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shutvik [7]
2 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]2 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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A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d
Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

E=U=\frac{1}{2}kA^2

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

U=\frac{1}{2}kx^2=0 because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

K=E=50.9 J

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

K=\frac{1}{2}mv^2

where m is the mass

Solving the equation for m, we find the mass:

m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

So the kinetic energy is

K=E-U=50.9 J-31.3 J=19.6 J

And so we can find the speed through the formula of the kinetic energy:

K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

K=E-U=50.9 J-31.3 J=19.6 J

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

U=\frac{1}{2}kx^2

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant  the system passes the equilibrium position, which is zero.

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