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Vlad1618 [11]
4 years ago
10

Which of these best describes how the kinetic energy of carbon dioxide molecules change as the carbon dioxide is heated? (2 poin

ts) Kinetic energy decreases. Kinetic energy increases. Kinetic energy increases, then decreases. Kinetic energy decreases, then increases.
Physics
2 answers:
Delvig [45]4 years ago
8 0

<u>Answer:</u> B.)  <em>Kinetic energy increases.</em>

kari74 [83]4 years ago
5 0

B) Kinetic energy increases

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Answer:

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Explanation:

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The ozone layer is an important component of Earth’s atmosphere because it helps to block UV radiation from reaching Earth. The
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This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ
soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is the velocity of the object

We notice that:

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  • The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

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brainly.com/question/6536722

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5 0
3 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
3 years ago
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