Active Optics.
Hope that helps, Good luck! (:
Answer:
The metalloids are located on the right side of the periodic table in a "step-like" arrangement.
All of the possible metalloids are:
boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)
A "screen" or even just a set of parallel bars are highly reflective to electromagnetic waves as long as the open spaces are small compared to the wavelengths.
"Grid" dishes work fine ... with less weight and less wind resistance ... for frequencies below about 3 GHz. (Wavelengths of at least 10 cm.)
(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)
To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.
The intensity depending on the power is defined as

Where
P = Power
r = Radius
Replacing the values that we have,


The definition of intensity tells us that,

Where,
Magnetic field
Permeability constant
c = Speed velocity
Then replacing with our values we have,

Re-arrange to find the magnetic Field B_0

Therefore the amplitude of the magnetic field of this light is
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e 





v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s