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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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The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
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