Which of the following is the time it takes to complete a cycle of motion

If all the stars in an elliptical galaxy traveled random directions in their orbits, the elliptical galaxy would be type

Lemur [1.5K]

The answer would be E7. Galaxies categorized as E0 look to be nearly perfect, while those registered as E7 seem much extended than they are widespread. It is worth noting, though, that a galaxy's look is connected to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and thin or the flattest of them all.

7 0

3 years ago

Why does urine sometime come out as a clear color?

o-na [289]

It is because you drink a lot of water. So the less water you drink the more yellow it is. It means you have healthy urine.

6 0

3 years ago

Read 2 more answers

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

Alinara [238K]

Answer:

The angular separation between the red and violet components of the spectrum that emerges from the glass is 1.72°

Explanation:

Given:

Refractive index for red light $n _{1} = 1.620$

Refractive index for violet $n' _{1} = 1.660$

Refractive index for air $n_{2} = 1$

Incident angle $\theta _{1} =$ 28.30°

According to the snell's law,

$n_{1} \sin \theta _{1} = n_{2} \sin \theta _{2}$

For red light,

$1.620 \times \sin 28.30 = 1 \sin \theta _{2}$

Where $\theta _{2} =$ transmitted angle for red light

$\theta _{2} =$ 50.18°

For violet light,

$1.660 \times \sin 28.30 = 1 \sin \theta' _{2}$

Where $\theta' _{2} =$ transmitted angle for violet light

$\theta' _{2} =$ 51.9°

Angular separation between red and violet light is given by,

$\theta ' _{2} - \theta _{2} = 51.9 -50.18 =$ 1.72°

Therefore, the angular separation between the red and violet components of the spectrum that emerges from the glass is 1.72°

6 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago

An object has a mass of 20 grams and a volume of 5 cubic centimeters what is the object density

Solnce55 [7]

20 g / 5 cm^3 = 4 g/cm^3 or 4000 kg / m^3.

3 0

3 years ago