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02drai

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**My answer**

Minimum power the eye can detect= 10e-17w

1% of 60w=0.6w

Therefore, the maximum distance to see the lit bulb,

Power output ÷ surface area of a sphere = power per meter square

0.6÷(4πr^2)=10e-17

0.6÷10e-17=(4πr^2)

(6X10^16)/4π=r^2

so r=sqrt(6X10^16)/4π)

Please help me with this. Thanks.

Minimum power the eye can detect= 10e-17w

1% of 60w=0.6w

Therefore, the maximum distance to see the lit bulb,

Power output ÷ surface area of a sphere = power per meter square

0.6÷(4πr^2)=10e-17

0.6÷10e-17=(4πr^2)

(6X10^16)/4π=r^2

so r=sqrt(6X10^16)/4π)

__r= 217080km however in answer is 140km.__Please help me with this. Thanks.