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Butoxors [25]
3 years ago
16

Visit this website and locate the element mercury, whose chemical symbol is Hg. Click on the square to read about the history, p

roperties, and uses of mercury. Name three uses for mercury.
Physics
2 answers:
lubasha [3.4K]3 years ago
7 0

Answer: Mercury is used for making thermometers, barometers, diffusion pumps, and other laboratory instruments. It can also be used to make mercury-vapor lamps, advertising signs, and mercury switches as well as pesticides, dental preparations, batteries, and catalysts.

~ :)

Irina-Kira [14]3 years ago
6 0
We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

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Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w
amm1812

Answer:

1) f= 8.6 GHz

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Explanation:

1)

  • Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
  • There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

        v = \lambda * f (1)

  • Replacing by the givens, and solving for f, we get:

       f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)

⇒     f = 8.6 Ghz (with two significative figures)

2)

  • Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

  • Solving for t, we get:

       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

       ⇒  t = 0.2 ms (with two significative figures)

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