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Klio2033 [76]
3 years ago
8

A 50kg box is pulled on by a rope. The rope applies a horizontal force of 180 N to the box. What is the normal force exerted on

the box by the floor?
Physics
1 answer:
Over [174]3 years ago
4 0
The normal force would be the opposing and equal force to gravity ...  that is why the box isn't floating or going through the ground ... this is Newton's third law of motion

 
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That's two different things it depends on:

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-- Pour 1 ounce of water into a wide dish, with a large surface area.
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=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
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<u>Experiment B:</u>

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
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-- Start the fan.
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<em>Show that it took longer to evaporate when the air </em>
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6 0
3 years ago
Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistan
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A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its
True [87]

Answer:

The angular velocity is  w = 1.43\  rad/sec

Explanation:

From the question we are told that

   The  mass of wooden gate  is m_g = 4.5 kg

    The  length of side is  L = 2 m

    The mass of the raven is  m_r = 1.2 kg

     The initial speed of the raven is u_r = 5.0m/s

     The final speed of the raven is   v_r = 1.5 m/s

From the law of  conservation of angular momentum we express this question mathematically as

       Total initial angular momentum  of both the Raven and  the Gate =  The Final angular momentum of both the Raven and the Gate  

The initial angular momentum of the Raven is m_r * u_r * \frac{L}{2}

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is  zero

Note: This above is the generally formula for angular momentum of  square objects

  The final angular velocity  of the Raven is  m_r * v_r * \frac{L}{2}

   The  final angular velocity of the Gate  is   \frac{1}{3} m_g L^2 w

Substituting this formula

  m_r * u_r * \frac{L}{2}  =   \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r * v_r * \frac{L}{2} -   m_r * u_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r *  \frac{L}{2} * [u_r - v_r]

Where w is the angular velocity

     Substituting value  

   \frac{1}{3} (4.5)(2)^2  w   =    1.2 *  \frac{2}{2} * [5 - 1.5]

     6w = 4.2

       w = \frac{6}{4.2}

            w = 1.43\  rad/sec

5 0
2 years ago
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