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m_a_m_a [10]
3 years ago
5

Key Questions and Terms Notes

Chemistry
1 answer:
Elanso [62]3 years ago
5 0

Answer:

Following are the solution to the given question:

Explanation:

  • Its interactions among light waves as well as different media were also called light products.
  • Even before sound rays hit a medium, the light rays bounce off.
  • Even at a similar angle with which the material initially struck its medium, sunlight reflecting.
  • When the light hits a mirror at such an angle of forty degrees, everything just rebonds at an angle of 40 degrees.
  • When light waves pass via a transparent medium, those who bend as well as change their direction.
  • Light is called transparent if it can pass via an object.
  • Whenever light waves strike a platform, this is neither re-elected nor refracted, nor expected to move.
  • The light of interacts with these materials are:

Reflection Mirror:

Water glass: refractive

Dark fabricate: Absorption 

  • Sound waves and waves of water are examples of matter waves.
  • Light waves and X-rays are electromagnetic waves.
  • There is no need for sound radiation to pass through a medium/requirement to communicate with other particles to travel like waves of matter.
  • No, even though sound waves didn't reach through the space vacuum.
  • Matter waves should interact with solid, liquid, or gas particulates to travel.

The light waves pass thru the air more quickly than sound waves.

Sound waves almost always go through a metal frame.

Light waves travel faster than electrical signals even though they don't have to crash in electromagnetic waves to particles.

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Usimov [2.4K]

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The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS
pashok25 [27]

Answer:

d. 103.3

Explanation:

In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:

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Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.

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