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3 years ago

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A wire of length L is wound into a square coil with 167 turns and used in a generator that operates at 60.0 Hz and 120 V rms val

ue in a 0.041-T magnetic field. What is the length L of the wire used to construct the coil

1 answer:

Mrac [35]3 years ago

8 0

Answer:

$171.43m$

Explanation:

First, define $emf$ ( electromotive force )-is the unit electric charge imparted by an energy source. In this case the generator.

The peak emf is:

$E_p_e_a_k=\sqrt2(E_m_a_x)$ = $\sqrt(2\times 120V)=170V$

Substituting $w=2\pi f$ and the value for peaf $E$ gives:

Total length= $4\sqrt {\frac{NE_p_e_a_k}{Bw}$ = $4\sqrt{\frac{167\times 170}{0.041T\times 2pi \times 60.0Hz}$

= $171.43m$

Hence, wire's length is 171.43m

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Vadim26 [7]

Answer:

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Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

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Here

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$M =V \rho$

Where,

V = Volume

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Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

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For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

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$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

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As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

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Step2247 [10]

Answer:

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Your question has been heard loud and clear.

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Thank you

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3 years ago

Read 2 more answers