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Marysya12 [62]
3 years ago
15

The magnetic field due to the loop always opposes the external magnetic field.(b) The flux due to the loop always has the opposi

te sign as the flux due to the external magnetic field.(c) The flux due to the loop always opposes the change in the flux due to the external magnetic field
Physics
1 answer:
JulsSmile [24]3 years ago
6 0

Answer: C

Explanation:

The flux due to the loop always opposes the change in the flux due to the external magnetic field. This occurs as a result of the current that is induced in a coil (due to a magnetic flux change through the coil) which will always be in such that it opposes the change that caused it. ... Any induced current in a coil will result in a magnetic flux that is opposite to the original changing flux. This was what  Faraday was explaining in the law of Induction

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Answer:

a

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b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

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So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

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We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

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