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Nezavi [6.7K]
2 years ago
7

The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the

resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W, what is the operating voltage of the bulb?
Physics
1 answer:
Alex787 [66]2 years ago
6 0

Answer:

The voltage operating is the 85,20563362

Explanation:

Power is the relation between Voltage and Current so knowing the resistance is going to be constant the equation of power can be just replacing in voltage terms:

1. P=I*V

using also Law OHM

2. V= I*R

I = \frac{V}{R}

Replacing 2 in the equation 1 so all the data are going to be in Voltage terms:

1. P= \frac{V}{R} *V

P= \frac{V^{2} }{R}  ⇒ R= \frac{V^{2} }{P}

R= \frac{110^{2} v }{60 w}

R= 201,6666667 Ω

So the resistance is constant so the current is going to be the same at the other Power 36 W:

P= \frac{V^{2} }{R}

V^{2} = R*P

V=\sqrt{R*P}

V=\sqrt{201,6666667*36 }

V=\sqrt{7260} ⇒V= 85,20563362

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There are two stages to the flight: acceleration stage and deceleration stage.

m₁ = 200 kg, mass of the rocket
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Sources:

https://wonderopolis.org/wonder/why-do-bridges-freeze-before-roads

http://wxguys.ssec.wisc.edu/2011/02/28/why-do-bridges-ice-before-the-road/

http://icyroadsafety.com/icybridges.shtml

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