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Umnica [9.8K]
2 years ago
12

| 1)

Physics
1 answer:
Zarrin [17]2 years ago
3 0

Answer:

  • <u><em>Option A) A gas was produced.</em></u>

Explanation:

Indeed, the production of a gas is an indication that a new substance was produced as result of the reaction of the matter that constitutes the antiacid tablet and the distilled water.

A chemical reaction is a change that yields the formation of new substances (products), due to the rearrangements of the atoms of the reactants.

In a very simple form, you can represent a chemical reaction by the general equation:

  • Reactants → Products

Which means that some elements or compounds, those in the reactant side of the equation, undergo a change in which some chemical bonds are broken, the atoms separate from the original substances and combine into a new arrangement in which they form one or more different substances (products).

The option B), that the table dissapeared is not a clear evidence of a chemical change because it could have been dissolved, which is a physical change.

Also, the option C), that there was a color change, cannot be cited as an evidence of a chemical reaction, since the color is a physical property.

As for the option D, that there was a change in mass, is not a viable option, since mass cannot change either with a chemical or physical change.

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Which of the following is not an example of work being done on an object?
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Identify the correct sequence for the formation and use of coal.
Zanzabum

The correct answer is - A. Plants store solar energy; the plants die; the plants are compressed; solar energy is released;

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An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1
goblinko [34]

Answer:

v"_{1} = v_{1} tanΘ

v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

Θ = tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )

Explanation:

Applying the law of conservation of momentum, we have:

Δp_{x = 0}

p_{x} = p"_{x}

m_{1}v_{1} = m_{2}v"_{2} cosΘ (Equation 1)

Δp_{y} = 0

p_{y} = p"_{y}

0 = m_{1} v"_{1} - m_{2} v"_{2} sinΘ (Equation 2)

From Equation 1:

v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

From Equation 2:

m_{2} v"_{2}sinΘ = m_{1} v_{1}

v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }

Replacing Equation 3 in Equation 4:

v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}

v"_{1}=v_{1}\frac{sinΘ}{cosΘ}

v"_{1}=v_{1}tanΘ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=\frac{v"_{1}}{v_{1}}

Θ=tan^{-1}(\frac{v"_{1}}{v_{1}})

7 0
3 years ago
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