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3 years ago

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Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i

ncreases when the plate area is decreased. The electric field is dependent on the charge density on the plates. The voltage of a connected charged capacitor remains the same when the plate area is decreased.

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

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A truck starts off 151 miles directly north from the city of Hartville. It travels due east at a speed of 41 miles per hour. Aft

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The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

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Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

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b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

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$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

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