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3 years ago

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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.]

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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Explanation:

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3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

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$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

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$\phi=7.74 rad$

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$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

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3 years ago

What is the moving kinetic energy of a 1500kg car moving at 25m/s?

EleoNora [17]

Answer:

<h2>468,750 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 1500 \times {25}^{2} \\ = 750 \times 625$

We have the final answer as

<h3>468,750 J</h3>

Hope this helps you

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