I need Steps Please.!!! Two astronauts (figure), each having a mass of 70.0 kg,

Question

I need Steps 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 kg m2/s (b) Calculate the rotational energy of the system kJ (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? kg m2/s (d) What are the astronauts' new speeds? m/s (e) What is the new rotational energy of the system? (f) 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

here,

m = 70 Kg

d = 11 m

v = 5.3 m/s

a)

magnitude of angular momentum = m * v * d

magnitude of angular momentum = 70 * 11 * 5.3

magnitude of angular momentum = 4081 Kg.m^2/s

b)

rotational energy of the system = 2 * 0.50 * m * v^2

rotational energy of the system = 70 * 5.3^2

rotational energy of the system = 1966 J = 1.97 kJ

c)

as there is no external torque acting on the system

the angular momentum will reamin constant

the final angular momentum will be 4081 Kg.m^2/s

d) let the new speed is v

70 * v * 5 = 4081

v = 11.66 m/s

the new speed is 11.7 m/s

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