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3 years ago

100 + 50 + BRAINLIST PLEASE HELP!!!

Physics

2 answers:

irina1246 [14]3 years ago

6 0

Answer:

Explanation:

1. What are the forces acting on the block when it is hanging freely from the spring scale? What is the net force on the block? What are the magnitudes of each of the forces acting on the block? Explain.

When a block is hanging freely, two forces are acting on it = tension force from the spring scale and gravity force on the block itself. The net force is zero as the block is not accelerating. The magnitudes of tension and gravity force are the same but in opposite directions.

2. What are the forces that act on the block when it is placed on the ramp and is held in place by the spring scale? What is the net force acting on the block? Explain. (Assume that the ramps are frictionless surfaces.)

There are three forces acting on the block when it is placed on the ramp and is held in place by the spring scale: as in 1, there are tension and gravity but there is a third force - reaction force from the ramp surface on the block that is perpendicular to the surface. Again the block is not moving so the net force is zero.

3. What is the magnitude of normal force acting on the block when it is resting on the flat surface? How does the normal force change as the angle of the ramp increases? Explain. (Assume that the ramps are frictionless surfaces.)

On flat surface, the normal force is equal to the gravity force of the block i.e. its weight. On a vertical surface, the normal force is equal to zero. For the angle of ramp, θ, the normal force = weight * cos θ.

SOVA2 [1]3 years ago

3 0

Answer:

Explanation:

1. block is hanging freely - freely means there is no net force

from the spring scale - spring force

block - weight

magnitude of weight = magnitude of spring force as there is no net force

2. same thing - held in place means there is no net force

from the spring scale - spring force

block - weight

ramp - normal reaction

3. when its resting, there is no net force

magnitude of normal reaction force = magnitude of block weight

normal force reaction force decreases as angle of ramp increases from zero

reaction = weight x sin (ramp angle)

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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:

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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
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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:

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

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