To determine what happens to the plane’s kinetic energy during the flights, we first need to recall the formula for kinetic energy (KE), which is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass (or weight) of the object and \(v\) is its velocity.
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Flight from Phoenix to Los Angeles:
- The plane carries 320 passengers. Assuming each passenger has a certain average weight, the total weight (mass) of the plane will be higher on this flight due to the extra passengers.
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Return Flight:
- On the return flight, the plane carries only 164 passengers. This means that the total weight (mass) of the airplane will be less on the return flight.
However, the speed of the airplane remains constant at 490 miles per hour during both flights.
Analysis of Kinetic Energy Changes:
- On the outgoing flight: The plane has a larger mass (due to 320 passengers) and is traveling at 490 mph.
- On the return flight: The plane has a smaller mass (due to 164 passengers) but is still traveling at 490 mph.
Since kinetic energy depends on both mass and the square of the velocity, we can analyze the change:
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Kinetic Energy on the Outgoing Flight: \[ KE_{\text{outgoing}} = \frac{1}{2} m_1 (490^2) \] where \(m_1\) is the mass with 320 passengers.
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Kinetic Energy on the Return Flight: \[ KE_{\text{return}} = \frac{1}{2} m_2 (490^2) \] where \(m_2\) is the mass with 164 passengers.
Conclusion:
Since \(m_2 < m_1\) (the mass is less on the return flight), the kinetic energy on the return flight will also be less:
\[ KE_{\text{return}} < KE_{\text{outgoing}} \]
Thus, the correct response is:
On the return flight, the plane has less kinetic energy.
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