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Aleks [24]
3 years ago
5

When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large

grains a positive charge. The small grains are lifted higher by the wind.?
Physics
1 answer:
diamong [38]3 years ago
4 0

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.                  

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During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=
Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

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6 0
2 years ago
You run away from a plane mirror at 1.80 m/s. At what speed does your image move away from you?
Vlad [161]

Answer:

3.60m/s

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2 years ago
The sun is located in
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Milky Way Galaxy, same one as you.
7 0
2 years ago
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Jennifer, who has a mass of 48.0 kg, is riding at 40.2 m/s in her red sports car when she must suddenly slam on the brakes. An a
Bogdan [553]

Answer:

F = 3445.714\,N

Explanation:

The system can be modelled appropriately by the use of the Principle of Momentum Conservation and Impact Theorem:

(48\,kg)\cdot (40.2\,\frac{m}{s} ) - F\cdot (0.56\,s) = 0\,\frac{kg\cdot m}{s}

The average force exerted on her:

F = 3445.714\,N

3 0
3 years ago
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5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
2 years ago
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