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melisa1 [442]
3 years ago
12

Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is w

ound around the core N2 = 13 times. Shortly after the switch has been closed, the flux through the second coil of wire changes from 2.1 Wb per turn to 2.7 Wb in each turn of wire over a time interval of 0.3 s. Find the magnitude of the emf induced in the coil.
Physics
2 answers:
Elena L [17]3 years ago
8 0

Answer:

emf=26\ V

Explanation:

Given:

no. of turns in the first coil, n_1=7

no. of turns in the second coil, n_2=13

the change in flux through the secondary coil, d\phi=2.7-2.1=0.6\ Wb

time taken for the change in flux, dt=0.3\ s

  • According to the Faraday's law there will be an emf induced in the coil associated with the rate of change in magnetic flux.

<u>This emf is mathematically given as:</u>

emf=n_2\times \frac{d\phi}{dt}

emf=13\times \frac{0.6}{0.3} (we take no. of turns of the second coil because the rate of change in flux is associated with the second coil)

emf=26\ V

SVETLANKA909090 [29]3 years ago
5 0

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux d\Phi =2.7-2.1=0.6Weber

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt

So emf induced in the coil will be equal to 6 volt

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Maksim231197 [3]

Answer: 72.66 AU=1.089(10)^{10} km

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit”:

T^{2}\propto a^{3} (1)  

Now, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:  

T^{2}=a^{3} (2)  

So, knowing T=619.36 years and isolating a from (2) we have:  

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a=72.66 AU T his is the distance between the dwarf planet and the Sun in astronomical units

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a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km

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C is the correct answer

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The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

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Explanation:

Data

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h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

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     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

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                                                                   no negative times.

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