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AVprozaik [17]
3 years ago
8

A horizontal disk rotates about a vertical axis through its center. Point P is midway between the center and the rim of the disk

, and point Q is on the rim. If the disk turns with constant angular velocity, which of the following statements about it are true? (There may be more than one correct choice.)
A. P and Q have the same linear acceleration.
B. The angular velocity of Q is twice as great as the angular velocity of P.
C. The linear acceleration of P is twice as great as the linear acceleration of Q.
D. The linear acceleration of Q is twice as great as the linear acceleration of P.
E. is moving twice Q as fast as P.
Physics
1 answer:
hjlf3 years ago
6 0

Answer:

The answer is explained below.

Explanation:

All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.

So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.

<em>Rp - R</em>

<em>Rq - 2R</em>

The linear acceleration is

α2 - Rα

So, the linear acceleration of Q is twice as great as the linear acceleration of P.

The speed of the particle when it is in the circular motion depends on the radius of the particle.

In this case, the speed of point Q is twice the speed of point P.

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Answer:

The induced emf  is  \epsilon  = 0.1041 \  V  

Explanation:

From the question we are told that

   The  radius of the circular loop is  r =  9.50 \ cm  =  0.095 \ m

     The  intensity of the wave is  I  =  0.0215 \ W/m^2

      The wavelength is  \lambda =  6.90\ m

Generally the intensity is mathematically represented as

         I  =  \frac{ c *  B^2  }{ 2 * \mu_o  }

Here  \mu_o is the permeability of free space with value  

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B is the magnetic field which can be mathematically represented from the equation as

          B  =  \sqrt{ \frac{ 2 *  \mu_o  *  I  }{ c} }

substituting values

          B  =  \sqrt{ \frac{ 2 *  4\pi *10^{-7} *   0.0215  }{ 3.0*10^{8}} }

          B  =  1.342 *10^{-8} \  T

The  area is mathematically represented as

       A =  \pi r^2

substituting values

       A =  3.142 *   (0.095)^2

       A = 0.0284

The angular velocity is mathematically represented as

        w =  2 *  \pi  *  \frac{c}{\lambda }

substituting values          

       w =  2 *  3.142   *  \frac{3.0*10^{8}}{ 6.90 }  

        w =  2.732 *10^{8} rad  \ s^{-1}  

Generally the induced emf is mathematically represented as

        \epsilon  =  N *  B  *  A  *  w * sin (wt )

At maximum induced emf  sin (wt)  =  1

    So

         \epsilon  =  N *  B  *  A  *  w

substituting values

         \epsilon  = 1  *    1.342 *10^{-8}   *  0.0284  *2.732 *10^{8}  

         \epsilon  = 0.1041 \  V  

         

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