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Bas_tet [7]
3 years ago
7

Can some one help me with number 2 urgent?!

Physics
1 answer:
babunello [35]3 years ago
6 0
A) I = 0
b) to the right side
c) to the left side
d) to the left side
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A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,
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P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2

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b) Rms value

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∴ Rms value of the electric field is 13.407 N/C

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E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C

∴ Peak value of the electric field is 18.96 N/C

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3 years ago
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