Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50 m and moves with a constant s
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Answer:
The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)
Explanation:
First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).
From the free body diagram we can build an equation with the sum of forces between the start and the planet.
We know that the force between two bodies due to gravity is given by the following equation:
in this case we will call:
M= mass of the star
m= mass of the planet
r = distance between the star and the planet
G= constant of gravitation.
so:
Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:
where the centripetal acceleration is given by:
where
Where T is the period, and is the angular speed of the planet, so:
or:
so:
so now we can do the sum of forces:
in this case we can get rid of the mass of the planet, so we get:
we can now solve this for so we get:
We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:
So let's see what's going on there, we'll call:
= mass of Star 1
= mass of Star 2
So:
we can get rid of all the constants so we end up with:
and let's flip the inequality, so we get:
This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
v=5.02 m/s
c.Work done by friction on the incline,