All categories

3 years ago

What is the farthest distance parallaxes can be used to measure star distances from Earth?

1 answer:

8 0

Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.

You might be interested in

An object with a mass of 78 kg is lifted through a height of 6 meters how much work is done

il63 [147K]

We got the following:
m = 78kg
h = 6m
g = 9.8 m/ $sec^{2}$ ≈ 10 m/ $sec^{2}$
----------------------------------------------------------------------------
A = ?

As we know A = Ep = mgh -> potential energy
so the answer would be A = 78 * 6 * 10 = 4680 Joule

5 0

3 years ago

The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around

zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0

3 years ago

Diego hypothesizes that the Earth wobbles on its axis in a pattern that is repeated every 20,000 years.

DedPeter [7]

Answer:

D. make a model

Explanation:

3 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Vladimir [108]

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time = 1 day = 86400 s

$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{ 86400 }$

$\omega = 7.272 \times 10^{-5}\ rad/s$

a) at equator

$v = R_E \omega$

$v = 6.38 \times 10^6\times 7.272 \times 10^{-5}$

v = 464 m/s

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)$

$a = 0.03374 m/s^2$

b) at a latitude of 61.0 ° north of the equator.

$R = R_E cos \theta$

$R = 6.38 \times 10^6\times cos 61^0$

$R = 3.093 \times 10^6 m$

$v = R \omega$

$v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}$

$v = 225 m/s$

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )$

$a = 0.01635 m/s^2$

4 0

3 years ago

Describe how the motion of a planet (like Mars) among the stars of the zodiac on the sky is qualitatively different from the mot

Advocard [28]

Explanation:

In the plane of the Solar System, called Ecliptic plane, Sun is at the center and all the planets including Earth are going around the Sun in their respective orbits. As seen form the Earth, Sun appear to go around it and in this process it crosses through 12 constellations. These constellations are called as Zodiacs. Now since all the planets move in almost the same plane, they will appear in the sky to be lying near the ecliptic.

Sun's own motion is not that evident to us because of which the impact of that motion is nullified. But, motion of planet is clearly visible and it also shows in their annual round around the Sun. Here, Mars also shows retrograde motion which means it will show a back and forth motion in the sky. Because of which Mars might be visible in the same zodiac for a longer Duration as compared to the Sun.

7 0

3 years ago