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3 years ago

What is the efficacy of a 60.0 W incandescent lightbulb that produces 830 lumens?

1 answer:

7 0

η = 13.8 lm/W. The luminous efficacy of a incandescent lightbulb that produces 830 lumens and consumed a power of 60 W is 13.8 lm/W.

The luminous efficacy of a light source is the relationship between the luminous flux (in lumens) emitted by a light source and the power (in watts). The luminous efficacy of a light source or luminous efficiency measures the part of electrical energy that is used to illuminate and is obtained by dividing the luminous flux emitted by the electrical power consumed. Luminous efficiency is expressed in lumens per watt (lm / W). It is given by the relation:

η = F / P. Where F is the luminous flux, and P is the power consumed by the light source.

The efficacy of a 60.0 W incandescent lightbulb that produces 830 lm is:

η = 830 lm / 60 W

η = 13.8 lm/W

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Answer:

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Explanation:

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Now, using 2nd equation of motion:

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So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

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Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

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t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

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Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

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Answer:

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Answer:

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