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Marizza181 [45]
2 years ago
12

G a person of mass 100 kg is riding an elevator which was initially moving up with a velocity of 3 m/s. over a distance of 4 m t

he elevator slows down to 1 m/s. find the "apparent" mass of the person during that time (as the elevator slows down).
Physics
1 answer:
andriy [413]2 years ago
3 0
E=mc² where c is speed of the light
3 m/s more andmore less than speed of the light. So mass of the person still 100 kg
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Answer:

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Explanation:

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     I = ∫ r² dm

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Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

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The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

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Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

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Answer:

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