To solve this problem we will apply the relation of Ohm's law, at the same time we will use the concept of resistance in a cable, resistivity and potential difference.

According to Ohm's law we have to

$V= IR$

Here,

V = Voltage

I = Current

R = Resistance

At the same time resistance can be described as

$R = \frac{\rho l}{A}$

Here,

$\rho$ = Resistivity of the material

l = Length of the specimen

A = Cross-sectional area

From the above expression we can write the current as,

$I = \frac{V}{R}$

$I = \frac{V}{\frac{\rho l}{A}}$

$I =\frac{VA}{\rho l}$

Replacing we have that,

$I = \frac{(0.8V)(0.4*10^{-6}m^2)}{(5.6*10^{-8}\Omega \cdot m)(1.5m)}$

$I = 3.809A$

Therefore the current in the wire is 3.809A

<em>Note: The value obtained for the resistivity of Tungsten was theoretically obtained and can be consulted online.</em>

5 0

3 years ago

A Ray of light in air is incident on an air to glass boundary at an angle of 30. degrees with the normal. If the index of refrac

elena55 [62]

Answer:

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1. So:

1 sin 30° = 1.52 sin θ

θ ≈ 19°

5 0

2 years ago

If the incoming wave strikes a barrier at a 90 degree angle, the wave energyis __________.bounced back at oncoming waves making

olga nikolaevna [1]

Answer:

bounced back at oncoming waves

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If the incoming wave strikes a barrier at a 90 degree angle, the wave energy is, in fact bouncing back at oncoming waves.

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Hope this helps!

3 0

3 years ago