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Shkiper50 [21]
3 years ago
8

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

ength is 2.0 mT inside the solenoid near its center when a certain current flows through the coils. If the coils of the solenoid are now pulled apart slightly, stretching it to 21 cm without appreciably changing the size of the coils, what does the magnetic field become near the center of the solenoid when the same current flows through the coils
Physics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

B_2 = 1.71 mT

Explanation:

As we know that the magnetic field near the center of solenoid is given as

B = \frac{\mu_0 N i}{L}

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

\frac{B_1}{B_2} = \frac{L_2}{L_1}

now plug in all values in it

\frac{2.0 mT}{B_2} = \frac{21}{18}

B_2 = 1.71 mT

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A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coeff
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Answer:

a

H  =212.6 \  J

b

v  =  7.647  \  m/s

Explanation:

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   The child's weight is  W_c  =  287 \ N

    The length of the sliding surface of the playground is  L =  7.20 \  m

    The coefficient of friction is  \mu =  0.120

      The angle is \theta = 31.0 ^o

      The initial  speed is  u =  0.559 \  m/s

Generally the normal force acting on the child is mathematically represented as

=>    N  =  mg  *  cos \theta

Note  m *  g  =  W_c

Generally the frictional force between the slide and the child is    

         F_f  =  \mu *  mg  *  cos \theta

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      F =m* g sin(\theta) - F_f

Here  F is the resultant force and it is represented as  F =  ma

=>   ma =   m* g sin(31.0)  - \mu *  mg  *  cos (31.0)

=>   a =  g sin(31.0)-  \mu *  g  *  cos (31.0)

=>  a =    9.8 *  sin(31.0) - 0.120 *  9.8  *  cos (31.0)

=>a =  4.039 \ m/s^2

So

   F_f  =  0.120  * 287  *  cos (31.0)

=> F_f  = 29.52 \  N

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     H  =  F_f  * L

=>  H  = 29.52 *  7.2

=>  H  =212.6 \  J

Generally from kinematic equation we have that

    v^2  =  u^2  +  2as

=>  v^2  =  0.559^2  +  2 * 4.039 * 7.2

=>  v  =  \sqrt{0.559^2  +  2 * 4.039 * 7.2}

=>  v  =  7.647  \  m/s

   

6 0
3 years ago
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