To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,
Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity
PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,
From the speed it is possible to use find the formula, so
Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is
Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
Answer:
you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.
whatever we are, you and I will always collide.
There you go! Let me know if it helped.
:)
I think the answer would be physics.
Answer:
Explanation:
Let 100 m/s be the velocity of projection.
So horizontal component
= 100 cos42
= 74.31 m /s
Vertical component = - 100 sin 42 . in upward direction
66.91 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = - 66.91 t + .5 x 9.8 x t²
4.9 t² - 66.91 t - 2.1 = 0
t = 13.685 s
Horizontal distance covered
= 13.685 x 74.31
= 1016.93 m
If angle of projction is 40°
So horizontal component
= 100 cos40
= 76.60 m /s
Vertical component = - 100 sin 42 . in upward direction
64.27 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = -76.60 t + .5 x 9.8 x t²
4.9 t² - 76.60 t - 2.1 = 0
t = 15.659 s
Horizontal distance covered
= 15.659 x 76.60
= 1199.49 m
So horizontal range is increased , if angle of projection is increased .
