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MA_775_DIABLO [31]
3 years ago
5

the light intensity of a source is 100 Candelas. The illuminance on a surface is 4 Lux. How far is the surface from the source

Physics
2 answers:
pychu [463]3 years ago
6 0

Answer:

The surface from the source is 5 m.

Explanation:

Given that,

Light intensity = 100 candelas

Illuminance = 4 lux

The formula of illumination is

According to illumination law,

The intensity of illuminance of the the light source is inversely proportional to the square of the distance from the source.

The formula of illumination is

E\propto\dfrac{1}{d^2}

E=\dfrac{I}{d^2}

Where, E = illuminance

I = luminous Intensity of the light

D = distance

We substitute the value into formula

d^2=\dfrac{I}{E}

d^2=\dfrac{100}{4}

d^2=25

d = 5\ m

Hence, The surface from the source is 5 m.

-Dominant- [34]3 years ago
5 0
The answer is: 5 meters 
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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
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(a) 3.1\cdot 10^7 J

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E_i = E_f\\K_i + U_i = K_f + U_f (1)

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U_i is the gravitational potential energy at the surface

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U_i is the final gravitational potential energy

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K_i = 5.0 \cdot 10^7 J

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At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

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U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

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(b) 6.3\cdot 10^7 J

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r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

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So now we can use eq.(1) to find the initial kinetic energy:

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The mechanical energy must be conserved, so E_i = E_f, so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
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8 0
3 years ago
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