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3 years ago

I NEED HELP PLEASE, THANKS! :)

Physics

1 answer:

UkoKoshka [18]3 years ago

7 0

Answer:

Explanation:

1) Hypermetropia (better known as Farsighted- this is why nearby objects seem blurry for him)

2) In such instances, image are typically formed farther from the near point

3) Such defects are quite common so there are common procedures such as using convex lens which can restore the sight to normal.

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The blacks holes will absorb

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3 years ago

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Answer:

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3 years ago

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3 years ago

For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red

barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

$z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1$

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

$z=\sqrt{\frac{c+v}{c-v}}-1$

- A red dwarf moving away from Earth at 39.1 km/s :

$z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}$

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

$z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}$

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

$z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}$

- A blue dwarf moving away from Earth at 25.9 km/s $z=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}$

- A red dwarf moving transversely at 14.1 km/s

$z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}$

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4 years ago

Describe the arrangement, motion, and energy of the molecules in the solid, liquid, and gas phases of water.

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3 years ago