Final answer:
The kinetic energy released during the explosion of a 6.0-kilogram object in space into three equal segments, with two segments moving at 20.0 m/s, is 800 J.
Step-by-step explanation:
The question deals with an explosion in space where a 6.0-kilogram object breaks into three segments, resulting in two segments moving at 20.0 m/s with a 60.0° angle between them.
Since the three fragments have equal masses and the system was initially at rest, conservation of momentum dictates that the velocity vector of the third segment must be equal and opposite to the vector sum of the first two. However, to find the kinetic energy released, we only need to consider the two segments observed since the total kinetic energy is the sum of the kinetic energy of all three segments.
The kinetic energy (KE) for each segment is given by the formula KE = ½mv^2, where 'm' is the mass of each segment and 'v' the speed. As the mass of each segment is 2 kg (6.0 kg / 3), and the speed is 20.0 m/s, the kinetic energy for one segment would be:
KE = ½ * 2 kg * (20.0 m/s)^2 = 400 J
Since there are two segments with the same kinetic energy, the total kinetic energy released in the explosion would be:
Total KE = 2 * 400 J = 800 J