A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

Where:

- Linear momentum of the lion, measured in kilograms-meters per second.

- Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

- Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

Vectorially speaking, the final velocity of the lion-gazelle system is:

Where:

, - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

, , - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If , , and , the final velocity of the lion-gazelle system is:

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.