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Vadim26 [7]
2 years ago
15

Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1=2.5cm and finis

hes at radius R2=5.8cm. The distance between the centers of neighboring spiral-windings is 1.6μm(=1.6×10−6m). Part A Determine the total length of the spiraling path. [Hint: Imagine "unwinding" the spiral into a straight path of width 1.6μm, and note that the original spiral and the straight path both occupy the same area.]
Physics
1 answer:
BabaBlast [244]2 years ago
5 0

Answer:

total length of the spiral is L is  5378.01 m

Explanation:

Given data:

Inner radius R1=2.5 cm

and outer radius R2= 5.8 cm.

the width of spiral winding  is (d) =1.6 \mu m =  1.6x 10^{-6} m

the total length of the spiral is L is given as

= \frac{(Area\ of\ the\ spiraing\ portion\ on\ the\ disk)}{d}

=\frac{\pi *(R_2)^2 - \pi*(R_1)^2}{d}

=\frac{\pi *(0.058)^2 - \pi*(0.025)^2}{1.6*10^{-6}}

= 5378.01 m

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Explanation:

A) The distance between the two successive compressions (or rarefactions) is actually called the wavelength of the longitudinal waves.

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6 0
2 years ago
What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s
Doss [256]

The kinetic energy of any moving object is

                           (1/2) (mass) (speed²) .

For the object you described, that's

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
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7 0
2 years ago
if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^
Zarrin [17]
This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:

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4 0
2 years ago
A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is
Luba_88 [7]

Answer:

The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

Explanation:

Given;

radius of the circular loop of wire = 0.5 m

current in circular loop of wire = 2 A

strength of magnetic field in the wire = 0.3 T

τ = μ x Bsinθ

where;

τ is the magnitude of the magnetic torque

μ is the dipole moment of the magnetic field

θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90

μ = IA

where;

I is current in circular loop of wire

A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²

μ = 2 x 0.7885 = 1.571 A.m²

τ = μ x Bsinθ =  1.571 x 0.3 sin(90)

τ = 0.4713 J

Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

4 0
2 years ago
You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of
Leni [432]

Answer:

Explanation:

Given

Speed while running towards east is v_1=3\ m/s

Distance traveled in east direction x_1=120\ m

For Another interval you  run with velocity

v_2=5\ m/s

x_2=240\ m

Total displacement=x_1+x_2

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Time for first interval

t_1=\frac{x_1}{v_1}=\frac{120}{3}

t_1=\frac{120}{3}=40\ s

Time for second interval

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total time t=t_1+t_2

t=40+24=64\ s

average velocity v_{avg}=\frac{x_1+x_2}{t}

v_{avg}=\frac{240}{64}=3.75\ m/s

Therefore average velocity is less than 4 m/s  

7 0
2 years ago
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