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6 October, 14:28

A small space telescope at the end of a tether line of length L moves at linear speed v about a central space station. What will be the linear speed of the telescope if the length of the line is changed to x*L? x = 2.8; v = 2 m/s?

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  1. 6 October, 14:50
    0
    The answer to the question is

    The linear speed of the telescope will be 5.6 m/s if the length of the line is changed to x*L where x = 2.8; and initial velocity v = 2 m/s

    Explanation:

    Speed = v₁ = ωL = 2 m/s

    When the line is changed to x*L where x = 2.8 the linear speed will be

    v₂ = 2.8 * L * ω = 2.8 * 2 = 5.6 m/s

    The linear speed varies with the angular speed following the relation v/r = ω where

    ω = angular speed

    v = linear speed and

    r = radius of the path of travel of the object at the vertex
  2. 6 October, 14:55
    0
    v' = 0.714 m/s

    Explanation:

    Solution:

    - Assuming no external torque is acting on the system then the angular momentum is conserved for the system.

    - The initial momentum angular Mi and final angular momentum Mf are as follows:

    Mi = Mf

    m*L*v = m*x*L*v'

    Where,

    m : mass of the telescope

    L : Length of teether line

    v: Initial speed

    v' : Changed speed.

    - Then we have:

    L*v = x*L*v'

    v' = v / x

    v' = 2 / 2.8

    v' = 0.714 m/s
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