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2 years ago

10

The amplitude of the voltage across an inductor can be greater than the amplitude of the generator EMF in an AC RLC circuit. Con

sider an RLC circuit with Epeak = 90 V, R = 28.9 capital omega, L = 1.31 H, and C = 2.86 µ F. Find the amplitude of the voltage across the inductor at resonance. V

1 answer:

Degger [83]2 years ago

4 0

Answer:

VL=2107.6v

Explanation:

at resonance xc=xL

1/wc = wL

z=R

because sqr(R^2+(xL-xc))

^

(xL=xc)

V/R=I

90/28.9=3.1142A

w=1/sqr(1.31×(2.86×10^-6))=516.63

xL= wL

xL= 516.63×1.31=676.785

VL=3.1142×676.785

VL=2107.6v

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A ______ is a closed loop containing a source of electrical energy and a load.

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Answer:

d. circuit

Explanation:

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Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

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Explanation:

Given masses:

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a)

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$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

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b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

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$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

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2 years ago

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and this is what the diver use during his diving

so correct answer will be

<u><em>It decreases her moment of inertia.</em></u>

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2 years ago

Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat

Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

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The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

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Lyrx [107]

Answer:

$\theta=4.64^{\circ}$

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2 years ago