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krok68 [10]
3 years ago
10

he plates of a parallel - plate capacitor are maintained with constant voltage by a battery as they are pulled apart. What happe

ns to the strength of the electric field between the plates during this process?
Physics
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

The strength of the electric field between the plates of the capacitor decreases during this process.

Explanation:

The strength of electric field between the plates of a capacitor is given as

E = (V/d)

where V = potential difference across the plates of the capacitor

And d = distance between the two places

So, if the potential difference across the plates of the capacitor stays constant, the electric field strength varies inversely as the distance between the plates.

Therefore, as the plates of the capacitor are pulled apart, with distance between them increasing (with constant voltage maintained across the plates), the strength of the electric field between the plates decreases.

Hope this Helps!!

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0
4 years ago
The planet Neptune is the farthest planet from the Sun. A satellite is orbiting around Neptune at a distance of 180 km. As the s
stich3 [128]
2,992² + r² = ( r + 180 )²
8,952,064 + r² = r² + 360 r + 32,400
360 r = 8,952,064 - 32,400
360 r = 8,919,664
r = 8,919,664 : 360
r = 24,776.844 km
d = 24,776.844  ·  2 = 49,553688 ≈ 49,554 km
Answer:
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4 years ago
You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a
iogann1982 [59]

Answer:

a) tangent ; b) 153.88 yds ; c) No , less than or equal to 45° ; D) 45°

Explanation:

Given the following ;

From the triangle sketch :

Base length = 50 yards

Angle of elevation = 72°

a. Which trigonometric ratio would you use to calculate how high the blimp will be above the 50 yard line?

Using trigonometry :

The height of the blimp will be calculated using :

Tangent :

Tan θ = opposite / Adjacent

B.) How high above the ground is the blimp?

Using :

Tan θ = opposite / Adjacent

Θ = 72° ; adjacent = 52, opposite = height(h)

Tan 72° = h / 50

h = 3.0776835 * 50

h = 153.88 yds

C.) In order to be able to read the advertisement on the side of the blimp the highest the blimp can be is 150 feet. Will the fans be able to read the advertisement?

1 yard = 3 Feets

153.88 yards = 3 * 153.88

= 461.65 feets

No, because the height of the blimp is 461.65 Feets which is greater than 150 Feets.

To make viewing possible, the angle of elevation should be:

50 yards is equivalent to (3 * 50) = 150 feets

Max imum Height of blimp = 150 Feets

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Tanθ = 150 Feets / 150 Feets

Tanθ = 1

θ = tan^-1(1)

θ = 45°

To make viewing advertisement possible, angle of elevation should not exceed 45°

d.)If height of blimp is 150 Feets, then the exact angle of elevation will be 45°

5 0
3 years ago
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