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SashulF [63]
4 years ago
14

A battery rated at 7.2 V and 10000 J is connected across a light bulb. Assume that the internal resistance of the battery is neg

ligible, and that the resistance of the bulb's filament is 100 Ω. Draw the associated circuit, and label all branch variables, following the Associated Variable Convention. What is the power into the bulb? What is the power into de battery? How long will the battery last in the circuit?

Engineering
1 answer:
Over [174]4 years ago
6 0

Answer:

b) Power to bulb = 0.5184 watt

Power into the battery = - 0.5184 watt

c) T = 19290.123 sec

Explanation:

Given data:

Voltage = 7.2 V

energy = 10,000 J

Resistance = 100 ohm

a) associated circuit is shown in attachment

b)Power = \frac{v^2}{R} = \frac{7.2^2}{100}

Power = 0.5184 watt

Power into the battery = - 0.5184 watt

c)

Power = \frac{E}{t}

T = \frac{E}{P}

T = \frac{10000}{0.5184}

T = 19290.123 sec

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Answer:

For 145 injuries, IR=82.86 // For 79 injuries, IR=45.14

Explanation:

As stablished by OSHA, incident rate (IR) is used as a measurement to compare a company's safety against the national average. It is calculated by multiplying the number of recordable cases by 200.000 and then dividing the result by the number of total labor hours of the company.

For 145 injuries:

IR= (145*200.000)/350.000 = 82.86

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