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Allisa [31]
3 years ago
15

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

diameter of the wire in the fuse?
Physics
1 answer:
Alborosie3 years ago
8 0

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

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Answer:

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Explanation:

Given data:

frequency of the light f = 5.09 x 10^{14} Hz

angle of incidence Θ_{i} = 55^{o}

index of refraction of material n_{2} = 1.66

A) To find material X

Given the index of refraction is 1.66 and hence the material is the flint glass

B) To calculate the speed of light in the material.

We know that the relation between index of refraction (n), velocity of light (c = 3 x 10^{8} m/s) and velocity of light is given by the equation:

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Hence,

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                                                          = \frac{3 x 10^{8} }{1.66}

                                                          = 1.807 x 10^{8} m/s

C) To calculate angle of refraction of light in medium X

We know that the Snell's law states that

                       n_{1} sin Θ_{i}  = n_{2} sin Θ_{r}

n_{1} = incident index

n_{2}  = refracted index

Θ_{i} = incident angle

Θ_{r} = refracted angle

In given problem, n_{1} = 1 since medium is air

Substituting the known values, we get

1 x  sin 55^{o}  = 1.66 x sin Θ_{r}

sin Θ_{r} = sin 55^{o} /1.66

           = 0.4935

Hence, angle of refraction of light in medium X Θ_{r} = sin^{-1} (0.4935)

                                                                                       = 29.57^{o}

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