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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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Answer:

a

The height is   H  = 6.74 \  m

b

The horizontal distance is  D  = 23.74 \  m

Explanation:

From the question we are told that

  The speed is  v  =  15 \  m/s

  The angle is  \theta  =  40^o

   The height of the cannon from the ground is  h  =  2 m

  The distance of the net from the ground is k  =  1 m

 

Generally the maximum height she reaches is mathematically represented as  

     H  =  \frac{v^2 sin^2 \theta }{2 *  g }  +  h

=>    H  =  \frac{(15)^2 [sin (40)]^2 }{2 * 9.8}  +  2

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Generally from kinematic equation  

    s = ut + \frac{1}{2} at^2

Here s is the displacement which is mathematically represented as

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    u  =  v sin (\theta)

So

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=>  -4.9t^2 + 9.6418t + 1 = 0

using  quadratic formula to solve the equation we have

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=>   D  =  15 cos (40) *  2.07

=>   D  = 23.74 \  m

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