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antiseptic1488 [7]
3 years ago
11

What happens when an electron emits a photon?

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

An electron emits a photon and falls from a higher energy level to a lower energy level.

Explanation:

A quantized form of electromagnetic radiation is called a photon.

Since the electromagnetic radiation has dual characteristics. The photons are the energy wave packets of the radiation.

When an electron is exposed to such radiation, it absorbs some of the energies of photons and goes to the excited state.

The excited state is only a temporary state. So, the electron releases some energy that comes to a lower energy level.

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Well i think the answer is impossible to find because there is no picture
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A golf club with mass 0.5 kg is swung at 23 m/s toward a stationary golf ball on a tee. The club barely slows down to 17 m/s, bu
mylen [45]

Answer:

I gotchu fam. 0.0435 kg

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)

11.5=8.5+69m2

3=69m2

0.0435 kg

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2 years ago
A man pulls a 10 kg box at constant speed across the floor with a rope. The tension in the rope is 120 N at an angle of 30°.
Murrr4er [49]

Answer:

98n

Explanation:

5 0
2 years ago
compare the numbers of each subatomic particle found in both diagrams and then list what makes hydrogen diffrent from hellium
Cloud [144]

The helium atom is found to contain more particles than the hydrogen atom.

<h3>What is a subparticle?</h3>

An atom is composed of particles and these particles are electrons protons and neutrons. The collective name for these particles is the subatomic particles.

Now we know that the helium atom contains more sub atomic particles than the atom of hydrogen. This is because, the helium atom contains two neutrons and two protons as well as two electrons. On the other hand, the hydrogen atom contains one neutron and one proton as well as one electron.

Thus, the helium atom is found to contain more particles than the hydrogen atom.

Learn more about helium atom:brainly.com/question/4945478

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7 0
1 year ago
Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun
allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction = u\cos\theta

and the speed in the vertical direction is = u\sin\theta upward.

For A:

The speed in the horizontal direction = u\cos75^{\circ}

and the speed in the vertical direction is = u\sin75^{\circ} upward.

For B:

The speed in the horizontal direction = u\cos15^{\circ}

and the speed in the vertical direction is = u\sin15^{\circ} upward.

Let t_A and t_B are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A = u\cos75^{\circ}\times t_A\cdots(i)

The range for the projectile B = u\cos15^{\circ}\times t_B\cdots(ii)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion v^2=u^2 +2a s.

Here, the final velocity v=0, the initial velocity u = u \sin \theta , h= vertical distance up to the highest point, and a= -g (as per sign convention).

So, s= \frac{u^2\sin^2 \theta}{2g}

For projectile A: The maximum height attained.

s_A= \frac{u^2\sin^2 75^{\circ}}{2g}

For projectile B: The maximum height attained.

s_B= \frac{u^2\sin^2 15^{\circ}}{2g}

As \sin^2 75^{\circ} > \sin^2 15^{\circ}, the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point =\frac {u\sin\theta}{g}

where g is the acceleration due to gravity.

So, the total time of flight

= 2 \times \frac {u\sin\theta}{g}

The total time of flight for A

=2 \times \frac {u\sin75^{\circ}}{g}

The total time of flight for A

=2 \times \frac {u\sin15^{\circ}}{g}

Now, from equations (i) and (ii),

The range for the projectile A =

u\cos75^{\circ}\times  \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}

The range for the projectile B =

u\cos15^{\circ}\times  \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.

Both the projectile have the same range.

Hence, option (A) is correct.

7 0
3 years ago
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