All categories

2 years ago

What do radio waves and gamma rays have in common?

Physics

2 answers:

Irina-Kira [14]2 years ago

5 0

Answer:

A- They are both electromagnetic waves

I took the test and this was the answer

Explanation:

Nastasia [14]2 years ago

4 0

Answer:

A - They are both electromagnetic waves.

Explanation:

Electromagnetic waves are waves consisting of periodic oscillations of electric and magnetic fields, that vibrate in a plane perpendicular to the direction of propagation of the wave (for this reason, they are said to be "transverse waves").

Electromagnetic waves, unlike mechanical waves, can travel through a vacuum, and do not need a medium to propagate. Their speed in a vacuum is a constant and it is called speed of light ( $c=3.0\cdot 10^8 m/s$ ).

Electromagnetic waves are classified, depending on their wavelength and frequency, into 7 different types - together they form the electromagnetic spectrum. The 7 types, listed from shortest to longest wavelength, are:

gamma rays

X-rays

ultraviolet radiation

visible light

infrared radiation

microwaves

radio waves

All these waves, despite having different properties, are all electromagnetic waves -so we see that both radio waves and gamma rays belong to this type of waves.

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A balanced relationship between the energy that you intake and your bodys energy output is the definition of _____.

Jet001 [13]

Whole body METABOLISM

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3 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

8 0

3 years ago

Which energy conversion takes place when a solar cell is used to light a street lamp?

allsm [11]

Answer:

the answer is for the question is B

8 0

2 years ago

Which is the force of repulsion between two positively-charged particles?

klio [65]

answer>>>>electric force <<<<by the way i don't like physics but i answer this for you ^-^

This is because centripetal force is just the net force of a circular motion. There are no attractive or repulsive forces here. This is not the case here. 
The gravitational force is a force reliant on mass and attraction of the masses. There are attractive forces here, but not really repulsive forces. 
The electric force is the only one that would make sense because it has to do with a relationship between charges and includes both repulsive and attractive forces.

4 0

2 years ago

Read 2 more answers

Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation

Artemon [7]

Answer:

L = μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

$E_{L}$ = - L dI / dt

Let's use Faraday's law

E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

- d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

n d Ф_B = L dI

we can remove the differentials

n Ф_B = L I

magnetic flux is defined by

Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

n B A = L I

the loop area is

A = π R²

we substitute

n B π R² = L I (1)

To find the magnetic field in the coil let's use Ampere's law

∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

s = 2π R

we solve

B 2ππ R = μ₀ I

B = μ₀ I / 2πR

we substitute in

n ( μ₀ I / 2πR) π R² = L I

n μ₀ R / 2 = L I

L = μ₀ n r / 2I

4 0

3 years ago