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Ksivusya [100]
3 years ago
7

Is it true the atoms in sulphur liquid are the same as the atoms in solid sulphur?

Physics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Yes, it is true that the atoms in liquid sulphur are the same as the atoms
in solid sulphur.

Also, if you can heat it past 832°F, it'll boil, and you'll have sulphur vapor.
Those atoms are also the same.

In any one of its three physical phases, you can also spell it as "sulfur"
without changing the nature of its atoms.


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Explanation:

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3. Describe the flow of one molecule of water through the water cycle, beginning in the ocean.
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molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.

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A notebook sitting on a desk has 19 j of potential energy. What is the total mechanical energy of the notebook?
Alex17521 [72]

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3 years ago
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A _______ is a series of waves that come ashore at interval 10-45 mintues​
Nostrana [21]

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3 0
3 years ago
A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h
Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

              P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

             

The pressure when the water comes out is the atmospheric pressure

           P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

           y₂-y₁ = 15 m

Now let's use the continuity equation

             v₁ A₁ = v₂ A₁

The area of ​​a circle is

             A = π r² = π (d/2)²

            v₁ π d₁²/ 4 = v₂ π d₂²/ 4

            v₂ = v₁ d₁² / d₂²

Let's replace

           P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

           P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

           v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

           d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

           d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

            v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2  -9.8 15/2

            v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

            v₁ = √ 2.2477 10⁸ /0.8710

             v₁ = 1,606 10⁴ m / s

4 0
3 years ago
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