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Annette [7]
3 years ago
11

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

form electric field. What is the magnitude of the electric field if the field points 31.6° above the horizontal?
Physics
1 answer:
Tresset [83]3 years ago
4 0

Answer:231.16 N/C

Explanation:

Given

Electric Flux=147 N-m^2/C

Area(A)=0.824 m^2

Given Field point above 31.6 ^{\circ}

Therefore angle between Area vector Electric Field =90-31.6=58.4^{\circ}

We know that Flux is given by

\phi =\vec{E}\cdot \vec{A}

\phi =EAcos\theta

147=E\times 0.824\times cos(58.4)

E=231.16 N/C

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3 years ago
Read 2 more answers
A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

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v(max)² = 0.036

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v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

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K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
2 years ago
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