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Aleonysh [2.5K]
3 years ago
13

What do radio waves and gamma rays have in common?

Physics
2 answers:
Reil [10]3 years ago
7 0

Answer:

The only difference between radio waves, visible light and gamma rays is the energy of the photons.'

Bothrays have photons, but in radio waves they are weaker.

Andrews [41]3 years ago
3 0

Answer:

Radio waves and gamma rays have a common speed when they propagate in a vacuum. This speed is:

3.10^{8} cm/s

Explanation:

Gamma rays are produced naturally by atomic nuclei during their natural radiative transformations, and their wavelength is less than  10^{-12} m.

Such radiation has the property of ionizing atoms and also has great penetrating power. For its absorption, it is very common to use thick lead walls.

Radio waves are those with the lowest frequency and, consequently, the longest wavelengths, ranging from 10 m to 10 km. They are widely used in radio and TV broadcasts. These waves were generated for the first time by the German physicist Heinrich Rudolf Hertz, whose experiment was perfected, eight years later, by the Italian Guglielmo Marconi.

The speed of radio waves and river ranges in a vacuum is the same, this is something they have in common, their approximate value is: 3.10^{8} cm/s

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An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate
Bumek [7]

Answer:

a)    n = 9.9       b)      E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

         Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

 In this case En = 19 eV let us reduce to the SI system

          En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

          n = √ (In 8 m L² / h²)

let's calculate

          n = √ (8  9.1 10⁻³¹ (1.4 10⁻⁹)²  30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

          n = √ (98)            n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

        In = (h² / 8mL2² n²

        In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹  (1.4 10⁻⁹)² 10²

        E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

      E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

      E₁₀ = 1.925 101 eV

      E₁₀ = 19.25 eV

the result with significant figures is

        E₁₀ = 19.25 eV

3 0
3 years ago
Why does moon appear to have alot of phases
Bingel [31]
It depends on the angle of the earth, and our point of view. A full moon would occur if we were to be right in front of it.
4 0
2 years ago
A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

4 0
2 years ago
If you were to be drawn into a black hole, what would happen? To the black hole, not to you.
love history [14]

Answer:

It grows

Explanation:

The blacks holes will absorb

Me hoizontally stretching me like a noodle by the spaghtification process,thus growing bigger.

7 0
3 years ago
An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.
ivann1987 [24]
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters


</span>
5 0
3 years ago
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