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Fed [463]
3 years ago
11

Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible

mass. The distance between any two adjacent masses is l. What is the rotational inertia I_cm of this system about its center of mass?
1) 1/2 ml^2
2) 3 ml^2
3) ml^2
4) 2 ml^2
5) 4 ml^2
6) 7 ml^2
7) 5 ml^2
8) 6 ml^2
Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer: 5m/L^2

Explanation:

Inertial I = mr^2 where r = distance from axis of rotation, while m is the mass of the object.

I = 2[m(1L/2)^2] + 2[m(3L/2)^2] = 2m×. 25/L^2+ 3m×2. 25/L^2= 0. 5m/l^2 +4. 5m/l^2

= 5m/l^2.

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Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
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Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

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The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

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The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

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