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3 years ago

What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

5 0

Complete Question

The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Answer:

The value is $P = 3.270960 *10^{6} \ W$

Explanation:

From the question we are told that

The electric field strength is $E = 4.0 \ M \ V/m = 4.0 *10^6 \ V/m$

The diameter is $d = 1.4 \ cm = 0.014 \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.014}{2}$

=> $r = 0.007 \ m$

Generally the cross-sectional area is mathematically represented as

$A = \pi r^2$

$A = 3.142 * (0.007)^2$

$A = 0.000154 \ m^2$

Generally the maximum power that can be delivered is mathematically represented as

$P = \frac{c * \epsilon_o * E^2 * A }{2}$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$\epsilon_o$ is the permittivity of free space with value $\epsilon_o = 8.85 *10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$P = \frac{3,0*10^8 * 8.85*10^{-12} * (4 *10^6)^2 * 0.00154}{ 2}$

$P = 3.270960 *10^{6} \ W$

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Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

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Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

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(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

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$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

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Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

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3 years ago

To learn to apply the concept of current density and microscopic Ohm's law. A "gauge 8" jumper cable has a diameter d of 0.326 c

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Answer:

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Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Option A is correct. Copper.

Explanation:

Given,

I = 30.0 A

d = 0.326 cm = 0.00326 m

E = 0.062 V/m

The current density for the wire is given as

J = (I/A)

where J = current density = ?

I = current = 30.0 A

A = Cross sectional Area of the wire = (πd²/4) = [π×(0.00326²)/4] = 0.0000083503 m²

J = (30 ÷ 0.0000083503) = 3,592,685.3 A/m²

On a microscopic scale, Ohm's law can be stated as

E = Jρ

where E = Electric field = 0.062 V

J = Current density = 3,592,685.3 A/m²

ρ = Resistivity of the material = ?

ρ = (E/J)

ρ = 0.062 ÷ 3,592,685.3 = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Hope this Helps!!!

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