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-Dominant- [34]
2 years ago
15

100 POINTS!! PLEASE HELP ASAP. WILL MARK BRANLIEST.

Physics
2 answers:
rewona [7]2 years ago
8 0

Answer:

the answer is equal to visible light , ultraviolet, x-rays, gamma rayst

svet-max [94.6K]2 years ago
4 0

The sequence of types of electromagnetic radiation is arranged from that with the least to the most energetic photons will be visible light , ultraviolet, x-rays, gamma rays.

<h3>What is electromagnetic radiation?</h3>

Electromagnetic radiation is a type of energy that is transmitted in incident photons as both magnetic and electric waves.

Electromagnetic radiation has a spectrum with varying wavelengths and frequencies, which gives it different properties.

Electromagnetic radiation is a type of energy that travels in packets of energy called photons as both electrical and magnetic waves.

Wave-particle duality is the name given to this phenomenon. This is a notion that defines how a wave behaves whether it behaves like particles, waves, or both particles and waves.

Hence the sequence of types of electromagnetic radiation is arranged from that with the least to the most energetic photons will be  visible light , ultraviolet, x-rays, gamma rays.

To learn more about the electromagnetic radiation, refer to the link;

brainly.com/question/10759891

#SPJ2

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A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.
Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected  is B = 0.0036T

Explanation:

From the question we are told that

    The charge on the particle is q = +2e

    The mass of the particle is  m = 6.64 *10^{-27} kg

    The potential difference is V_a  = 1.8 kV = 1.8 *10^{3} V

    The potential difference between the two parallel plate is  V_b = 120 V

    The separation between the plate is  d = 8 mm =  \frac{8}{1000} =  8*10^{-3}m

   

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam  after the region having a potential difference of 1.8kV

               KE_b  =  PE_b

Generelly

              KE_b = \frac{1}{2} m v^2

And      PE_b = q V_a

 Equating this two formulas

              \frac{1}{2} mv^2 = q V_a

making v the subject

           v = \sqrt{\frac{q V_a}{2 m} }

Substituting value  

           v = \sqrt{\frac{ 2* 1.602 *10^{-19}  * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }

           v = 41.65*10^4 m/s

Generally the electric field between the plates is mathematically represented as

                 E = \frac{V_b}{d}  

Substituting value  

                 E = \frac{120}{8*10^{-3}}              

                E = 15 *10^3 NC^{-1}

the magnetic field  is mathematically evaluate    

                     B = \frac{E}{v}

                   B = \frac{15 *10^{3}}{41.65 *10^4}

                    B = 0.0036T

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3 years ago
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