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Varvara68 [4.7K]
3 years ago
14

Interactive Solution 22.43 provides one model for solving this problem. The maximum strength of the earth's magnetic field is ab

out 6.9 x 10-5 T near the south magnetic pole. Suppose we want to utilize this field with a rotating coil to generate 54.1-Hz ac electricity. What is the minimum number of turns (area per turn = 0.024 m2) that the coil must have to produce an rms voltage of 170 V?
Physics
1 answer:
laila [671]3 years ago
5 0

Answer:

427097

Explanation:

V = Voltage = 170 V

B = Magnetic field = 6.9\times 10^{-5}\ T

f = Frequency = 54.1 Hz

A = Area = 0.024\ m^2

Number of turns is given by

N=\dfrac{V\sqrt{2}}{BA2\pi f}\\\Rightarrow N=\dfrac{170\sqrt{2}}{6.9\times 10^{-5}\times 0.024\times 2\pi\times 54.1}\\\Rightarrow N=427096.933536\approx 427097\ turns

Number of turns is 427097

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Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

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     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

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Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

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Here it is given that the object oscillates 20 times in 10 seconds.

So f = \frac{20}{10} = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

T = \frac{1}{f}

T= \frac{1}{2} =0.5 s

<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>

7 0
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