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worty [1.4K]
3 years ago
11

Longitudinal waves transfer energy ___________ to the direction of the wave motion.

Physics
2 answers:
andriy [413]3 years ago
6 0

The correct answer to the question is parallel.

EXPLANATION:

Before answering the question, first we have to understand a longitudinal wave.

A longitudinal wave is a type of mechanical wave in which the particle vibration is parallel to the direction of wave propagation.

In this type of wave, one will find compression and rarefaction . Compression is the high pressure region where particles are closely aggregated to each other while rarefaction is the low pressure region where particles are far away from each other. Hence, a longitudinal wave is a series of compression and rarefaction.

The transfer of energy occurs from particle to particle. On the other hand, we may say that transfer of energy takes place from compression to rarefaction, and this process is repeated.

Hence, the longitudinal waves transfer energy parallel to the direction of the wave motion.


viva [34]3 years ago
4 0
Longitudinal waves transfer energy parallel to the direction of the wave motion
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0.5

Explanation:

Data provided in the question:

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A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the
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Answer:

(a):  The magnitude of the electric force on the small sphere = \dfrac{q\sigma}{2\epsilon_o}.

(b): Shown below.

Explanation:

<u>Given:</u>

  • m = mass of the small sphere.
  • q = charge on the small sphere.
  • L = length of the silk fiber.
  • \sigma = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

\rm E = \dfrac{\sigma}{2\epsilon_o}.

<em>where,</em>

\epsilon_o is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

\rm F_e=qE.

Thus, the magnitude of the electric force on the small sphere is given by

\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

\rm \tan \theta = \dfrac{F_e}{W}.

<em>where,</em>

  • \rm F_e = electric force on the sphere, acting along left.
  • \rm W = weight of the sphere, acting vertically downwards.

<em />

\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .

g is the acceleration due to gravity.

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