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34kurt
3 years ago
11

In a thunderstorm, the air must be ionized by a high voltage before a conducting path for a lightning bolt can be created. an el

ectric field of about 1.0 × 106 v/m is required to ionize dry air. what would the break- down voltage in air be if a thundercloud were 1.60 km above ground? assume that the electric field between the cloud and the ground is uniform.
Physics
1 answer:
nydimaria [60]3 years ago
4 0
We know that the electric field is equal to 1 E 6 V/m

The distance between the thundercloud and the ground is 1.6km = 1600m

Electric field = Voltage/distance

This means that the breakdown voltage must be equal to

V = 1 E 6 V/m * 1600m = 1.6 E 9 V = 1.6 GV
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figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un
Vlad [161]

Answer:

E = 307667  N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then  E = 0.007384/2.4 * 10^(-8)  N/C

E = 307667  N/C

8 0
3 years ago
How long would this acceleration last
Aleksandr-060686 [28]
It would last as long as the applied force continued, or until the accelerating object hit something.
6 0
3 years ago
Extrusive rocks forms beneath earth's surface true or false
Verizon [17]
The answer is false your welcome
8 0
4 years ago
Find the poing of center of gravity<br><br>plz show the steps...​
olasank [31]

Answer:

C

Explanation:

For a uniformly distributed mass, the center of gravity is also the geometric center.  For this shape, the center is at point C.

7 0
3 years ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
3 years ago
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