<u>Answer</u>
3.44 m/s
<u>Explanation</u>
The motion apply the equations of Newton's law of motion. The ball is acceleration is -9.8 m/s² (acceleration due to gravity. It is negative because the ball is going against gravity, so it is decelerating).
The first equation of Newton's law of motion is;
V = U + at
Where V is the final velocity, U is the initial velocity, a is acceleration and t the time taken.
V = 25 + (-9.8 × 2.2)
= 25 - 21.56
= 3.44 m/s
It’s my guess but from my opinion i would say yes
the answer in my opinion would be A
(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
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.