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olga nikolaevna [1]

4 years ago

7

Which type of device enables its users to see around corners?

2 answers:

Dmitriy789 [7]4 years ago

5 0

Answer:

We can use optical fibers.

Explanation:

The optical fibers are the optical devices which is based on the total internal reflection.

The optical fiber has two layers, one is called core and the other is called cladding.

The refractive index of core is more than the cladding.

As a ray of light falls on the core at more than the critical angle for that pair of media(core and cladding), it suffers the total internal reflection at the interface.

fenix001 [56]4 years ago

4 0

C: fiber optics

Explanation:

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The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

$R = \rho \frac{L}{A}$

Where,

R= Resistance of the conductor

$\rho =$ Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

$A_m = 9.274*10^{-3}m^2$

Using the equation from Resistance we have that at the muscle the value is:

$R_m = \rho \frac{L}{A}$

$R_m = \frac{13*(0.4)}{9.274*10^{-3}}$

$R_m = 560.70\Omega$

At the same time we can make the same process to calculate the resistance of the fat, then

$A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2$

$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

$R_f = \rho \frac{L}{A}$

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

$I= \frac{V}{R}$

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

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The higher frequency waves will be radioactive material, while the electromagnetic radiation generated by circuits will have a much lower frequency.

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What color does the star alpha centauri appear to be?

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B.yellow-white is the answer.

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