The last two parts please Two astronauts (figure), each having a mass of 70.0 kg

Question

The last two parts please

Two astronauts (figure), each having a mass of 70.0 kg, are connected by a d 11.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.30 m/s. CM (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum of the two-astronaut system. 4081 kg m2s (b) Calculate the rotational energy of the system. 1.97 k) (c) By pulling on the rope, one astronaut shortens the distance between them to 5.00 m. What is the new angular momentum of the system? 4081 kg m2/s (d) What are the astronauts' new speeds? 11.66 m/s (e) What is the new rotational energy of the system? 0.98315X You know the speed and mass of each astronaut. How do you calculate the kinetic energy? k (t) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope? ki

Explanation / Answer

part(e)

new rotational kinetic energy = 2*(1/2)*m*vnew^2

Kinetic energy = 2*(1/2)*70*11.66^2 = 9.516 kJ <<<----------ANSWER

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f)

chemical energy converted = KEnew - KEold

chemical energy = 9.516 - 1.97 = 7.546 kJ <<<---------ANSWER

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