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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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Induced emf, \epsilon=9.79\times 10^7\ volts

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We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

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48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

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First convert km/h to m/s

(97 × 1000)/3600

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26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

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O = 26.944^2 - 2 × 3.5 S

726 = 7S

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The westbound train

Convert km/h to m/s

(127×1000)/3600

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35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

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300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

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3 years ago
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