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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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2 years ago
A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m
nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

  • Speed of the truck, u = 80 km/h
  • Distance, d = 32 km
  • Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h

94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0
2 years ago
If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, what is the airplane's final velocity?
givi [52]
If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, the final velocity can be calculated using simple vector addition. In this case, the planes velocity is positive (+330 km/h) and head wind has a negative component (-18.0 km/h). Vector addition yields +330 km / h + (-18.0 km /h) = 312 km / h. 
8 0
3 years ago
What happens when electromagnetic waves cause a disturbance in electric
Leviafan [203]

Answer:

c

Explanation:

just answered it

7 0
2 years ago
A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?
lesantik [10]
<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>
  • Half life is the time taken for a radioactive isotope to decay by half of its original amount.
  • We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
  • A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

<em>1/4 = 1 × (1/2)^n</em>

<em>1/4 = (1/2)^n </em>

<em>n = 2 </em>

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0
3 years ago
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