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Tju [1.3M]
4 years ago
6

an airplane is traveling 735 KM/H and direction 41.5 west of north find the component of the velocity vector in the notherly and

in Westerly directions how far north and how far west has the plane traveled after 3.00h
Physics
2 answers:
Alchen [17]4 years ago
5 0

Answer:

v_x = 550.5 km/h Along North

v_y = 487 km/h Along West

d_{west} = 1461 km Along West

d_{north} = 1651.5 km Along North

Explanation:

Velocity of Airplane is given as

v = 735 km/h at 41.5 degree West of North

now we know that

Along North Direction

v_x = v cos41.5

v_x = 550.5 km/h

Along West direction

v_y = 735 sin41.5

v_y = 487 km/h

Now distance moved by the plane in 3.00 h is given as

Along west

d_{west} = 487 \times 3

d_{west} = 1461 km

Along North

d_{north} = 550.5 \times 3

d_{north} = 1651.5 km

damaskus [11]4 years ago
4 0
The. correct answer is 53.1
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Explanation:

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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:

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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}

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3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2

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Show work/ no links
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