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schepotkina [342]
3 years ago
10

An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant

when the elevator is accelerating downward at a rate of 2.0 , what is the mass of the suspended object
Physics
1 answer:
Phantasy [73]3 years ago
4 0

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>T</em> - <em>mg</em> = - <em>ma</em>

where

• <em>T</em> = 25 N, the tension in the string

• <em>m</em> is the mass of the object

• <em>g</em> = 9.8 m/s², the acceleration due to gravity

• <em>a</em> = 2.0 m/s², the acceleration of the elevator-object system

Solve for <em>m</em> :

25 N - <em>m</em> (9.8 m/s²) = - <em>m</em> (2.0 m/s²)

==>   <em>m</em> = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg

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3 years ago
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca
zlopas [31]

Answer:

The value is  I = 0.0932 ML ^2  

Explanation:

From the question we are told that

  The rotational inertia about one end is I_R =  \frac{1}{3} ML^2

   The location of the axis of rotation considered is d =  0.4 L

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  0.4 M

Generally the length of the rod from the its beginning to the axis of rotation consider is

      k = 1 - 0.4 L = 0.6L

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    m  =  1- 0.4 M = 0.6 M

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     I_{R1} =  \frac{1}{3} (0.6 M )(0.6L)^2

    I_{R1} =  \frac{1}{3} (0.6 M )L^2 0.6^2

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     I_{R2} =  \frac{1}{3} (0.6 M )(0.6L)^2

=> I_{R2} =  \frac{1}{3} (0.4 M )(0.4L)^2

=>  I_{R2} =  \frac{1}{3} (0.4 M )L^2 0.4^2

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  I =   \frac{1}{3} (0.6 M )L^2 0.6^2 +   \frac{1}{3} (0.4 M )L^2 0.4^2

=>   I =  \frac{1}{3} ML ^2  [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]

=>   I = 0.0932 ML ^2  

8 0
3 years ago
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