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Nat2105 [25]
3 years ago
13

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

yringe tightly. 3. Pull the plunger back to double the volume of gas in the syringe. Which best describes the purpose and outcome of the demonstration? This is a demonstration of Charles’s law. As the volume increases, the temperature decreases, and the marshmallow will freeze. This is a demonstration of Charles’s law. As the volume increases, the temperature increases, and the marshmallow will melt. This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. This is a demonstration of Boyle’s law. As the volume increases, the pressure increases, and the marshmallow will shrink.
Physics
2 answers:
lana [24]3 years ago
7 0
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. 
</span><span>
 Keisha follows the instructions for a demonstration on gas laws.
 1. Place a small marshmallow in a large plastic syringe.
 2. Cap the syringe tightly.
 3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change.  Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
MrRissso [65]3 years ago
5 0

Answer:

C) This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

Explanation:Took the quiz on edg

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Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.
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A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
pav-90 [236]
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

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Consider the vertical motion of projectile,

When the projectile return to the ground we have

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Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

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\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}

\\ \rm\Rrightarrow u=\dfrac{190}{-14}

\\ \rm\Rrightarrow u=13.6cm

Real

5 0
2 years ago
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