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amid [387]
3 years ago
12

Which of the following explains why international travelers need to use special adapters when plugging a device designed to oper

ate in the U.S. into a wall outlet in Europe?
The increased voltage causes the device's resistance to increase, and the device won't work.
The increased voltage causes the device's resistance to decrease, and the device will over-heat.
The increased voltage causes an increase in power usage, and the device will over-heat.
The increased voltage causes a decrease in power usage, and the device won't work.
Physics
1 answer:
Darina [25.2K]3 years ago
8 0

Answer:

The answer is: The increased voltage causes an increase in power usage, and the device will over-heat.

Explanation:

First, we must consider the variables of the electrical system that will allow us to respond. In this case, power, current and voltage, which are related by

P=VI

Where P=Power, V=Voltage, I=Current.

In the equation it can be observed that power is directly proportional to the system voltage. Thus, if the voltage increases as in this case, the power will also increase, which overheats the device and can cause damage to it.

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A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.
luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

n_1 sin \theta_1 = n_2 sin \theta_2

where we have:

n1, n2: index of refraction of the first and second medium

\theta_1, \theta_2; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

n_1 > n_2

Re-arranging Snell's law we find

sin \theta_2 = \frac{n_1}{n_2} sin \theta_1

since we have

\frac{n_1}{n_2}>1

this implies

sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2

and again it can be rewritten as

\lambda_2 = \frac{n_1}{n_2} \lambda_1

where

\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

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Please help I'm having a mental breakdown
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Answer:

there it is fella

And that's according to ma observation

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