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Lelu [443]
1 year ago
8

to determine the mass of the central object, we must apply newton's version of kepler's third law, which requires knowing the or

bital period and average orbital distance (semimajor axis) for at least one star. we could consider any of the stars shown in the figure, so let's consider the star with the highlighted orbit (chosen because its dots are relatively easy to distinguish). what is the approximate orbital period of this star?
Physics
1 answer:
mixer [17]1 year ago
6 0

The approximate orbital period of this star is 13 years.

<h3>What is Kepler's third law?</h3>

The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.

T² ∝ a³

The time it takes for one rotation to complete depends on how closely the planet orbits the sun. With the use of the equations for Newton's theories of motion and gravitation, Kepler's third law assumes a more comprehensive shape:

P² = 4π² /[G(M₁+ M₂)] × a³

where M₁ and M₂ are the two circling objects' respective masses in solar masses.

Learn more about Kepler's third law here:

brainly.com/question/1608361

#SPJ1

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A wildebeest and chicken participate in a race over a 2.00km long course. the wildebeest travels at a speed of 16.0m/s and chick
Nezavi [6.7K]

Answer:

(a)  The distance of the chicken from the finish line is 62.5 m

(b) The stationary time of the wildebeest is 675 s

Explanation:

Given;

total distance traveled by wildebeest and chicken, d = 2 km = 2000 m

speed of the wildebeest, v_w = 16 m/s

speed of the chicken, v_c = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

Time for chicken to finish the race without stopping, 2000/2.5 = 800 s

(b) for how long in time (in s) was the wildebeest stationary?

t(stationary) = t(chicken) - t(wildebeest)

t = 800s - 125 s

t = 675 s

(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?

The time taken for the wildebeest to run 1.6 km (1600 m) is given by;

t = 1600 / 16 = 100 s

The total time spent by the wildebeest before it resumed the race = stationary time + 100s

t (total) = 675 s + 100 s = 775 s

Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m

Thus, the distance of the chicken from the finish line = 2000 m -  1937.5 m

the distance of the chicken from the finish line = 62.5 m

7 0
3 years ago
Question 4 of 10
ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0
3 years ago
An airplane during departure has a constant acceleration of 3 m / s².
Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>

t

3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>

8

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

6 0
2 years ago
A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?
ladessa [460]
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
7 0
3 years ago
What is the correct path of sperm cells through the male reproductive system?
irakobra [83]

Answer:

What is the correct path of sperm cells through the male reproductive system?

Epididymis, seminiferous tubules, urethra, vas deferens

<u>Seminiferous tubules, epididymis, vas deferens, urethra </u>

Urethra, seminiferous tubules, epididymis, vas deferens

Seminiferous tubules, vas deferens, epididymis, urethra

Hope this helps :)

Have a great day !

5INGH

Explanation:

8 0
2 years ago
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