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kow [346]
3 years ago
13

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.
Physics
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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<h3>Answer: any path that allows electrons to flow</h3>

An electrical circuit is a path in which electrons from a voltage or current source flow. ... The part of an electrical circuit that is between the electrons' starting point and the point where they return to the source is called an electrical circuit's "load".

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The object distance for a concave lens is 8.0 cm, and the image distance is 12.0 cm. The height of the object is 4.0 cm. What is
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Two identical stones are thrown from the top of a tall building. Stone 1 is thrown vertically downward with an initial speed v,
Molodets [167]

If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

v= \frac{1}{2}gt

Where

g = Gravitational acceleration

t = time

As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.

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5 0
3 years ago
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
Bess [88]

Answer:

9.51

Explanation:

The distance s is given by:

s(t) = \sqrt{151^2 + (vt)^2}

The change in distance is given by the time derivative of s:

\frac{ds}{dt} = \frac{v^2t}{\sqrt{151^2 + (vt)^2}}

For the time t you solve the equation of distance x for time:

x = vt => t = \frac{x}{v}

Plugging in for t:

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6 0
3 years ago
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
Cerrena [4.2K]

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}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

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}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\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

5 0
3 years ago
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