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3 years ago

Light year is a unit of?

1 answer:

3 0

Light year is the unit of distance. It is the distance that an object travels in one year with the speed of light.

In short, Your Answer would be "Distance"

Hope this helps!

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What happens when a magenta light is shone on a green surface?

Sedbober [7]

Answer:

When Magenta light is shown on a green surface, it looks black.

Explanation:

It absorbs the Magenta light and also reflects none of the light.

5 0

3 years ago

Someone please help me answer this !!!

Elan Coil [88]

Answer:

I think its C sorry if it's wrong

7 0

3 years ago

There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is

Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0

3 years ago

A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?

dybincka [34]

Answer:

Radius of the circle will be 2.5 m

Explanation:

We have given velocity of particle moving in the circle v = 5 m/sec

Acceleration of particle in the circle $a=10m/sec^2$

We have to find the radius of the circle

We know that acceleration is given by $a=\frac{v^2}{r}$

So $10=\frac{5^2}{r}$

$r=\frac{25}{10}=2.5m$

So radius of the circle will be 2.5 m

3 0

3 years ago

If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what

marin [14]

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of ( $p('') = 0.01$ ):

$d(pc) = \frac{1}{0.01}$

$d(pc) = 100$

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

7 0

3 years ago