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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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motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

E = mc^2

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

E = mc^2

4 0
3 years ago
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A Navy vessel is traveling due north during wartime. A torpedo has been launched by an enemy directly toward the stern (rear) of
Anika [276]

Answer:

The correct option is;

B) No, the Navy vessel is slower

Explanation:

The speed of some torpedoes can be as high as 370 km/h. The average speed of a fast Navy vessel is approximately 110 km/h

Therefore, the torpedoes travel approximately 3 times as fast as the (slower) Navy vessel, such that the torpedo covers three times the distance of the Navy vessel in the same time and therefore, if the Navy vessel and the torpedo continue in a straight line (in the same direction) due north the vessel can not outrun the torpedo

Therefore, no the Navy vessel travels slower than a torpedo.

8 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh
lianna [129]

Answer: 3.21 N

Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}

\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg

For weight, we will multiply by g=9.8 m/s^{-2}

weight= 0.328\times9.8=3.21\hspace{1mm}N

Hence, the rock would weigh 3.21 N.

3 0
3 years ago
Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface
tatyana61 [14]

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

3 0
3 years ago
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