Kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is mass and \( v \) is speed.
If the hawk doubles its speed, let's denote the initial speed as \( v \). The initial kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 = 5.1 \text{ J} \]
When the speed is doubled (i.e., the new speed is \( 2v \)), the new kinetic energy will be:
\[ KE_{\text{final}} = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \left(\frac{1}{2}mv^2\right) = 4 \cdot KE_{\text{initial}} = 4 \cdot 5.1 \text{ J} = 20.4 \text{ J} \]
Therefore, the best prediction for the hawk's kinetic energy at the end of the chase is:
20.4 J
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