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alexgriva [62]
1 year ago
7

Fill in the blanks in the following paragraph to correctly identify phenomena that help provide evidence for the particle nature

of light
Physics
1 answer:
ValentinkaMS [17]1 year ago
4 0

An electron absorbs energy and emits electromagnetic radiation based on its transition.

<h3>What is an Electron?</h3>

This is defined as a sub-atomic particle which has a negative charge. It's involved in a chemical reaction which results in the formation of new substances.

Classical physics predicted that at a high enough temperature, ultraviolet light would be emitted. instead white light was emitted resulting in the ultraviolet radiation.

Read more about Electron here brainly.com/question/13998346

#SPJ1

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A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
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Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

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