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3 years ago

When the light on the moon is increasing, what is it called

Physics

2 answers:

nataly862011 [7]3 years ago

6 0

Answer:

Waxing Gibbous phase occurs when the Moon is mostly lit and the illuminated portion is egg-shaped (gibbous) with the eastern edge shaded. The amount of illuminated area visible is increasing from one day to the next which is what is meant by "waxing"

hope this helps

have a good day :)

Explanation:

alexandr402 [8]3 years ago

6 0

Answer:

Hey mate.......

Explanation:

This is ur answer....

<em>Waxing Gibbous phase occurs when the Moon is mostly lit and the illuminated portion is egg-shaped (gibbous) with the eastern edge shaded. The amount of illuminated area visible is increasing from one day to the next which is what is meant by "waxing".</em>

Hope it helps!

Brainliest pls!

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What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup

Fantom [35]

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

5 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

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3 years ago

Rills, gullies, streams, rivers and tributaries are all caused by

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3 years ago

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denis23 [38]

Answer:

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Explanation:

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