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3 years ago

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You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 130-turn coil that has

an area per turn of 0.82 m2. What should be the magnitude of the magnetic field in which the coil rotates

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

B = 0.129 T

Explanation:

Given,

frequency, f = 60 Hz

maximum emf = 5200 V

Number of turns, N = 130

Area per turn = 0.82 m²

We know,

ω = 2 π f

ω = 2 π x 60 = 376.99 rad/s

now, Magnetic field calculation

$B =\dfrac{\epsilon_{max}}{NA\omega}$

$B =\dfrac{5200}{130\times 0.82\times 376.99}$

B = 0.129 T

Hence, the magnetic field is equal to B = 0.129 T

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Dafna11 [192]

Answers:

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b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

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$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

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With this in mind, let's begin with the answers:

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Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

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b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

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Finally:

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