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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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4 0
3 years ago
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

m = mass of lead bullet

c = specific heat of lead = 128 J/(kg °C)

L = Latent heat of fusion of lead = 24500 J/kg

T_{i} = initial temperature = 27.4 °C

T_{f} = final temperature = melting point of lead =  327.5 °C

v = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}

3 0
3 years ago
A wheelbarrow is a complex machine that combines which simply marchines?
koban [17]
It's most likely the combination of a bucket and the wheel. 
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3 years ago
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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

Force due to charge Y

F_Y = k\dfrac{-27Q}{(0.60-x)^2}

Since both forces are equal and opposite,

F_X = -F_Y

k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}

\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}

2(0.60-x)^2 = 3x^2

2(0.36-1.20x+x^2) = 3x^2

0.72-2.40x+2x^2 = 3x^2

x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

x = -2.7 or x = 0.27

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

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