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sineoko [7]
4 years ago
11

he pressure inside a bottle of Champagne is 4.5 atm higher than the air pressure outside. The neck of the bottle has an inner ra

dius of 1.0 cm. What is the frictional force on the cork due to the neck of the bottle?
Physics
1 answer:
Effectus [21]4 years ago
4 0

Answer:

F = 141.3 N

Explanation:

Pressure inside the Champagne bottle = 4.5 atm

         1 atm = 10⁵ Pa

        4.5 atm = 4.5 x 10⁵ Pa

inner radius of the bottle = 1 cm = 0.01 m

Area of the bottle = π r²

                             = π x 0.01²

                             = 3.14 x 10⁻⁴ m²

Force acting on the cork due to excess pressure

  F = P A

  F =  4.5 x 10⁵  x 3.14 x 10⁻⁴

  F = 141.3 N

The frictional force on the cork due to the neck bottle will also be equal to F = 141.3 N

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
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