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Mila [183]
3 years ago
15

A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What wi

ll be the compression if the same block collides with the spring at a speed of 2v?
Physics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

4.0 cm

Explanation:

For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v

Since m and k are constant since its the same spring x ∝ v

If our speed is now v₁ = 2v, our compression is x₁

x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x

x₁ = 2x

Since x = 2.0 cm, our compression for speed = 2v is

x₁ = 2(2.0) = 4.0 cm

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The magnitude of the hiker’s displacement is 2.96 km

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the final displacement of the hiker, = y = 1.4 km

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The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total
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Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass m travelling at a velocity \vec{v}. The momentum \vec{p} of this object would be:

\vec{p} = m \cdot \vec{v}.

For the law of conservation of momentum, consider two objects: object \rm a and object \rm b. Assume that these two objects collided with each other.

  • Let m_{\rm a} and m_{\rm b} denote the mass of the two objects.
  • Let \vec{v}_{\rm a}(\text{initial}) and \vec{v}_{\rm b}(\text{initial}) denote the velocity of the two object right before the interaction.
  • Let \vec{v}_{\rm a}(\text{final}) and \vec{v}_{\rm b}(\text{final}) denote the velocity of the two objects right after the interaction.
  • The momentum of the two objects right before the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}), respectively.
  • The momentum of the two objects right after the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}), respectively.

The sum of the momentum of the two objects would be:

  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) right before the collision, and
  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}) right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}).

4 0
3 years ago
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In the thermal equilibrium, the change in temperature is said to be zero in between the bodies. Thermal equilibrium is reached when both objects have the same temperature.

<h3>What is thermal equilibrium?</h3>

Thermal equilibrium is easily explained by the zeroth law of thermodynamics. If any two-body is at thermal equilibrium there is no change in the temperature of the body.

According to zeroth law if body A is in thermal equilibrium with body B and body B is in thermal equilibrium with C . So body A and C are also in thermal equilibrium.

In the thermal equilibrium, the net heat transfer is said to be zero in between the bodies.

Hence option A IS RIGHT. Thermal equilibrium is reached when both objects have the same temperature

To learn more about the thermal equilibrium refer to the link;

brainly.com/question/2637015

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