answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tangare
9 days 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:
inna [2.7K]9 days ago
7 0
The accurate choice is option C. <span>This serves as an illustration of Boyle’s law. As the volume expands, the pressure diminishes, resulting in the marshmallow increasing in size.
</span><span>
Keisha is following the instructions for a demonstration of gas laws.
1. Insert a small marshmallow into a large plastic syringe.
2. Secure the syringe with a cap.
3. Pull back the plunger to double the gas volume in the syringe.

This activity occurs at a constant temperature, as there is no indication of any temperature change. Thus, as the plunger is drawn back, the volume doubles, leading to a reduction in pressure. Hence, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow expands.
Sav [2.8K]9 days ago
5 0

Response:

C) This is an illustration of Boyle’s law. As the volume increases, pressure decreases, which causes the marshmallow to enlarge.

Explanation: I took the quiz on edg

You might be interested in
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Softa [2668]

Answer:

 Baseball mass:  m_b=3.992*10^{14}kg  

Explanation:

 Circumference of a baseball is calculated using 2πr = 23 cm

 Thus, the radius comes out to be 3.66 cm, which equals 3.66*10^{-2} m

 The mass density of the baseball matches that of a neutron or proton.

 Proton mass = 10^{-27} kg  

 Proton diameter = 10^{-15} m

 Proton radius =  5*10^{-16} m

 Volume of the baseball is \frac{4}{3} \pi r^3

 Now by substituting all values into the mass per unit volume equation for the baseball, we get:  

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       Therefore, the baseball mass amounts to m_b=3.992*10^{14}kg              

5 0
26 days ago
Read 2 more answers
What is the least possible initial kinetic energy in the oxygen atom could have and still excite the cesium atom?
inna [2740]
K=E[(m+M)/M] Kmin=4.4
8 0
17 days ago
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [2968]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
1 month ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [2826]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
1 month ago
A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa
Maru [2979]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

Thus, the distance of the trip in miles is:

d=16.77706389mi

(b)

Next, we will compute the man's speed during the journey:

v=\frac{d}{t}

Before that, we must convert minutes to hours:

16min*\frac{1h}{60min} =2.666666667h

The resulting speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

Consequently:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

Thus, it can be concluded that the driver was speeding

8 0
1 month ago
Other questions:
  • An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha
    7·1 answer
  • On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo
    13·1 answer
  • An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
    6·1 answer
  • Read each scenario below. Then select the answer that best completes each sentence.
    6·2 answers
  • A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule w
    7·1 answer
  • Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.
    12·1 answer
  • (a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
    12·1 answer
  • Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5
    13·1 answer
  • A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
    12·1 answer
  • When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!