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scoundrel [369]
3 years ago
8

The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre

sent on the Sun? Hint: Use the small-angle formula: angular diameter (in arc seconds) 2.06 ✕ 105 = linear diameter distance . (Note: The average distance to the Sun can be found in Celestial profile: The Sun.)
Physics
1 answer:
mestny [16]3 years ago
4 0

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

       tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

   

        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

       θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

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Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
3 years ago
What charge does an atom have when electrons are lost from the atom?
maw [93]
Negative charge (it has more protons than electrons)
6 0
3 years ago
If 478 watts of power are used in 14 seconds, how much work was done?
sergiy2304 [10]
" 478 watts " means 478 joules per second.

In 14 seconds, the total work or energy is (14 x 478) = <em>6,692 joules</em>
7 0
3 years ago
One of the great triumphs of spectroscopy was when astronomers identified a new element in the sun (one that was only later foun
Nezavi [6.7K]

Answer:

When Helium is identified by astronomers is one of the great triumphs of spectroscopy.

Explanation:

Janssen managed to do this great triumphs on August 18, 1868 . Janssen was the first person to introduce the helium, an element that never seen before on Earth, in the solar spectrum. At that time, he didn’t know that what he’d seen—he just think that it was something new. In the mid of 1800, the spectroscope instrument is introduced in astronomy.

Later on we heard that all helium in Universe has been created by the fusion of hydrogen nuclei.

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brainly.com/question/14677550

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6 0
2 years ago
Electrons are ejected from sodium metal by any light that has a wavelength shorter than 544 nm. What is the kinetic energy of an
wel

Answer:

KE=2.3 x 10⁻¹⁹ J

Explanation:

Given that

λ = 544 nm

λ' = 485 nm

The kinetic energy KE given as

KE= E - Ф

Where

E=\dfrac{hC}{\lambda'}

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h= 6.626 x 10⁻³⁴

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Now by putting the values

KE=\dfrac{hC}{\lambda'}-\dfrac{hC}{\lambda}

KE=\dfrac{34.34\times 10^{-34}\times 3\times 10^8}{485\times 10^{-9}}- \dfrac{34.34\times 10^{-34}\times 3\times 10^8}{544\times 10^{-9}}

KE=2.3 x 10⁻¹⁹ J

This is kinetic energy.

6 0
3 years ago
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