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Natasha2012 [34]
2 years ago
6

A loop of wire in the shape of a rectangle rotates with a frequency of 219 rotation per minute in an applied magnetic field of m

agnitude 6 T. Assume the magnetic field is uniform. The area of the loop is A = 4 cm2 and the total resistance in the circuit is 7 Ω.
Find the maximum induced emf.
e m fmax =_______

Find the maximum current through the bulb.
Imax=__________

Physics
1 answer:
bazaltina [42]2 years ago
6 0

Answer:

Emax = 0.055V

Imax = 7.86mA

Explanation:

See attachment below.

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A new conveyor system at the local packaging plant will utilize a motor powered mechanical arm to exertion average force of 890N
valkas [14]

Answer:

power =( 890 N x 12 m ) / 22 s=

=   485 Watts

Explanation:

4 0
2 years ago
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
A ball starts from rest and accelerates at a constant rate of 1.0m/s to a final velocity of
den301095 [7]

Answer:

Time taken to reach final velocity = 5.5 second

Explanation:

Given:

Initial velocity (Starting from rest)(u) = 0 m/s

Acceleration of ball (a) = 1 m/s²

Final velocity (v) = 5.5 m/s

Find:

Time taken to reach final velocity

Computation:

Using first equation of motion;

v = u + at

where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

5.5 = 0 + (1)(t)

5.5 = t

Time taken to reach final velocity = 5.5 second

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Answer C. 1.scientists learn by using the law of superposition B
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Answer:The answer is A

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