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Scilla [17]
2 years ago
12

The radial velocity method preferentially detects: Choose one: A. all of the above described planets equally well. B. small plan

ets far from the central star. C. small planets close to the central star. D. large planets close to the central star. E. large planets far from the central star.
Physics
1 answer:
Zina [86]2 years ago
5 0

The radial velocity method preferentially detects large planets close to the central star

  • what is the Radial velocity:

The radial velocity technique is able to detect planets around low-mass stars, such as M-type (red dwarf) stars.

This is due to the fact that low mass stars are more affected by the gravitational tug of planets.

When a planet orbits around a star, the star wobbles a little.

From this, we can determine the mass of the planet and its distance from the star.

hence we can say that,

option D is correct.

The radial velocity method preferentially detects large planets close to the central star

Learn more about radial velocity here:

<u>brainly.com/question/13117597</u>

#SPJ4

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The attraction between the Sun's gravity and Earth's gravity holds Earth in is ?
soldi70 [24.7K]
The attraction between the suns gravity and earths gravity holds earth in it's orbit
5 0
3 years ago
A 9,900 kg object feels a gravitational force of 12 N due to a 52,000 kg object. What is the distance between the
aleksley [76]

Answer:

0.05 m

Explanation:

From the question given above, the following data were obtained:

Mass of first object (M1) = 9900 kg

Gravitational force (F) = 12 N

Mass of second object (M2) = 52000 kg

Distance apart (r) =?

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Thus, we can obtain the distance between the two objects as shown below:

F = GM1M2/r²

12 = 6.67×10¯¹¹ × 9900 × 52000 /r²

Cross multiply

12 × r² = 6.67×10¯¹¹ × 9900 × 52000

Divide both side by 12

r² = (6.67×10¯¹¹ × 9900 × 52000)/12

Take the square root of both side

r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]

r = 0.05 m

Therefore, the distance between the two objects is 0.05 m

3 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high (formula:pe=mgh)
yKpoI14uk [10]

The potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735.75 J.

<h3>What is potential energy?</h3>

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

The potential energy is mainly depending upon the height of the object. when the cyclist is at the highest position, the height is maximum. Therefore, the potential energy is also maximum.

The potential energy is found as;

PE=mgh

PE=25 kg× 9.81 m/s² ×3 m

PE= 735.75 J.

Hence, the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735.75 J.

To learn more about the potential energy, refer to the link;

brainly.com/question/24284560

#SPJ1

5 0
2 years ago
How did the author describe the claws of uwang and salagubang?​
Anna007 [38]

Answer:

Describes them as sharp and dangerous.

Explanation:

The author describes the claws of uwang and salagubang as large, sharp, dangerous and scary. This creates a certain suspense in the story, as it leaves the reader apprehensive about the imminent danger that these claws pose to those who know them. However, despite their appearance, the claws are not as deadly as they seem because they can only hurt people who are pinched by them.

7 0
3 years ago
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