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IgorC [24]
3 years ago
6

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?
a. 78 V
b. 72 V
c. 66 V
d. 90 V
e. 85 V
Physics
1 answer:
laila [671]3 years ago
3 0

Answer:

c. 66 V

Explanation:

p =IV

I =P/V

P1/V1=P2/V2

60/110=36/V2

0.55 = 36/V2

V2 =36/0.55 = 65.5V

V2 = 66V

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Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

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Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

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n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

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V = l w x

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Taking the 1st derivative:

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Set dV/dx = 0 to get maxima:

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Divide by 12:

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So the largest volume is:

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