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3 years ago

Which phenomena support only the wave theory of light? Check all that apply.

Physics

2 answers:

eduard3 years ago

6 0

<h3>Answer;</h3>

Diffraction
interference

<h3>Explanation;</h3>

Light may have both wave or particle properties.

According to wave theory of light, light behaves like a wave. Light is an electromagnetic wave which means it does not require a material medium for transmission. Just like electromagnetic waves light possess both magnetic field and electric fields.

Light waves displays a transverse type of a wave in which it oscillates in a similar direction as that of the wave travel. Due to these characteristics of a wave light can undergo diffraction and also interference.

FrozenT [24]3 years ago

6 0

Answer:

Diffraction and interference

Explanation:

Light has a property that it exhibit both wave like and particle like behavior. It is an electromagnetic wave. It do not require any medium for its propagation.

Light shows diffraction and interference both. In interference of light wave, two waves combine to form an interference pattern. While in diffraction of light light bends through the obstacles.

So, the correct options are "diffraction and interference".

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What is the energy level of hydrogen?

barxatty [35]

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

$E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...$

8 0

2 years ago

16x^2y^2-25a^2b^2 factorize the expression

SIZIF [17.4K]

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

8 0

2 years ago

If the second ball has a mass of 2.4kg and a constant acceleration of a⃗ 2= 2.8m/s2 j^ , what must the mass of the first ball be

kompoz [17]

Hope this helps you!

8 0

2 years ago

3. What do we call the ONLY part of the electromagnetic spectrum that we can

otez555 [7]

Answer:

Visible Light

wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns

1 A unit = 10^-10 m

1 mμ = 10^-9 m

6 0

2 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

2 years ago