What is the kinetic energy of 25,000 L of water (18.02 g/mol) traveling at 5.8 mph? Assume that the density of water is 1.0 g/mL.(1 mi= 1.609 km,1 J =1 kg m²s ^- ² )

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First, convert the volume of water to mass. Since the density of water is 1.0 g/mL, 25,000 L (or 25,000,000 mL) of water has a mass of 25,000,000 g. Converting this to kilograms gives 25,000 kg. Next, convert the velocity from mph to m/s: $5.8 \, \text{mph} = 5.8 \times 1.609 \, \text{km/h} = 9.3322 \, \text{km/h} = 9.3322 \times \frac{1000}{3600} \, \text{m/s} = 2.5923 \, \text{m/s}$. The kinetic energy (KE) is given by the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass in kilograms and $v$ is the velocity in meters per second. Substituting the values gives $KE = \frac{1}{2} \times 25000 \times (2.5923)^2 \approx 84,000 \, \text{J} = 84 \, \text{kJ}$.

What Is the Kinetic Energy of 25,000 L of Water