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3 years ago

The illuminance On a surface is six lux and the surface is 4 meters from the light source what is the intensity of the source

Physics

1 answer:

Liula [17]3 years ago

5 0

Answer : 96 candelas

Explanation : Luminous intensity is a quantity of light that is emitted by a light source with a solid angle in a particular direction.

We know the formula of luminous intensity

Luminous intensity = illuminance x distance from light source

$L_{v} = E_{v}\times d^{2}$

$L_{v} = 6\times 16\ candelas$

$L_{v} = 96\ candelas$

So, the intensity of the source will be 96 candelas.

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Complete Question

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Explanation:

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The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

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The electric field change is mathematically defied as

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Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

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$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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