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Gre4nikov
3 months ago
9

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

The distance d between both lines is 4.10 m. What is the magnitude of the electric field at a point equidistant from the lines, with a distance 2d from both lines?

Physics
1 answer:
Sav [3.1K]3 months ago
6 0

Answer:

The strength of the electric field at an equidistant point from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to find the angle

Using angle formula

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to determine the magnitude of the electric field at a point equidistant from the lines

Using electric field formula

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Plugging the values into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Thus, the magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

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Maru [3345]

0.04m²

Explanation:

Known values:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot =?

Solution:

Pressure is defined as the force applied per unit area of an object

  Pressure = \frac{force}{area}

To determine the area;

        Area = \frac{force }{pressure}

    Area = \frac{40000}{250000} = 0.16m²

The force exerted by all four feet amounts to 0.16m²

thus, the area for each foot is \frac{0.16}{4} = 0.04m²

Learn more:

Pressure

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2 months ago
An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Maru [3345]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

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-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

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7 0
3 months ago
A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o
Ostrovityanka [3204]

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

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   Em₀ = U = mg y

Bottom

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    Emo =Em_{f}

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

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Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

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7 0
3 months ago
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