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katrin2010 [14]
3 years ago
10

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

Physics
1 answer:
Alex17521 [72]3 years ago
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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f = 1/2π√(k/m)    m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²

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