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padilas [110]
3 years ago
14

The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is dou

bled without altering the charge on the plates, what is the new potential difference between the plates? Answer in units of V.
Physics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

= 796 V.

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500000000

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Creating plasma can be dangerous because of the high amount of ___ needed to create it.
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As air is heated, the particles speed up and spread out (becoming less dense). Hot air balloon pilots can change the balloon’s a
Jobisdone [24]

Answer:

Turn the heater on  

Explanation:

There are two main forces involved in a balloon flight

The downward force is the total weight of the balloon: the air it contains, the gas bag, the basket, the passengers, etc.

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3 years ago
Will mark as brainliest if correct!!!!!!!!!!!
Aneli [31]

Answer:

The answer is A.

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The diagram shows the light ray bending away from the normal. Light rays bend away from the normal when their speed increases. This means that in the diagram, the light ray moves from a medium in which light has a lower speed to a medium in light has a higher speed. The only choice where the speed of light increases from A to B is answer A. So that has to be the answer.

7 0
3 years ago
Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
3 years ago
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