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padilas [110]
3 years ago
7

What would be the best method for a scientist to use in order to obtain a precise measurement of star’s distance from Earth?

Physics
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

D.). Radial Method - is the stars, and planets, that orbit around a common center of mass, The star will move towards, and away from earth. Light is shifted, as the star motion changes.

C.). Redshift Method - is described as, how light shifts toward shorter, or longer wavelengths; as objects in space, move closer, or farther away from us. If the star is moving away from us, The light it gets stretched a little, which makes it appear more red than it really is.

B.). Blueshift Method - is the opposite of redshift, its' a decrease in wavelength, due to its' motions towards Earth. If a star is moving closer to us, the light it gives off gets squeezed together, which makes it appear bluer than it actually is.

A.). Triangulation Method - is a method used to determine distances through parallax; The distance equals baseline divides by parallax. It is used to calculate stellar distances by using Earth's orbital diameter, as baseline, the angular displacement of a star is measured.

Therefore, Your answer: Letter Choice (A), Triangulation Method.

Hope that helps, (Answer: Letter Choice (A), Triangulation Method is used to obtain a precise measurement of a star's distance from Earth.).

Have a great day!!!!! : )

elena-s [515]3 years ago
6 0

Answer:

triangulation method

Explanation:

The triangulation method is used in Astronomy to measure distances of planets and nearby stars. To measure distances from distant stars in our galaxy, or other galaxies, we have to resort to indirect methods. For this reason, we can conclude that the triangulation method is the best method among those cited for a scientist to use to obtain an accurate measure of the distance between a star to Earth

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Yes

Explanation:

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Mg I_2 + Br_2 \rightarrow Mg Br_2 + I_2

is balanced, because the number of atoms of each element is the same on the reactant side and on the product side. In fact:

- Mg: one atom on the left, one on the right

- I: 2 atoms on the left, and 2 on the right

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So, the reaction is balanced.

3 0
3 years ago
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A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V
sladkih [1.3K]

Answer:

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

Given;

Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²

Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F

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Capacitance of the capacitor after the addition of dielectric material:

Cequ. = Ck

Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

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Additional charge = Q₂ - Q₁

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Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

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