Answer:
11.9 years
Explanation:
We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:
Using the Earth as reference, we can re-write the law as
where
Te = 1 year is the orbital period of the Earth
re = 1 AU is the average distance of the Earth from the Sun
Tj = ? is the orbital period of Jupiter
rj = 5.20 AU is the average distance of Jupiter from the Sun
Substituting the numbers and re-arranging the equation, we find:
Answer:
the answer would be "using more heat" btw
Explanation:
As simply put as I believe is possible, it's the immediate space-time co-oridinate in which the observer is not in motion. All measurements by the observer are in relation to that 'place'
<u>Answer:</u>
<em>The outside edge of a spinning compact disc moves with a higher velocity than the inner track of the disc.</em>
<u>Explanation:</u>
Here the compact disc undergoes rotational motion about a fixed axis which is its centre in this case. The particles in rotational motion have angular velocity which is given by the equation
ω = ∅/t
Where θ is the angular displacement and t is the time.
The transnational speed of a particle which is in circular motion is given by the equation
v = rω
r is the distance of the point from the rotation centre
The transnational speed of the particles is merely determined by their distance from the centre in this case. It is due to the equality of angular velocity of all the points.
The distance of the outer edge of the compact disc from its rotational centre is larger than the distance of inner edge from the rotational centre. Thus the farther edge of a spinning disc moves faster than the nearer edge.
