Answer:
The answer is Dependent Variable
Answer:
Decrease the voltage,and do not change the resistance,the current will also decrease
Explanation:
Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current
Kinetic Energy,K.E=1/2MV²
mass,m=16kg
velocity,v=4m/s
K.E=1/2×16×4²
=128kgm²/s²
=128 Joules
I am not so sure about this it is too difficult
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.