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Pani-rosa [81]
4 years ago
11

Question 1 (1 point)

Physics
1 answer:
TiliK225 [7]4 years ago
3 0

Answer:

The resultant velocity of the plane relative to the ground is;

150 kh/h north

Explanation:

The flight speed of the plane = 210 km/h

The direction of flight of the plane = North

The speed at which the wind is blowing = 60 km/h

The direction of the wind = South

Therefore, representing the speed of the plane and the wind in vector format, we have;

The velocity vector of the plane = 210.\hat j

The velocity vector of the wind = -60.\hat j

Where, North is taken as the positive y or \hat j direction

The resultant velocity vector is found by summation of the two vectors as follows;

Resultant velocity vector = The velocity vector of the plane + The velocity vector of the wind

Resultant velocity vector = 210.\hat j  + (-60.\hat j) =  210.\hat j  - 60.\hat j = 150.\hat j

The resultant velocity vector = 150.\hat j

Therefore, the resultant velocity of the plane relative to the ground = 150 kh/h north.

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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
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