )

= -  6.42 x 10⁻⁹ rad / s²

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our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

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Answer:

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