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77julia77 [94]
3 years ago
8

A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery a

nd the plate separation is increased. What happens to the potential difference between the plates? A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased. What happens to the potential difference between the plates? The potential difference between the plates decreases. More information is needed to answer this question. The potential difference between the plates increases. The potential difference between the plates stays the same.
Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

The potential difference between the plates increases

Explanation:

As we know that the capacitance of the capacitor is given by:

q = CV         (1)

where

q = charge

C = capacitance

V = Voltage or Potential Difference

Also, the capacitance of a parallel plate capacitor is given as:

C = \frac{\epsilon_{o}A}{D}           (2)

where

\epsilon_{o} = permittivity of free space or vacuum

A = Area of the plates

D = Separation distance between the plates

Now, from eqn (1) and (2):

V = \frac{qD}{A\epsilon_{o}}

Now, from the above eqn  we can say that:

Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.

Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.

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An irrational number is a real number that cannot be written as a simple fraction.

4 = 4/1 = rational

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Answer:

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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of
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Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

h=(u\sin60)\times t-\dfrac{1}{2}gt^2

Put the value into the formula

h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2

h=41.6\ m

(b). We need to calculate the maximum height of the ball

Using formula of height

h_{max}=\dfrac{(u\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\sin60)^2}{2\times 9.8}

h=41.67\ m

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

v=u-gt

v=u\sin\theta-gt

Put the value into the formula

v_{y}=33\times\sin 60-9.8\times3.0

v_{y}=-0.82\ m/s

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=33\times\cos60

v_{x}=16.5\ m/s

We need to calculate the ball's impact speed

Using formula of velocity

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(16.5)^2+(-0.82)^2}

v=16.52\ m/s

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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