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
Explanation:
From the question we are told that
The electric field strength is
The diameter is
Generally the radius is mathematically represented as
=>
=>
Generally the cross-sectional area is mathematically represented as
Generally the maximum power that can be delivered is mathematically represented as
Here c is the speed of light with value
is the permittivity of free space with value
Answer:
The speed is 6.1m/s with an angle of 56.3 degrees north of east.
Explanation:
We meed to apply the Linear momentum theorem:
The magnitude of the velocity is given by:
and the angle:
Answer: Explanation:
A number used to multiply a variable. Example: 6z means 6 times z, and "z" is a variable, so 6 is a coefficient. ... Example: In ax^2 + bx + c, "x" is a variable, and "a" and "b" are coefficients.
Answer:
Explanation:
This problem is based on conservation of rotational momentum.
Moment of inertia of rod about its center
= 1/12 m l² , m is mass of the rod and l is its length .
= 1 / 12 x 4.6 x .11²
I = .004638 kg m²
The angular momentum of the bullet about the center of rod = mvr
where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .
5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet
According to law of conservation of angular momentum
5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .
.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12
.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12
.238 x 10⁻³ v = 55.8375 x 10⁻³
.238 v = 55.8375
v = 234.6 m /s