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alina1380 [7]
2 years ago
9

A cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

No other charges are nearby. (ii) Through how many of the cube's faces is the electric flux zero? Choose from the same possibilities as in part (i).
Physics
1 answer:
laiz [17]2 years ago
7 0

Answer:

The electric flux is zero through four cube surfaces given that a cubical gaussian surface surrounds a long, straight, charged filament that passes perpendicularly through two opposite faces.

Explanation:

Assuming the charged filament is quite long and you are not near the edges, the two opposing sides that the filament travels through have no flux. If the charge filament is long, which you may assume is indefinitely long, then there is the equal amount of charge on the left and right of where you are, therefore the electric field has no preference for left or right. This implies that the electric field can only travel in or out of the filament. No field lines run through the two faces of the cube that the filament goes through if the electric field is not moving left or right. There are electric field lines on the four sides of the filament.

To learn more about cubical gaussian surface and electric flux. Click brainly.com/question/13003911

#SPJ4

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The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
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The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

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Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

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