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Butoxors [25]
2 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]2 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]2 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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According to the law of conservation of momentum:

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'

m1 = mass of first object
m2 = mass of second object
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v2 = Velocity of the second object before the collision
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v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know. 

m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
m2 = 235kg      v2 = -13m/s     v'2 = ?

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')
-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
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