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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

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Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

  • initial position of the man = 0
  • direction of the man's first displacement = backward
  • time of first motion, t₁ = 6 seconds
  • velocity of this first displacement = v₁
  • time without any motion (<em>zero movement</em>) = 6 seconds
  • direction of the second displacement = forward
  • velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

  • <em>graph B</em>, and
  • <em>graph D</em>.

The difference between graph B and D;

  • in graph B there is a uniform motion for 6 seconds
  • in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

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This is because both areas are the same.

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Read 2 more answers
9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku
Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

  • v₁ = -(\frac{\sqrt{2} }{2} × 3.5)·i + (\frac{\sqrt{2} }{2} × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

  • v₂ =  (\frac{\sqrt{2} }{2} × 4.0)·i - (\frac{\sqrt{2} }{2} × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-(\frac{\sqrt{2} }{2} × 3.5) - (\frac{\sqrt{2} }{2} × 4.0))·i + ((\frac{\sqrt{2} }{2} × 3.5 + 12.5) + (\frac{\sqrt{2} }{2} × 4.0 + 15))·j

  • v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

Red Skull's velocity relative to Captain America,  v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

  • Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

  • The direction is arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}

Therefore;

  • Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

  • |v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

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