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geniusboy [140]
3 years ago
14

A 9.38 microFarad capacitor is measured to have a reactance of 260.0 Ohm. At what frequency (in Hz) is it being driven?

Physics
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

The frequency is 65.25 Hz.

Explanation:

Given that,

Capacitor = 9.38 μF

Reactant = 260.0 ohm

We need to calculate the frequency

Using formula of reactant

X_{c}=\dfrac{1}{2\pi fC}

Where, f = frequency

C = capacitor

X_{c} =reactant

Put the value in to the formula

260.0=\dfrac{1}{2\times\pi\times f\times9.38\times10^{6}}

f=\dfrac{1}{2\times\pi\times9.38\times10^{-6}\times260.0}

f=65.25\ Hz

Hence, The frequency is 65.25 Hz

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Answer:

the ball's velocity was approximately 0.66 m/s

Explanation:

Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.

Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.

Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:

0.7 = (1/2) g t^2

solve for t:

t^2 = 1.4 / g

t = 0.3779  sec

which we can round to about 0.38 seconds

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horizontal distance covered = vi * t

0.25 = vi * (0.38)

solve for vi:

vi = 0.25/0.38  m/s

vi = 0.65798  m/s

Then the ball's velocity was approximately 0.66 m/s

4 0
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Here,
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