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Hitman42 [59]
3 years ago
14

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

tar. What is the planet’s orbital period in Earth years? Round your answer to the nearest whole number. __ Earth years
Physics
2 answers:
nadezda [96]3 years ago
7 1

Answer:

16 years just got it wrong for choosing 4.6 believe me if u want

Explanation:

kondaur [170]3 years ago
4 1

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

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A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum
OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0
3 years ago
Consider a weather balloon floating in the air. There are three forces acting on this balloon: the force of gravity is FG, the f
Andreyy89
Hey I just got home from work
4 0
3 years ago
What is the amount of charge a capacitor can store per volt of potential difference?
sladkih [1.3K]

The amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

<h3>What is the capacitance?</h3>

The capacitance has to do with the amount of charge that is stored by a capacitor. We know that a capacitor is a device that can be used to store electric charges. We have in it, two capacitors separated by a dielectric material.

Hence, the amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

Learn more about capacitor:brainly.com/question/17176550

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4 0
2 years ago
Read 2 more answers
A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.​
Ulleksa [173]

Explanation:

ac = v^2/r

= (7.0 m/s)^2/(0.75 m)

= 65 m/s^2

3 0
3 years ago
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