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ra1l [238]
3 years ago
12

A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr

ic field of magnitude 73.9 N/C that is directed at 20 ∘ from the plane of the sheet. Find the magnitude of the electric fluxthrough the sheet.
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

6.07 Nm²/C

Explanation:

L = length of the flat sheet = 0.400 m

w = width of the flat sheet = 0.600 m

Area of the rectangular sheet is given as

A = L w

A = (0.400) (0.600)

A = 0.24 m²

E = magnitude of the electric field in the region = 73.9 N/C

θ = Angle of the electric field relative to normal to the plane of sheet = 90 - 20 = 70 deg

Magnitude of magnetic flux through the sheet can be given as

Φ = E A Cosθ

Inserting the values

Φ = (73.9) (0.24) Cos70

Φ = 6.07 Nm²/C

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89.5 Hz

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a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
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Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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