What do you think that astronomers mean when they use the term observable universe?
1 answer:
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Answer:
a)
b) I =
c) N*m
d) a =
e) W= 0.25 rad/s
Explanation:
a) we know that:
where is the ubicaton of the center of mass,
the mass of the first rocket,
its distances with the rocket 1,
the mass of the second rocket and
its distance with the rocket 1. So, replacing values, we get:
So, the center of mass is at 60m from the rocket 1.
b) we know that:
where I is the moment of inertia, is the mass of the rocket 1,
its distance from the center of mass,
the mass of the second rocket and
the distance between the rocket 2 and the center of mass. So, replacing values, we get:
I =
c) We know that:
T = Fr
where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:
Replacing values, we get:
50,000N(60m)+50,000N(30m) = T
N*m
d) We know that:
T = Ia
where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:
solving for a:
a =
e) Finally, using:
W = at
where W is the angular velocity, a is the angular aceleration and t is the time.
Then, replacing values. we get:
W = ()(30s)
W = 0.25 rad/s