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Vlad1618 [11]
2 years ago
8

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

lock and the incline is 0.425. The acceleration of gravity is 9.8 m/s 2 . Find the work done by the force of gravity. Answer in units of J.
Physics
1 answer:
Veronika [31]2 years ago
8 0

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

\theta = Angle of slide = 20.7°

\mu = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J

The work done by the force of gravity is 23.52092 J

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Answer:

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3 years ago
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

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     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

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

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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

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k = 1.4

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Which best describes the transition from gas to liquid?

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