The largest land predator is the male polar bear, which has a mass of around 500 kg. If he runs with a kinetic energy of:30,250 J. What is his velocity? El depredador terrestre más grande es el oso polar macho, que pesa alrededor de 500 kg. Si corre con una energía cinética de 30,250 J. ¿Cuál es su velocidad? 60.5m/s 24m/s 11m/s 100m/s

To find the velocity of the male polar bear, we can use the formula for kinetic energy:<br /><br />\[ KE = \frac{1}{2} m v^2 \]<br /><br />where:<br />- \( KE \) is the kinetic energy,<br />- \( m \) is the mass,<br />- \( v \) is the velocity.<br /><br />Given:<br />- \( KE = 30,250 \mathrm{~J} \)<br />- \( m = 500 \mathrm{~kg} \)<br /><br />We need to solve for \( v \). Rearrange the formula to solve for \( v \):<br /><br />\[ v^2 = \frac{2 \cdot KE}{m} \]<br /><br />Substitute the given values into the equation:<br /><br />\[ v^2 = \frac{2 \cdot 30,250 \mathrm{~J}}{500 \mathrm{~kg}} \]<br /><br />\[ v^2 = \frac{60,500 \mathrm{~J}}{500 \mathrm{~kg}} \]<br /><br />\[ v^2 = 121 \mathrm{~m}^2/\mathrm{s}^2 \]<br /><br />Now take the square root of both sides to find \( v \):<br /><br />\[ v = \sqrt{121 \mathrm{~m}^2/\mathrm{s}^2} \]<br /><br />\[ v = 11 \mathrm{~m}/\mathrm{s} \]<br /><br />Therefore, the velocity of the male polar bear is \( 11 \mathrm{~m}/\mathrm{s} \).<br /><br />The correct answer is:<br />\( 11 \mathrm{~m} / \mathrm{s} \)