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Pavel [41]
3 years ago
10

Bradley is standing on the top of a mountain holding a soccer ball. Which of these best describes what happens when he throws th

e ball over the edge?
A)
The kinetic energy increases and so does the total amount of energy.


B)
The kinetic energy decreases and the total amount of energy increases.


C)
The potential energy decreases and the total amount of energy also decreases.


D)
Both the kinetic and potential energy change, but the total energy remains constant.
Physics
1 answer:
Simora [160]3 years ago
7 0
Which one was it, please tell me :( :)
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Answer:

Heat energy does not cool. Objects cool. Energy is not an object. Now, what happens when an object cools is that it gives off some of its thermal energy by one or more mechanisms: radiation, conduction, or convection. In radiation the energy escaped the object as electromagnetic waves - you see the red hot poker slowly dim as it cools, in conduction the energy is transferred Mechanically by the atoms of the hot object being in physical contact with those of the cool object and in convection the hot material moves to a cooler region where it gives up its heat energy by either radiation or conduction.

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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

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