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AysviL [449]
3 years ago
14

What's the difference between transmitted and absorbed waves

Physics
2 answers:
ipn [44]3 years ago
8 0

Answer:

The light wave could be absorbed by the object, in which case its energy is converted to heat. The light wave could be reflected by the object. And the light wave could be transmitted by the object. ... When this occurs, objects have a tendency to selectively absorb, reflect or transmit light certain frequencies.

Explanation:

harkovskaia [24]3 years ago
7 0

Answer:

The light wave could be absorbed by the object, in which case its energy is converted to heat. The light wave could be reflected by the object. And the light wave could be transmitted by the object. ... When this occurs, objects have a tendency to selectively absorb, reflect or transmit light certain frequencies.

Explanation:

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To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

q_1 = -Q_0

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As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be q_3 = +Q is placed at a distance x from q_1

F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}

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F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}

The condition of equilibrium is

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\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}

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To find the magnitude of q_3 we use F_{1,2} = F_{1,3}

\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}

Q = 0.43Q_0

The magnitude of the third charge must be 0.43 the first charge Q_0

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Explanation:

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