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Gre4nikov
17 days 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 [2.2K]17 days 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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Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

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Average Acceleration for the interval from 3s to 8s

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Acceleration can be defined as the first derivative of velocity concerning time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}

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Yuliya22 [2420]

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C. vx

F. ax

G. ay

Clarification:

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The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

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28 days ago
Select the areas that would receive snowfall because of the lake effect.
Keith_Richards [2256]

Result:

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Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

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25 days ago
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Answer:

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This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

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