A second important difference between comets coming from the Kuiper Belt and from the Oort cloud is represented by their different characteristic periods.
In fact, short period comets are thought to generate in the Kuiper belt and have rather predictable orbits with short periods (up to 200 years). There are two major families of short period comets: the Jupiter family with periods of less than 20 years and the Halley family with periods form 20 to 200 years. That's short
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Answer: 56.718 min</h2>
Explanation:
According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
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In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is 
is the mass of Mars
is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
The net force of the object is equal to the force applied minus the force of friction.
Fnet = ma = F - Ff
12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107.
Answer:
Force is classified as a push or a pull
Explanation:
