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2 months ago

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A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

nitude of the electric flux through the surface is 78 N · m2/C. What is the angle (less than 90) between the direction of the electric field and thenormal to the surface?

1 answer:

Softa [3K]2 months ago

4 0

Answer:

57.94°

Explanation:

We understand that the formula for flux is

$\Phi =E\times S\times COS\Theta$

where Ф represents flux

E indicates electric field

S denotes surface area

θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

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i) Refer to the attached diagram

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The problem does not provide full details: consult the attachments for the complete text.

a)

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By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

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b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

$W=mg$

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

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ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

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4 0

2 months ago