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Vadim26 [7]
3 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]3 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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An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w
adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 ×10^{-2} m

diameter d = 0.074 cm = 74 ×10^{-5} m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×10^{-5}×5 ×10^{-2}

area = 1162.3892  ×10^{-5} m²

so here power emitted  is express as

power emitted  = E × σ × area × (temperature)^4

put here all value

power emitted  = 0.300× 5.67 × 1162.3892  ×10^{-5}  × (3068)^4

power emitted = 1.75 W

5 0
3 years ago
The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
mihalych1998 [28]
I believe this is known as wave period.
hope this helps!
7 0
3 years ago
Which statement about velocity is true?
LekaFEV [45]

Answer:

Velocity has both speed and direction. Speed is constant. ... Speed is measured over time.

Explanation:

8 0
3 years ago
To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r
sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}

a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

6 0
3 years ago
1. What layer of the Earth is<br> flowing, solid and plastic?
melisa1 [442]

Answer:

The mantle

Explanation:

The mantle is a plastic solid of varying densities which allow convection currents to flow molten rock towards the earth's surface resulting in volcanic activity, tectonic plate movement, earthquakes, and movement of continents.

5 0
3 years ago
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