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velikii [3]
2 years ago
13

Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the

center of the milky way galaxy allows us to determine ______.
Physics
1 answer:
Ugo [173]2 years ago
5 0

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

#SPJ4

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Please help me fr fr !
Furkat [3]
The answer is A because it describes both toy cars
6 0
3 years ago
6
choli [55]

Answer:

B. Acceleration in the direction of motion speeds you up

Explanation:

Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.

3 0
3 years ago
2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
1 year ago
Which of the following would NOT produce a current? A. Moving a bar magnet inside a coil of wire O B. Moving a bar magnet toward
sineoko [7]

Answer:

c

Explanation:

7 0
3 years ago
A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
Likurg_2 [28]
Skskdidododododoorrororo
3 0
3 years ago
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