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ruslelena [56]
4 years ago
13

What is the phase difference between the oscillation of the electric field and the oscillation of the magnetic field?

Physics
1 answer:
love history [14]4 years ago
5 0

Answer: Zero

Explanation:

In electromagnetic wave, the magnetic and electric field are perpendicular to each other as well as perpendicular to the propagation wave. Both are always in a phase.

When the electric field is minimum, the magnetic field is also minimum and when the magnetic field is maximum, the electric field is also maximum. They are both in the same phase. Therefore, the phase difference between the oscillation of the electric field and that of the magnetic field is zero.

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Answer:

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3 years ago
The opening to the windpipe is called the _____.
Nina [5.8K]
Glottis is the organ that is considered to be the opening to the windpipe. It consists of vocal chords wherein voice modulation usually takes place. The production of the sounds of the organ can be mainly attributed to the "buzzing" vibration inside the vocal chords wherein they vibrate.
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3 years ago
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A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
3 years ago
Two particles with oppositely signed charges nC are placed at two of the vertices of an equilateral triangle with side length 3
babymother [125]

The magnitude of the electric field at the third vertex of the triangle is determined as zero.

<h3>Electric field at the third vertex of the triangle </h3>

The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;

E = E(13) + E(23)

E = (kq₁)/r² + (kq₂)/r²

where;

  • q1 is positive charge
  • q2 is negative charge

E =  (kq₁)/r² - (kq₂)/r²

E = 0

Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

5 0
2 years ago
Which type of energy is released when a bond between atoms is broken
Murrr4er [49]

Answer: gases

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dont use this this is a bad explanation

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