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ser-zykov [4K]
2 years ago
12

An object has 1200 J of gravitational potential energy when it is dropped. Right when the object hits the ground how much kineti

c energy does it have?
Physics
1 answer:
Fittoniya [83]2 years ago
5 0

Answer:

The kinetic energy is 1200 J

Explanation:

The Principle of Conservation of energy states that "energy is neither created nor destroyed, it is transformed".

This means that energy can be transformed from one form to another, but the total amount of energy always remains constant, that is, the total energy is the same before and after each transformation.

The mechanical energy of a body or a physical system is the sum of its kinetic energy and the potential energy. According to the Principle of Conservation of Energy for mechanical energy, the total mechanical energy that a body possesses is constant at every instant of time.

Since mechanical energy is equal to the sum of kinetic energy and gravitational potential energy that a body possesses, the only way to stay constant is that:

  • when the kinetic energy increases the gravitational potential energy decreases,
  • when gravitational potential energy increases, kinetic energy decreases.

Due to the Principle of Conservation of Energy you can say that the gravitational potential energy is converted to kinetic energy.  So Gravitational potential energy at the top = kinetic energy at the bottom

<u><em>The kinetic energy is 1200 J</em></u>

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an object has a momentum of 250 kg m/s what would be the momentum of an object that has half of the mass and going at the same v
Romashka-Z-Leto [24]

The momentum of the second object is 125 kg m/s

Explanation:

The momentum of an object is given by

p=mv

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m is the mass of the object

v is its velocity

For the object in this problem,

p = 250 kg m/s

And its mass is m and its velocity is v.

The second object has a mass of m' = \frac{m}{2} and same velocity v, so its momentum is

p'=m'v = (\frac{m}{2})v=\frac{1}{2}(mv)=\frac{1}{2}p

So, the second object has a momentum that is half of the momentum of the first object, therefore it is:

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3 0
3 years ago
A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33
Anuta_ua [19.1K]

Answer:

\frac{v_{2}}{v_{1}}=2.

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

\bar{K}=\frac{3k_{B}T}{2}

Now, we know that \bar{K} = (1/2)m\bar{v}^{2}

Before the absorption we have:

(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2} (1)

After the absorption,

(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2} (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}

\frac{v_{2}}{v_{1}}=\sqrt{4}

Therefore the ratio will be \frac{v_{2}}{v_{1}}=2

I hope it helps you!

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