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liberstina [14]
4 years ago
12

Dissolved in many gasolines are cleaning agents and anti-knock additives. These cleaning agents and additives are an example of

Physics
1 answer:
lorasvet [3.4K]4 years ago
3 0

Solutes

Explanation:

These cleaning agents and additives are examples of solutes.

  • Solutes are substances that dissolves in other substances called solvents to form a solution.
  • The solutes are the cleaning agents and anti-knock additives.
  • The solvents are the gasoline.
  • When solutes dissolves in solvents they form solutions.
  • A solution is a mixture that is homogeneous.
  • For a solute dissolve in a solvent, they must both have the same property.
  • Like substances will only dissolve in like ones.

learn more:

Heterogeneous mixtures brainly.com/question/1446244

#learnwithBrainly

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If you were making an electric device and needed a piece that would easily transmit an electric charge and also not react with o
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Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of options. However, one material that is good candidate for conducting electricity without reacting with other materials is metallic vanadium dioxide. This is because of the inability of this electrical conductor to conduct heat (an unusual property for all other electrical conductors) and thus makes it difficult for it to react with other materials (since an increase in temperature increases possibility of a reaction).

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An indicator of average kinetic energy is _____. pressure volume diffusion temperature
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rosijanka [135]

The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

<h3>What are the components of velocity?</h3>

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

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Learn more about components of velocity:brainly.com/question/14478315

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7 0
2 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
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There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

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