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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liberstina [14]

Answer: Doppler Effect

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3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).