1answer.
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
pantera1 [17]
2 years ago
7

Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2

= (1/3)E1. In what ways could the particle move?
Select all that apply.

a.) It could remain at rest at x3.
b.) It could oscillate between x1 and x5.
c.) It could undergo small oscillations around x2.
d.) It could undergo small oscillations around x3.
e.) It could undergo small oscillations around x4.

Please explain why, I have an exam coming up and want to understand the concept.

Physics
1 answer:
wolverine [178]2 years ago
3 0

The particles can undergo small oscillations around x₂.

The given parameters;

  • <em>initial energy of the particles = E₁</em>
  • <em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

  • The maximum position the particle can oscillate is x₅
  • The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

You might be interested in
Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an
Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

W = F\cdot \Delta s

Where:

W - Work, measured in joules.

F - Force exerted on the object, measured in newtons.

\Delta s - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

W_{net} = W_{Joel} - W_{Jerry}

W_{net} = F\cdot \Delta s - F\cdot \Delta s

W_{net} = (F-F)\cdot \Delta s

W_{net} = 0\cdot \Delta s

W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

W_{net} = W_{Joel}

W_{net} = F\cdot \Delta s

Hence, the work done by Joel is greater than the work done by Jerry.

7 0
3 years ago
An elevator is rising up a vertical shaft in a skyscraper. Between the 12th and 20th floors, it travels at a constant speed. Dur
zaharov [31]
3. The upward force is greater
.................,
8 0
3 years ago
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
arsen [322]

magnetic field due to a finite straight conductor is given by

B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)

here since it forms an equilateral triangle so we will have

\theta_1 = \theta_2 = 60 degre

also the perpendicular distance of the point from the wire is

r = \frac{a}{2\sqrt3}

now from the above equation magnetic field due to one wire is given by

B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)

B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)

B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

5 0
3 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
Rank the five terrestrial worlds in order of size from smallest to largest: Group of answer choices Mercury, Venus, Earth, Moon,
dimaraw [331]

Answer: Moon, Mercury, Mars, Venus, Earth

Explanation:

7 0
2 years ago
Other questions:
  • Which statement correctly describes a hypothesis?
    6·1 answer
  • Please hurry need ASAP!!! When a pendulum swings, at which point is kinetic energy highest?
    9·1 answer
  • In electroplating, the object to be plated is which part of an electrolytic cell? cathode or anode
    8·2 answers
  • Dalton proposed the first atomic model. Which of these statements accurately reflected his thinking at that time?. . . . A.. Ato
    15·2 answers
  • Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp
    12·1 answer
  • At what distance is the electrostatic force between two protons equal to the weight of one proton?
    10·1 answer
  • Water is initially present in a state where its molecules are far apart. During a change of state, its molecules slow down. Whic
    11·2 answers
  • An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is
    11·1 answer
  • A swimmer pushes backwad on water with their hand...what force allows the swimmer to move forward?
    5·1 answer
  • How do you build a sticky piston door​
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!