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julsineya [31]
3 years ago
9

What phenomena provides evidence of electric fields in the atmosphere

Physics
1 answer:
diamong [38]3 years ago
5 0
The phenomena<span> of </span>atmospheric<span> electricity are of three kinds. ..... In the Earth-</span>ionosphere cavity, the electric field<span> and conduction current in the lower </span>atmosphere<span> </span>
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The light bulb converts electrical energy into light and ____. A) chemical B) electromagnetic C) heat D) nuclear
morpeh [17]
The light bulb converts electrical energy into light and heat. 


Good luck! (:
7 0
3 years ago
Read 2 more answers
Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range
Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive  or attractive only

Y: Very small range

Z:  Repulsive and attractive

8 0
3 years ago
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
a car going 22 m/s accelerates to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration of the ca
Maksim231197 [3]
Using the formula

a\: = \frac{v - u}{t}

a = \frac{35 m/s \: - \: 22 m/s}{5s}

a = 2.6 {m/s^2}
5 0
3 years ago
A train increase its speed steadily from 10m/s to 20m/s in 1minutes A what is the average speed during this time in m/s B how fa
Ipatiy [6.2K]

If it increased its speed steadily at a constant rate, then the average speed for the minute was

(1/2)(10m/s + 20m/s) = 15 m/s .

Rolling at an average speed of 15 m/s for 1 minute (60 seconds), it travels

(15 m/s) (60 sec) = 900 meters

3 0
3 years ago
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