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Artemon [7]
3 years ago
7

An train car of total mass of 13000 kg is pulling into a station and slowing down with an acceleration of -4.56 m/s^2. Before th

e train pulls into the station, a group of 7 stowaway jump off the train. Each stowaway has a mass of 66 kg. If the braking force on the train remains the same, what will be the acceleration of the train once the stowaways jump out of the train?

Physics
1 answer:
lys-0071 [83]3 years ago
4 0
<h3>Answer:</h3>

-4.73 m/s²

<h3>Explanation:</h3>

Each jumper represents about 1/2 of 1% of the mass of the train, so together they total about 3.5% of the mass of the train. If the mass decreases by 3.5% and the braking force remains the same, the acceleration (magnitude) will increase by about 3.5%.

The acceleration will be about 0.17 m/s² more (in magnitude) than it was, or about -4.73 m/s².

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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

6 0
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Answer:

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Answer:

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