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:
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A closed circle means the number is included and an open circle means its not.
Answer:
2,500 Joules (J) or Newton Meter (N M)
Explanation:
Work = Force x Distance
The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.
Work = 5m x 500n
Work = 2,500 Joules or Newton-Meters.
Therefore 2,500 Joules or Newton Meters of work is done on an object.
Answer:
The center of mass can be calculated by taking the masses you are trying to find the center of mass between and multiplying them by their positions
Answer:
a = 1.41 m/s²
Explanation:
Given that
mass ,m= 41 kg
F₁ = 65 N , θ = 59°
F₂ = 35 N ,θ = 32°
The component of Force F₁
F₁x= F₁cos59° i
F₁x= 65 x cos59° i = 33.47 i
F₁y= - F₁ sin 59° j
F₁y= - 65 x sin 59° j = - 55.71 j
The component of Force F₂
F₂x= F₂ sin 32° i
F₂x= 35 x sin 32° i = 18.54 i
F₂y= F₂ cos 32° j
F₂y= 35 x cos 32° j = 29.68 j
The total force F
F= 33.47 i + 18.54 i - 55.71 j + 29.68 j
F= 52.01 i - 26.03 j
The magnitude of the force F
F=58.16 N
We know that
F= m a
a= Acceleration
m=mass
58.16 = 41 x a
a = 1.41 m/s²
