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Chapter 6: Problem 27

What's the kinetic energy of a \(2.4 \times 10^{5}-\mathrm{kg}\) airplane cruising at \(900 \mathrm{km} / \mathrm{h} ?\)

Short Answer

Expert verified
The kinetic energy of the airplane cruising is \(7.5 \times 10^{9} \mathrm{J}\).

Step by step solution

01

Convert speed to appropriate unit

Firstly, convert the given speed from kilometers per hour(km/h) to meters per second(m/s) since the standard unit of speed in physics is m/s. The conversion factor is \(1 \mathrm{km/h} = 0.27778 \mathrm{m/s}\), so the speed would therefore be \(900 \mathrm{km/h} \times 0.27778 \mathrm{m/s} = 250 \mathrm{m/s}\).
02

Substitute into kinetic energy formula

Use the kinetic energy equation \(KE = \frac{1}{2} m v^2\). Substitute \(m = 2.4 \times 10^{5} \mathrm{kg}\) and \(v = 250 \mathrm{m/s}\) into the equation. Hence, \(KE = \frac{1}{2} \times 2.4 \times 10^{5} \mathrm{kg} \times (250 \mathrm{m/s})^2\).
03

Calculate kinetic energy

Calculate the value of the kinetic energy based on the given values. This gives \(KE = 7.5 \times 10^{9} \mathrm{J}\), where Joule(J) is the unit of energy. The airplane's kinetic energy while cruising is thus \(7.5 \times 10^{9} \mathrm{J}\).

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