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Setler [38]
3 years ago
8

If a 200 turn, 10-3 m2 cross-sectional area coil is immersed in a magnetic field such that the plane of the coil is perpendicula

r to the field. If the magnetic field increases by 6 T/s, then how much voltage is induced
Physics
1 answer:
Stels [109]3 years ago
8 0

Answer:

<h3>1.2 volt induced in coil.</h3>

Explanation:

Given:

Number of turns N =  200

Cross sectional area A = 10^{-3} m^{2}

Rate of increasing magnetic field \frac{dB}{dt} = 6 \frac{T}{s}

From the faraday's law,

Induced emf is given by,

   \epsilon = -N\frac{d \phi}{dt }

Where \phi = magnetic flux

  \phi = AdB \cos(0)          ( because angle between normal coil and field is zero)

Where A = area of coil

Put the value of \phi in above equation,

Here we neglect minus sign

   \epsilon = NA\frac{dB}{dt}

   \epsilon = 200 \times 10^{-3} \times 6

   \epsilon = 1.2 V

Therefore, 1.2 volt induced in coil

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Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

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But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

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v = 7.2/2.3

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4 0
3 years ago
Which of Newton's laws explains why satellites need very little fuel to stay in oribit?
Angelina_Jolie [31]

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4 0
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VikaD [51]
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Hope this helps :)
3 0
3 years ago
What is the elevation of point B?<br> What about F?
kolbaska11 [484]

Explanation:

Since I can only do this by observation, the elevation of F is approximately 850km and the elevation of B is 925km.

7 0
3 years ago
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
3 years ago
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