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Aleonysh [2.5K]
4 years ago
13

What do radio waves and gamma rays have in common?

Physics
2 answers:
Reil [10]4 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]4 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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The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resi
gulaghasi [49]

Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

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7 0
3 years ago
A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

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3 years ago
ABCD next 3 letters??????
jasenka [17]

Answer:EFG

Explanation:AYEEE

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Read 2 more answers
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