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Softa [21]
3 years ago
14

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:

Explanation:

The guy wire is making a right angled triangle  with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

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The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

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A student, Daniela, is arguing that since a meteor is weightiess in space, if it crashes
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2 years ago
In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

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I=2.6363\ kg.m^2

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3 years ago
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