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Zielflug [23.3K]
3 years ago
10

A 1.40-kg ball tied to a string fixed to the ceiling is pulled to one side by a force F→ . where L = 1.40 kg. What is the tensio

n in the string just before the ball is released and allowed to swing back and forth?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

T=13.72N

Explanation:

The tension before the ball is released have no angle is in rest at the same axis of the weight so:

∑F=0

Using Newton law in this case the ball is tied so tension before become to swing is

∑F=FN-T=0

T=F_{N}

T=m*g

T=1.40Kg*9.8\frac{m}{s^2}

T=13.72N

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dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

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At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

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Answer:

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The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

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