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kondor19780726 [428]
3 years ago
12

Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni

tude of the maximum acceleration is typically 1.3 m/s2 , but the driver will usually maintain a constant acceleration that is less than the maximum. A train travels through a congested part of town at 4.0m/s . Once free of this area, it speeds up to 13m/s in 8.0 s . At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed.
What is the final speed?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Mkey [24]3 years ago
3 0
During the first phase of acceleration we have:
v o = 4 m/s;  t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?;  v o = 13 m/s; a = 1.125 m/s² ;  t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s 
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Answer:

The induced emf is  |\epsilon|=0.0261 V

Explanation:

From the question we are told that

     The number of turn is  N = 100

      The diameter of the coil is  d = 2.0 cm = 2.0 *10^{-2} m

      The uniform magnetic at initial is B_i = 0.50 T

      The uniform magnetic at initial is B_f = 1.50 T

      The time taken is t = 0.60s

      The angle the magnetic field makes with vertical is  \theta = 60^o

Generally induced emf is mathematically represented as

     \epsilon = -N \frac{d \o}{dt}

where d \o is the change  magnetic flux

Magnetic flux is mathematically represented as

       \O = \= B \cdot \= A

          = BA cos \theta

Substituting this above  

       \epsilon = -N \frac{d (BA cos \theta)}{dt}

      \epsilon = -N A \frac{d (B cos \theta)}{dt}

  Where B is the magnetic field and A is the area which is mathematically evaluated as

        A = \frac{\pi  d^2}{4}

Substituting values

        A = \frac{\pi (2.0 *10^{-2})^2}{4}

           A= 3.142*10^{-4}m^2

From the equation of emf

          \epsilon = -N A \frac{d (B cos \theta)}{dt}

dB = B_2 -B_1

       So

             |\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}

substituting values

            |\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}

                |\epsilon|=0.0261 V

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