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Sedaia [141]
3 years ago
15

A [email protected] capacito

r is charged by a 170.0-V power supply, then disconnected from the power and connected in series with a 0.210-mH inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time t = 0 ms (the moment of connection with the inductor); (c) the energy stored in the inductor at t = 1.35 ms.
Physics
1 answer:
3241004551 [841]3 years ago
5 0

Answer:

Part a)

f = 70.9 Hz

Part b)

U = 111 J

Explanation:

As we know that the capacitor is of capacitance

C = 24 mF

V = 170 Volts

now the maximum charge on it

Q = CV

Q = (24 mF)(170 V)

Q = 4.08 C

Part a)

Oscillation frequency of the charge is given as

f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}

f = 70.9 Hz

Part b)

Now the equation of charge oscillation is given as

q = Q cos(\omega t)

q = 4.08 cos(2\pi(70.9) t)

now current in the circuit is given as

i = \frac{dq}{dt}

i = (4.08)(2\pi(70.9)) sin(2\pi(70.9) t)

now at t = 1.35 ms we have

i = 1028.36 A

So the energy stored in inductor is given as

U = \frac{1}{2}Li^2

U = \frac{1}{2}(0.210 \times 10^{-3})(1028.36)^2

U = 111 J

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3 years ago
An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele
Citrus2011 [14]

Answer:

The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

\tau=I\times\alpha

Here, I = mk²

\tau=mk^2\times\alpha

Put the value into the formula

\tau=3\times(0.2)^2\times0.5

\tau=0.06\ N-m

\tau=6.0\times10^{-2}\ N-m

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4 years ago
If a gun is sighted to hit targets that are at the same height as the gun and 75 m away at the same height, how low, as a positi
Vinvika [58]

Answer:

y=-1.66 m

Explanation:

   We know that

Range R

R=\dfrac{u^2sin2\theta }{g}

Here given that

u= 275 m/s

R=75 m

Now by putting the values

R=\dfrac{u^2sin2\theta }{g}

35=\dfrac{275^2sin2\theta }{9.81}

θ=0.13°

Now horizontal component of velocity u will be u cosθ.

Horizontal component = 275 cos0.13 = 274.99 m/s

So the time required to cover 180 m in horizontal direction t

180 = 274.99 x t

t=0.65 sec

Now vertical component of velocity u will be u sinθ.

Horizontal component = 275 sin0.13 = 0.62 m/s

So now vertical displacement y will be

y=u_yt-\dfrac{1}{2}gt^2

y=0.62 \times 0.65-\dfrac{1}{2}\times 9.81\times 0.65^2

y=-1.66 m

5 0
3 years ago
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