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3 years ago

8

What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per

second (m/s) and assume acceleration from gravity is 10 m/s2.

1 answer:

gizmo_the_mogwai [7]3 years ago

6 0

speed = 40 m/s

Explanation:

Since the object is dropped, V0y = 0.

Vy = V0y - gt

= -(10 m/s^2)(4 s)

= -40 m/s

This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.

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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

pickupchik [31]

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:

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

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

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Answer:

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we can calculate the amount of heat energy required , using this formula

Q = mcθ

where.

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat capacity (units $Jkg^{-1} C^{-1}$ )

θ = change in temperature (Celcius,C or Kelvin K)

Assume Specific heat capacity (c) of water = $4200Jkg^{-1} C^{-1}$

mass =0.1 kg

$Q=0.1 kg*4200Jkg^{-1} C^{-1}*85C\\=35700J\\=35.7kJ$

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