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Veseljchak
3 months 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:
Sav [3.1K]3 months ago
5 0

Answer:

The overall length of the spiral, designated as L, is calculated to be 5378.01 m

Explanation:

Provided information:

Inner radius R1=2.5 cm

and outer radius R2= 5.8 cm.

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

The total length of the spiral can be computed 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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If a radio wave has a period of 1 μs what is the wave's period in seconds
Softa [3030]

Answer:

10^{-6} s


The period of a wave is the duration it takes to complete one full oscillation, such as from one peak to the next trough.

Since the period is expressed in microseconds, it needs to be converted into seconds.

The conversion is:

1\mu s=10^{-6} s


Accordingly, the wave's period in seconds is 10^{-6} s
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7 0
4 months ago
A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36
ValentinkaMS [3465]

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.

5 0
4 months ago
A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
kicyunya [3294]

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

Now, the potential for the smaller droplets is given as:

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

7 0
3 months ago
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