Question

a 1 kg hawk flies at a speed of 12 m/s. how much kinetic energy does the hawk have? remember : $e_k = \frac{1}{2}mv^2$ $e_k = ? j$

Explanation:

Step1: Identify given values

Mass $m = 1\ \text{kg}$, speed $v = 12\ \text{m/s}$

Step2: Substitute into kinetic energy formula

$$E_K = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (12)^2$$

Step3: Calculate the final value

$$E_K = \frac{1}{2} \times 1 \times 144 = 72$$

Answer:

$72$ J