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Tpy6a [65]
3 years ago
10

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

anged?
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

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This is an example of which scientific practice?
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8 0
2 years ago
Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a
xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, h_b.

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, h_d. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be h_d and the distance travelled by the ball measured from the top be h_b.

It follows that h_d+h_b=9.8.

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

h_b=0\times t + 0.5\times 9.8t^2 (since v_{b0} =0.)

h_b=4.9t^2

Dart:

h_d=17.8\times t - 0.5\times9.8t^2 (the acceleration is opposite to the displacement, hence the negative sign)

h_d=17.8\times t - 4.9t^2

But

h_b+h_d =9.8

17.8\times t - 4.9t^2+4.9t^2 =9.8

17.8\times t = 9.8

t = 0.55

The height of the collision is the height of the dart above the ground, h_d.

h_d=17.8\times t - 4.9t^2

h_d=17.8\times 0.55 - 4.9\times(0.55)^2

h_d=9.79 - 1.48225

h_d=8.3

8 0
4 years ago
Which statements are part of the safety protocol for this lab experiment? Check all that apply.
tekilochka [14]

Answer:

Always wear safety goggles when performing an experiment.

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8 0
3 years ago
Read 2 more answers
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